筛素数表文稿

筛素数筛素数 反素数序 反素数 素数判 快速幂取 辗转相除法求 欧拉函 拓展欧几里 Miller_rabin+ 高斯消 矩阵快速 组合数学八皇 组合数学错 组合数学斐波那 组合数学卡特兰 组合数学立方之 组合数学系列递 组合数学系列容斥+逆 组合数学系列鸽巢原 组合数学系列置换 组合数学Polya定理与Burnside引 组合数学盒子装小球问 费马小定 异面直线公垂线的垂足坐标计算公式推 点到直线距 凸包散点求三角形最大面 dp问题 dp问题 dp问题01背 dp问题完全背 dp问题多重背 dp问题最大括号匹 dp问题区间最大字段和 dp问题树形dp举 dp问题状压dp举 线段树系列经典问 字符串KMP算 回文子串O(n)manacher算 图论最短 图论最小生成 图论最大 图论匈牙利算法二分图匹 经典博弈纳什博 java快速 2筛素数表constintmaxn2600;//大小自定义，非素数均为trueboolvis[maxn];筛素数表constintmaxn2600;//大小自定义，非素数均为trueboolvis[maxn];void{memset(vis,0,sizeof(vis))vis[0]=true;vis[1]=for(inti=2;i*i<maxn;i++)if(!vis[i])for(intj=i*i;j<maxn;j+=i)vis[j]=true;}反素问题描n,n+1,…,mn1m10，1,3,5,4,26,9,7,8,就是一个反素数序列。它也是按字典顺序的第一个这样的序列。我们可以扩展此定义，度d反素数序列是相邻d个整数之和为合数的这种序列。所以，之前序列是一个度2反素数序列，但是不是度3的，因为子列5,4,2和为11。这些数按字典顺31,3,5,46,2,10879输入格输入包含多组输入集。每个输入集包含同一行的三个整数，nmd。n,md≤n<m≤1000和2≤d≤10。以一行000（此行不需处理）代表输入的结束。3对于每个输入集，输出一行以逗号隔开的整数列，此整数列组成对于每个输入集，输出一行以逗号隔开的整数列，此整数列组成一个度反素数序列（不要入任何空格并且输出不要跨越多行）如果不存在反素数序列，则输出“Noanti-primesequenceexist输入样121315406000输出样Noanti-primesequenceexists.usingnamespacestd;constintMAXP=1100000;boolisPrime[MAXP];intinta[1005],ans[1005],n,m,d,OK;boolvst[1005];voidprimeList()memset(isPrime,true,sizeof(isPrime));isPrime[1]=false;for(inti=2;i<=MAXP;{ifprime[++prime[0i;/prime[0]0for(intj=1,k;(j<=prime[0])&&(k=i*prime[j])<={isPrime[k]=false;if(i%prime[j]==0)}}}++ji4boolcheck(intnum)boolcheck(intnum){intsum=ans[num],i,k;if(num-d+1>=1)k=num-d+k=for(i=num-1;i>=k;i--{sum+=ans[i];return1;}return}voiddfs(intpos)if(check(pos-1))if(pos==m-n+{OK=}for(inti=n;i<=m;{if(!vst[i]){ans[pos]=i;vst[i]=1;dfs(pos+vst[i]=0;if(OK)}}}个数intmain()intk;while(scanf("%d%d%d",&n,&m,&d))if(n==0&&m==0&&d==k=1;for(inti=n;i<=m;i++)a[k++]=i;memset(vst,0,memset(ans,0,OK=if(OK==0){printf("Noanti-primesequence5}for(inti=}for(inti=1;i<=m-n;i++)printf("%d,",ans[i]);printf("%d\n",ans[m-n+}return}反素对于任何正整数x，其约数的个数记做g(x)。例如g(1)=1，g(6)=4如果某个正整数x满足：对于任意i(0ix)，都有g(i)g(x)，则称x而反素数表就是对应上面反素数所建立的一张表，这张表好处多多，例如给你一个n的找出1~~n范围内，谁的因子数之和最多！usingnamespace{inti,j;{{}}intmax=0;{{}}{6{}}return}以上为反素数表{}}return}以上为反素数表，下面为对应的因子个数，如1-2中，2的因子个数最多，有21-4中，4为3个；1-664个。1-1164个。1-19中，126个。反素数的搜任给一个数n，可以得到1~n中，最大的SampleInputSampleOutputCase#1:840int64int64voiddfs(int64cur,int64cnt,int64power,int64{int64i;{}int64p=prime[sub];{7}}int{intt,T;{printf("Case#%d:}return}素}}int{intt,T;{printf("Case#%d:}return}素数判定intisprime(long//传入待判定数intisprime(long{return0;longlongj=3;while(j<=(long{}return}快速longlongmodular(longlonglonglongmodular(longlonglonglong{longlongd=1,t=a;{}return}8publicstaticintgcd(intpublicstaticintgcd(inta,intbreturnb==0?a:gcd(b,a%}//计算a，b的最大公约数，返回1，LCM最小intLCM(intm,int{intm1=m;intn1=n;inttemp;while(n1%m1!={temp=n1%m1;n1=m1;m1=}intans=n/m1*m;returnans;}//计算m，n的最小公倍数欧拉函数定义：φφ(x)=x(1-1/p1)(1-1/p2)(1-1/p3)(1-1/p4)…..(1-其中p1,p2……pn为x的所有质因数，x是不为0的整数。φ(1)=1（唯一和1互质的数就是1本身）求出m的欧拉函数值phi[m]，可知区间[1,mphi[m]个数与m1)。单项欧拉函数：publicstaticintn){intres=for(inti=2;i<Math.sqrt(n);{if(n%i==0)res=res/i*(i-1);while(n%i==0)n=n/}}if(n>res=res/n*(n-1);returnres;}各项欧拉函数：intphinewint[1000005];/for(inti2;i1000000;iif(phi[i]==0){9for(intj=i;j<=1000000;j+={for(intj=i;j<=1000000;j+={if(phi[j]==0)phi[j]=j;phi[j]=phi[j]/i*(i-}}}拓展欧几里,x=(x0*n/gcd（a，b）)%b;if(x<0)通解可以表示为x=x0+b/d*ty=y0-a/dint64x,dextend_gcd(a,b,x,y);/int64extend_gcd(int64a,int64b,int64&x,int64&y)int64t,ret;if(b==0){x=y=return}ret=extend_gcd(b,a%b,x,===t-a/b*return}//返回值为a，b两系数的最大公约数，计算出一组解Miller_rabin判素数法，pollard_rho分解质例举三道模版题题意：判断所有质数i<=k，2^i-1是否是质数，不是的话就要将它分解质因数输出来。SampleInputSample23*89=2047=(2^11)-47*178481=8388607=(2^23)-233*1103*2089=536870911=(2^29)-usingnamespacelonglongfactor[100],fac_top=usingnamespacelonglongfactor[100],fac_top=-//计算两个数的longlongran[1000000],ID;longlongRand(){returnrand();}longlonggcd(longlong{longlongc;{}return}longlongmuti_mod(longlong{longlongexp=a%n,res=0;{{res+=exp;if(res>n)res-=n;}exp<<=1;exp-=b,longlong}return}longlongmod_exp(longlong{longlongexp=a%m,longp,longlong{exp=muti_mod(exp,exp,m);}return}//miller-rabin法测试素数,{exp=muti_mod(exp,exp,m);}return}//miller-rabin法测试素数,time测试次boolmiller_rabin(longlongn,{if(n==2)returnlonglonga,u=n-1,x,y;intt=0;{}{axi=0;i<times;Rand()%(n-1)+mod_exp(a,u,for(intj=0;j<t;{y=muti_mod(x,x,if(y==1&&x!=1&&x!=returnfalse;x=}if(y!=1)return)}return}longlongpollard_rho(longlongn,int{longlongx,y,d,i=1,k=2;x=Rand()%(n-1)+1;y=x;{x=(muti_mod(x,x,n)+c)%n;d=gcd(y-x,n);if(1<d&&if(1<d&&d<n)returnd;if(y==if(i==k){y=x;k<<=}}}voidfindFactor(longlong{if(miller_rabin(n,6)){factor[++fac_top]=n;}longlongp=n;while(p>=n)p=pollard_rho(p,k--);}{for(inttt=0;tt<1000000;tt++)intk;{for(inti=11;i<=k;{longlongn=(1ll<<i)-1;{for(intj=0;j<fac_top;j++)printf("%lld*printf("%lld=%lld=(2^%d)printf("%lld=%lld=(2^%d)-}}}return}2.给出两个数a，b的最大公约数，最小公倍数，输出最小的3601215usingnamespacestd;longlong=-//计算两个数的longlonggcd(longlong{returnb;longlongc;{}return}longlongmuti_mod(longlong{longlongexp=a%n,res=0;{{res+=exp;if(res>n)res-=n;}exp<<=1;exp-=long}return}longlongmod_exp(long}return}longlongmod_exp(longlonga,longlongp,long{longlongexp=a%m,res=1;{exp=muti_mod(exp,exp,m);}return}//miller-rabin法测试素数,time测试boolmiller_rabin(longlongn,long{longlonga,u=n-1,x,y;intt=0;}for(int{a=rand()%(n-1)+1;x=mod_exp(a,u,n);for(intj=0;j<t;j++){y=muti_mod(x,x,if(y==1&&x!=1&&x!=n-1)returnfalse;//mustnotx=}if(y!=1)return}return}longlongpollard_rho(longlongn,int{//找出一个longlongx,y,d,i=1,k=2;x=rand()%(n-1)+1;y=x;{x=(muti_mod(x,x,n)+c)%n;d=gcd(y-x,n);if(1x=(muti_mod(x,x,n)+c)%n;d=gcd(y-x,n);if(1<d&&d<n)returnif(y==x)if(i==k){y=x;k<<=}return}}voidfindFactor(longlongn,int{//二分找出所有质因子，存入factorif(miller_rabin(n,10)){factor[++fac_top]=n;}longlongp=n;while(p>=n)p=pollard_rho(p,k--);//k值变化，防止死循}longlonga[1000];longlongm,minx,ans;intcmp(constvoid*a,const*b){return*(longlong*)a-*(longvoiddfs(longlongs,longlongnum,longlong{{}}}voidlonglongi;{}{}longlongi;{}{}}int{longlongs,t,n;{{}fac_top=-1;if(ans>n/ans)ans=n/ans;}return}Sample25Sample#include<cstdio>#include<ctime>#include<cstring>#include<iostream>constintS=20;usingnamespacestd;typedeflonglongLL;#definemaxn10000#include<cstring>#include<iostream>constintS=20;usingnamespacestd;typedeflonglongLL;#definemaxn10000inttot;muti_mod(LLa,LLb,LL LLret=0;ifif(ret>=c)ret-if(a>=c)a-=c;}return}pow_mod(LLx,LLn,LLmod){//返回x^nmodc非递归(n==1)returnx%mod;intbit[64],k=0;while(n){}LLfor(k=k--){if(bit[k]==1)}return}boolcheck(LLa,LLn,LLx,LLt){//a为基，n-1=x*2^t，检n是不是合for(inti=1;i<=t;i++){if(ret==1&&last!=1&&last!=n-1)return1;}if(ret!=1)return1;return0;}boolMiller_Rabin(LLLLx=n-while((x&1)==0)x>>=1,t++;boolflag=1;if(t>=1&&LLx=n-while((x&1)==0)x>>=1,t++;boolflag=1;if(t>=1&&(x&1)==1){for(intk=0;k<S;k++){if(check(a,n,x,t)){flag=1;break;}}}if(!flag||n==2)return}gcd(LLa,LLif(a==0)returnif(a<0)returngcd(-a,b);while(b){LLt=a%b;a=b;}returnPollard_rho(LLx,LLwhile(1){LLd=gcd(y-x0,x);ifd!=x){return}if(y==x0)returnx;if(i==k){}}}voidfindfac(LLn){//递归进行质因iffactor[tot++]=n;}LLwhile(p>=n)p=Pollard_rho(p,rand()}%(n-intintt;while(t--LLif(!Miller_Rabin(n))printf("Prime\n");tot=0;LLans=n;for(inti=0;i<tot;i++)ansLLif(!Miller_Rabin(n))printf("Prime\n");tot=0;LLans=n;for(inti=0;i<tot;i++)ans=min(ans,factor[i]);}}}数，数到m的那个人又出列；//n=9，m=publicclassMain{publicstaticvoidmain(String[]{Scannerintf[]=newint[n+1];for(int}个退出的人的序号:注意程序只关注最后f[n],f[1~n-1]为中间程序无意义。}}//38657241//386572411importjava.util.Scanner;publicclassMainpublicstaticvoidmain(String[]{Scannerscan=newn=scan.nextInt();//n个人围成环，标号1~nm=scan.nextInt();//报到m号被杀死f[]=newint[n+1];i=2;i<=n;i++){}intans=(k-m+1+f[n])%Sample3405classMainstaticvoidmain(String[]args)Scannerscan=newScanner(System.in);while(scan.hasNext()){intk=scan.nextInt();if(k==0)elseelse1358657,2504881,System.out.println(a[k-}}}}importpublicclassMainpublicstaticmain(String[]{Scannerscan=newintresult[]=newint[14];intk=scan.nextInt();while(k!=0){for(intj=1;;{intm=j*(k+pp=rest=2*(inti=0;i<k;i++)(p+m-1)%rest;{if(p<k)}if(rest==}{p=rest=2*for(inti=0;i<k;i++)p=(p+m-1)%rest;if(p<k)}if(rest==k){}}}k=}}}}}}k=}}}结论m=5。题目意思大概n个小孩围成一个圈，每个小孩手里有张卡片，记录着一个数字。开始从第k个孩子，该孩子离开圈子，然后告诉别人他手里的数字，接下来便从位于该孩子数之和。例如第6个跳出的孩子能得到(1,2,3,6)四个糖果。usingnamespace#defineL(rt)(rt<<1)#defineR(rt)(rt<<1|1)constintusingnamespace#defineL(rt)(rt<<1)#defineR(rt)(rt<<1|1)constintintstructint date{intcharvoidbuild(intl,intr,intrt){tree[rt].l=l;returnintmid=(l+r)>>1;}update(intkey,intrt){tree[rt].sum--;returntree[rt].l;returnupdate(key-}voidfor(inti=1;i<=n;i++){}}}}}intinti,k,mod;intpos=0;if(boy[pos].val>0)//kk}printf("%s}return}usingnamespacestd;constintintx[MAXN];//intx[MAXN];//inlineintgcd(inta,int{intt;{}return}inlineintlcm(inta,int{}var+1,分别为0到Gauss(intequ,int{for(inti=0;i<=var;{}for(k=0;k<equ&&col<{枚举当前处理的行for(i=k+1;i<equ;{}{for(j=k;j<var+1;j++)}{}for(i=k+1;i<equ;{枚举要删去的行.{LCM=lcm(abs(a[i][col]),abs(a[k][col]));ta=LCM/abs(a[i][col]);tb=for(j=col;j<var+1;{a[i][j]=a[i][j]*ta-}}}}for(i=k;i<equ;{if(a[i][col]!=0)return-}这样的行，即说明没有形成严格的上三角阵if(k<var){if(k<var){for(i=k-1;i>=0;i--){i行一定不会是(0,00)的情况，因为这样的行是在第k行到第equ行.i行一定不会是(00aa0的情况，free_x_num0用于判断该行中的不确定的变元的个数，如果超过1个，则无法求解，它们仍然为不确定的变元.for(j=0;j<var;{free_index}(a[i][j]!=0&&free_x[j])free_x_num++,(free_x_num1continue;无法求解出确定的变元.变元，且该变元是确定的temp=for(j=0;j<var;{}free_x[free_index]=0;//该变元是确定的.}returnvarkvark个}Xn-1Xn-2for(i=var-1;i>=0;i--{temp=for(j=i+1;j<var;{if(a[i][j]!=0)temp-=a[i][j]*}if(tempa[i][i0return2说明有浮点数解，但无整数解.x[i]=temp/a[i][i];}return}int{inti,j;while(scanf("%d%d",&equ,&var)int{inti,j;while(scanf("%d%d",&equ,&var)!={memset(a,0,sizeof(a));for(i=0;i<equ;i++){for(j=0;j<var+1;{scanf("%d",}}intfree_num=if(free_num1printf("无解elseif(free_num>0){printf("无穷多解自由变元个数为%d\n"free_num);for(i=0;i<var;i++){i+if(free_x[iprintf("x%d是不确定的\n",elseprintf("x%d:%d\n",i+1,x[i]);}}{for(i=0;i<var;{printf("x%d:%d\n",i+}}}return}矩阵快速幂#includetypedef{int64int64powermod(int64x,int64int64int64powermod(int64x,int64n,int64p)//{int64res=1;{x=(x*x)%p;}return}Matrixmulti(Matrixx,Matrixy,int64p)//{inti,j,k;int64sum;for(i=0;i<2;i++)for(j={sum=for(k=0;k<2;k++)res.matrix[i][j]=sum%p;}return}MatrixMat_powermod(Matrixx,int64n,int64p)//x，n次幂，p{inti,j;for(j=0;j<2;j++){}{x=multi(x,x,p);}returnx=multi(x,x,p);}return}int{int64int64int64power,result;Matrixres;{keep1=(powermod(a,(p-1)/2,p)+1)%p;{res.matrix[0][0]=res.matrix[0][1]=res.matrix[1][0]=res.matrix[1][1]=power=(res.matrix[0][0]+res.matrix[0][1])%(p-}res.matrix[0][1]=(2*a*b)%p;res.matrix[1][0]=2%p;res.matrix[1][1]=(a+b)%p;result=1;result=(result*keep1)%p;}return} 当n>=k1122112323311223.1235813f(1)=1f(2)=1;f(n)=f(n-1)+f(n-2)atalan例如n=5时，有5种剖分方法。publicclassMain{publicstaticvoidmain(String[]Scannerscan={longf[]=new===for(inti=4;i===for(inti=4;i<=10000;i++){j++for(intj=2;j<=i-1;){f[i]+=f[j]*f[i-j+}}while{intn=}}1234511125SampleinputOutputforsampleinput310=1^3+1^3+…..+1^3;10=1^3+1^3+2^3importpublicclassuva11137intmaxn=dp[]=newvoid{dp[0]=for(inti=1;intt=ii<=21;i++)i*for(intj=t;j<maxn;j++)dp[j]+=dp[j-}}}publicstaticvoidmain(String[]args)Scannerscan=newScanner(System.in);while(scan.hasNext()){intn=scan.nextInt();}}}组合Scannerscan=newScanner(System.in);while(scan.hasNext()){intn=scan.nextInt();}}}组合数学中系列问题B-ConnectedSampleSample112340ForeachtestcaseoutputtheansweronasingleTheinputcontainsseveraltestcases.Eachtestcasecontainsanintegern,denotingthenumberofvertices.Youmayassumethat1<=n<=50.Thelasttestcaseisfollowedbyonezero.AnundirectedgraphisasetVofverticesandasetofE∈{V*V}edges.Anundirectedgraphisconnectedifandonlyifforeverypair(u,v)ofvertices,uisreachablefromv.Youaretowriteaprogramthattriestocalculatethenumberofdifferentconnectedundirectedgraphwithnvertices.Forexample,thereare4differentconnectedundirectedgraphswith3方案数为2^(C(n-k-1,2))，所以这样的方案数共有C(n-1,k)*F(k+1)*2^(C(n-k-1,2))种。这种思路的优点在于简单明了，但是缺点在于编程复杂，要算高精度的加、减、乘，2的幂也考虑拿1时点2的情况，设此时2所在连通块k各点，这k个点以及剩下的n-k个分别处在一个连通块中，其方案数为F(k)*F(n-k)，点需在除去点2的点中k-1个点成连通块，故方案数为C(n-2,k-1)，而这总共个点与剩下除去点n-k-1个点必须2^k-1,故这样的情才能连通，即这k个点与点1至少有1条边连通，这样的方案数F(k)*F(n-k)*C(n-2,k-1)*(2^k-1)种。担，只需要高精度加和乘，2的幂只到50。我的程序就是靠思路二解决的。classMainpublicstaticvoidmain(String[]args)Scannerscan=new=newBigInteger[51];f[1]=BigInteger.valueOf(1);f[2]=for(inti=3;i<=50;i++){for(intj=1;j<i;j++){f[i]=f[i].add(f[j].multiply(f[i-j]).multiply(C(i-}}intn=scan.nextInt();while(n!=0){14n=}}publicstaticBigIntegerC(longx,longy)forn=}}publicstaticBigIntegerC(longx,longy)for(intc====1;p<=x;p++)a=if(p==b=if(p==x-y)c=a;}return}}A-TheBossonSampleForeachtestcase,outputanintegerindicatingthemoneythebosscansave.Becausetheanswerissolarge,pleasemoduletheanswerwith1,000,000,007.Case1:Case2:ThefirstlinecontainsanintegerTindicatingthenumberoftestcases.(1 1000)Eachtestcase,thereisonlyoneintegern,indicatingthenumberofemployeesinACM.(1 OnMars,thereisahugecompanycalledACM(AhugeCompanyonMars),andit'sownedbyayoungerboss.DuetonomoonsaroundMars,theemployeescanonlygetthesalariesper-year.arenemployeesinACM,andit'stimeforthemtogetsalariesfromtheirboss.Allemployeesarenumberedfrom1ton.Withtheunknownreasons,iftheemployee'sworknumberisk,hecangetk4Marsdollarsthisyear.SotheemployeesworkingfortheACMareveryrich.BecausethenumberofemployeesissolargethatthebossofACMmustdistributetoomuchmoney,hewantstofirethepeoplewhoseworknumberisco-primewithnnextyear.Nowthebosswantstoknowhowmuchhewillsaveafterthedismissal.Sample1~n中与n4次方之和，即Sample1~n中与n4次方之和，即S=a1^4+a2^4+……;a1,a2……且与n则1=1;;ab≡1(modmbam逆,记作amax≡1(modm)的解.做辗转相除法求得整数b,k5b+7k=1,b57逆计算如下:7=5+2,5=2×2+1.245回代1=5-2×2=5-2×(7-53×5-2回代1=5-2×2=5-2×(7-53×5-2×7,5^做辗转相除法求得整数b,k21b+73k=1,则b2173逆21=10*2+1回代1=21-21^-PROB:HDUusingnamespacestd;LL//30MODb){LLret=1;}return}LLans=n;returnans;}LLans=n;returnans;LLans=n;returnans;}intvoidPrime(){ for(inti=2;i<=10000;i++){}}void//30MOD}n){LLi=idx;i<fact.size();i++){tret=((ret-}}int}LLsum=((Sum(n)-}return}G-FindaSamplereturn}G-FindaSampleSampleN个数，问其中是否存在M个数使其满足M个数的和是N的倍数，如果有多组解，随意输出一组即可。若不存在，输出0。223512341Incaseyourprogramdecidesthatthetargetsetofnumberscannotbefounditshouldprinttotheoutputthesinglenumber0.Otherwiseitshouldprintthenumberofthechosennumbersinthefirstlinefollowedbythechosennumbersthemselves(onaseparatelineeach)inarbitraryorder.Iftherearemorethanonesetofnumberswithrequiredpropertiesyoushouldprinttotheoutputonlyone(preferablyyourfavorite)ofthem.ThefirstlineoftheinputcontainsthesinglenumberN.EachofnextNlinescontainsonenumberfromthegivenset.TheinputcontainsNnatural(i.e.positiveinteger)numbers(N<=10000).Eachofthatnumbersisnotgreaterthan15000.Thisnumbersarenotnecessarilydifferent(soitmayhappenthattwoormoreofthemwillbeequal).Yourtaskistochooseafewofgivennumbers(1<=few<=N)sothatthesumofchosennumbersismultipleforN(i.e.N*k=(sumofchosennumbers)forsomenaturalnumberk).如果余数出现0，自然就是n的倍数。也就是说，n个数中一定存如果余数出现0，自然就是n的倍数。也就是说，n个数中一定存在一些数的和是n的倍数。本题的思路是从第一个数开始一次求得前i（iN）项的和关于N的余数sum，并依次记sum0；则从第一项到第isum在前边已经出现过，假设在第j（j<i）项出现sumj+1项到第iclassMainlonga[]=newlong[16000];longsum[]=newlong[16000];staticvoidmain(String[]args)throws{}PrintWriterstaticvoidrun()throwsIOException=newStreamTokenizer(newBufferedReader(newInputStreamReader(System.in)));out=newPrintWriter(new}publicstaticlongnextLong()throwsreturn(int)in.nval;}{publicstaticvoidsolve()throwsIOExceptionlongn=booleanflag=sum[0]=for(inti=1;i<=n;i++)a[i]=}for(inti=1;i<=n;i++)sum[i]=(sum[i-1]+a[i])%n;if(sum[i]==0){for(intk=1;k<=i;}}elsefor(intj=1;j<i;{if(sum[i]==sum[j])out.println(i-for(intk=j+1;k<=i;flag=true;}}}if}}}TimeLimit:2000MSMemoryLimit:65536KB64bitIOFormat:%I64d&EveryyearthereisthesameproblematHalloween:Eachneighbourisonlywillingtogiveacertaintotalnumberofsweetsonthatday,nomatterhowmanychildrencallonhim,soitmayhappenthatachildwillgetnothingifitistoolate.Toavoidconflicts,thechildrenhavedecidedtheywillputallsweetstogetherandthendividethemevenlyamongthemselves.Fromlastyear'sexperienceofHalloweentheyknowhowmanysweetstheygetfromeachneighbour.Sincetheycaremoreaboutjusticethanaboutthenumberofsweetstheyget,theywanttoselectasubsetoftheneighbourstovisit,sothatinsharingeverychildreceivesthesamenumberofsweets.Theywillnotbesatisfiediftheyhaveanysweetsleftwhichcannotbedivided.Yourjobistohelpthechildrenandpresentasolution.TheinputcontainsseveraltestThefirstlineofeachtestcasecontainstwointegerscandn(1≤c≤n≤100000),thenumberofchildrenandthenumberofneighbours,respectively.Thenextlinecontainsnspaceseparatedintegersa1,...,an(1≤ai≤100000),whereairepresentsthenumberofsweetsthechildrengetiftheyvisitneighbouri.ThelasttestcaseisfollowedbytwoForeachtestcaseoutputonelinewiththeindicesoftheneighboursthechildrenshouldselect(here,indexicorrespondstoneighbouriwhogivesatotalnumberofaForeachtestcaseoutputonelinewiththeindicesoftheneighboursthechildrenshouldselect(here,indexicorrespondstoneighbouriwhogivesatotalnumberofaisweets).Ifthereisnosolutionwhereeachchildgetsatleastonesweetprint"nosweets"instead.Notethatifthereareseveralsolutionswhereeachchildgetsatleastonesweet,youmayprintanyofthem.Sample4512375367112513170Sample3523//“nosweets”。可以的话输出任一个满足条件的即可。即要求选择的糖果数目之和能被c整出：1.sum[i02.sum[i]的值已经在前面出现过的情况时，就可以知道有解，并sum[i]数值的后一个位置开始→sum[i]的位置，这个貌似可以解出#includeusingnamespaceintsum[100001];intuse[100001];intc,n;boolis;{inti,j,k;{{{}}{{}{for(k=use[sum[i]]+1;k<=i+1;k++)printf("%d%c",k,k==i+1?'\n':'');}}}return}D-CowSortingFarmerJohn'sN(1≤N≤10,000)cowsarelineduptobemilkedintheevening.Eachcowhasaunique"grumpiness"levelintherange1...100,000.SincegrumpycowsaremorelikelytodamageFJ'smilkingequipment,FJwouldliketoreorderthecowsinlinesotheyarelinedupinincreasingorderofgrumpiness.Duringthisprocess,theplacesofanytwocows(notnecessarilyadjacent)canbeinterchanged.Sincegrumpycowsarehardertomove,ittakesFJatotalofX+YunitsoftimetoexchangetwocowswhosegrumpinesslevelsareXandY.PleasehelpFJcalculatetheminimaltimerequiredtoreorderthecows.Line1:AsingleLine1:Asingleinteger:Lines2..N+1:Eachlinecontainsasingleinteger:linei+1describesthegrumpinessofcowi.Line1:Asinglelinewiththeminimaltimerequiredtoreorderthecowsinincreasingorderof231231:Initial213:Afterinterchangingcowswithgrumpiness3and1123:Afterinterchangingcowswithgrumpiness1and2大致题意：FarmerJohn有N头牛(1N10000)N头牛都很各应，各自有一个不同的脾气L(1L100000)，这N头牛按脾气指数是无序排列，指数越大的越容易破坏farmersum+(len2)*minsumminlen1)*smallestusingnamespacestd;constintmaxn=10000+10;inta[maxn],c[100010];intvis[maxn],constintinf=1<<30;intMIN,MAX,sum;intwork(){intk,t,i,j,ans=for(i=1;i<=n;{ans=0;j=i;k=0;{if(t>t=vis[j]=1;j=vis[j]=1;j=ans+=min((k-2)*t,t+(k+1)*MIN);sum+=ans;}}return}{inti,j;intans;sum=0;MIN=inf,MAX=memset(vis,0,memset(c,0,sizeof(c));for(i=1;i<=n;i++){scanf("%d",&a[i]);if(a[i]<MIN=if(a[i]>MAX=}for(j=1;j<=MAX;c[jc[jc[j1求应有的位置ans=work();printf("%d\n",ans);return0;}题解：黑书P247,无聊的排序”。（完全一样5.Polya定理与Burnside引理6.组合数学N个小球放入Mnm个箱子里，有多少种不同的放法（不一定是球和箱子，也可能是其他的元素与其他的放置位置,例如NM不难看出一共8类情况--的公式是把n个球排成一排，（一种方法）n-1m-1个小棍，放到空是球减1里面挑M-1个箱子做组合就有n+m个球，C（n+m-1,m-1)，多了M个元素而已11的N2S（n,k)=S(n-1,k-1)+k*S(n-1,k)，含义是第NK个数等于他上一排的上一个位置数字加上一排的同样位置数字的K倍同理S(4,2)=S(3,1)+2*S(3,2)=1+2*3=7, 个相同箱子，每个箱子至少一个，则看三角形的第7行，第4个数字多少。6，N不同球，M同箱，允许空的时候（5的基础上允许空箱）。明显是N个球 看三角形就知道第8行前3个数字的和1 127+球不同，盒不同，非空，公式M!*S(N,M)=3!*S(8,3)=6*966=5796TimeLimit:1000MSMemoryLimit:（K表示）5，1，11，5，1t（0t20）M和N，以空格分开。对输入的每组数据M和N，用一行输出相应的K。SampleInput17intcount=0;intvoidDFS(intk,ints,int intvoidDFS(intk,ints,int {if(s==n){return;}inti;for(i=k;i>=0;i--){if(t+i<=m{DFS(i,s+1,}{}}}int{inti;{{DFS}}return}所有的小于p的质数就可以了。事实上，测试小于p的平方根的所有质数就可以了。AMt1*ABXm-Xa,Y-Ya,Z-Za)=t1*(Xb-Xa,Yb-Ya,Zb-M(t1(Xb-Xa)+XaAMt1*ABXm-Xa,Y-Ya,Z-Za)=t1*(Xb-Xa,Yb-Ya,Zb-M(t1(Xb-Xa)+Xa,t1(Yb-Ya)+Ya,t1(Zb-Za)+ZaN(t2(Xd-Xc)+Xc,t2(Yd-Yc)+Yc,t2(Zd-Zc)+Zcdoubledis(P&a,P&b,P{if(b.x!={doublek(b.ya.y(b.xa.x);fabs((pp.ya.y)k*(pp.xa.x))sqrt(1k*k}returnfabs(pp.x-}#include<stdio.h>#include<stdlib.h>#include<math.h>inttop;doubletypedef#include<stdio.h>#include<stdlib.h>#include<math.h>inttop;doubletypedefstruct{intx;intTPointvoidswap(TPointpoint[],inti,int{TPointtmp;tmp=point[i];point[j]=tmp;}doublemulti(TPointp1,TPointp2,TPoint{return(p1.x-p0.x)*(p2.y-p0.y)-(p2.x-p0.x)*(p1.y-}{return(p1.x-p2.x)*(p1.x-p2.x)+(p1.y-p2.y)*(p1.y-}intcmp(constvoid*a,constvoid{TPoint*c=(TPoint*)a;TPoint*d=(TPointdoublek=multi(*c,*d,point[0]);if(k<0)return1;elseif(k==0&&distance(*c,point[0])>=distance(*d,point[0]))return1;elsereturn-}{//Graham扫描求{//Graham扫描求inti,u=0;for(i=1;i<=n1;i++){if((point[i].y<(point[i].y==point[u].y&&point[i].x<point[u].x))u=i;}//p[1]p[n1]按按极角排序，可采用快速排序qsort(point+1,n-1,sizeof(point[0]),cmp);for(i=0;i<=2;i++)stack[i]=i;top=2;for(i=3;i<=n-while(multi(point[i],point[stack[top]],point[stack[top-1]])>=0){top--;if(top==0)}stack[top]=i;}}int{doubletriangleArea(inti,intj,intk);voidPloygonTriangle();inti,while(scanf("%d",&n)&&n!=1){for(i=0;i<scanf("%d%d",&point[i].x,&point[i].y);if(n<=2){}if(n==printf("%.2lf\n",triangleArea(0,1,2));}}return}void{double}return}void{doubletriangleArea(inti,intj,intk);inti,j,k;max=-1;for(i=0;i<=top-2;i++){k=-for(j=i+1;j<=top-1;j++){if(k<=j)k=j+1;area=triangleArea(stack[i],stack[j],stack[k]);if(area>max)max=area;while(k+1<=area1=triangleArea(stack[i],stack[j],stack[k+1]);if(area1<area)break;if(area1>max)max=area1;area=area1;}}}}doubletriangleArea(inti,intj,int{//已知三角形三个顶点的坐标，求三角形的面doublel=fabs(point[i].x*point[j].y+point[j].x*+point[k].x*point[i].y-point[j].x*-point[k].x*point[j].y-point[i].x*point[k].y)/2;returnl;}DpLISProblem据是不大于30000的正整数,用空格分隔)Sample8389207155300299170158据是不大于30000的正整数,用空格分隔)Sample8389207155300299170158SampleOutputimportpublicclassMainpublicstaticvoidmain(String[]{Scannerscan=newScanner(System.in);whilenum==newn=max[]=new(intia[i]=0;i<n;i++)=}(inti=for1;ij=n;i++)j<i;j++)a[i]&&if(a[j]+1>max[i]=max[j]+if(num<max[i])num=}}}}}importif(num<max[i])num=}}}}}importjava.util.Scanner;publicclassMain{publicstaticvoidmain(String[]{inta[]=newint[n];for(inti=0;i<n;i++)a[i]=scan.nextInt();}publicstaticintLIS(inta[],intn){intnum[]=newint[n];inti,j;{{}}intmax=0;returnmax;}模版3（推荐nlogn）publicclassMainpublicstaticvoid{Scannerscan=n==newint[n+b[]=newint[n+1];(inti=n==newint[n+b[]=newint[n+1];(inti=1;i<=n;{a[i]=}k=b[1]=(inti=2;i<=n;i++)if(a[i]b[k])//>=为最长不降子序列b[++k]=a[i];elseintj=while(a[i]<=b[j])b[j+1]=}}}}DP20classMain{publicstaticvoidmain(String[]args){Scannerin=newStringstr1=in.next();Stringstr2=int[][]DP=newint[str1.length()+1][str2.length()+1];inti;DP[i][0]=0;DP[0][i]=0;j=1;j<=str2.length();j++){if(str1.charAt(i-1)==str2.charAt(j-1))}}}}j=1;j<=str2.length();j++){if(str1.charAt(i-1)==str2.charAt(j-1))}}}}412326467{{classinta,b;staticvoidmain(String[]args)throwsInputStreamReaderis=newInputStreamReader(System.in);BufferedReaderin=newBufferedReader(is);String[]ss=in.readLine().split("");p=newint[a+1];q=newint[a+1];for(inti=1;i<=a;i++){ss=in.readLine().split("");}d=newint[b+1];for(inti=1;i<=a;i++)for(intj=b;j>=p[i];j--)d[j]=Math.max(d[j-p[i]]+q[i],d[j]);}}E-完全背包裸题Memory64bitIOFormat:%I64d&}}E-完全背包裸题Memory64bitIOFormat:%I64d&大，越幸福），b表示湫湫吃这种食物会吸收的卡路里量。[Technical101n<=a,b<=m<=37953791510676SampleOutputusingnamespace6SampleOutputusingnamespace{intt;{inti,j,aim;intka[105],val[105];for(i=0;i<t;i++)for(i0;i<t;i++){for(j=ka[i];j<=aim;{dp[j]=max(dp[j],dp[j-}}}return}Sample101201000100000SampleCollection#1:Can'tbedivided.00000SampleCollection#1:Can'tbedivided.Collection#2:Canbeusingnamespacestd;voidbag01(intv,int{for(inti=aim;i>=v;i--)}voidbag11(intv,int{for(inti=v;i<=aim;i++)}voidbagmuti(intm,intv,int{{intk=1;{}}}{intcas=0;{inttol=1*n[1]+2*n[2]+3*n[3]+4*n[4]+5*n[5]+6*n[6];{printf("Collection#%d:\n",(++cas));{}}{intcas=0;{inttol=1*n[1]+2*n[2]+3*n[3]+4*n[4]+5*n[5]+6*n[6];{printf("Collection#%d:\n",(++cas));{printf("Can'tbe}{for(inti=1;i<=6;if(dp[aim]==aim)printf("Canbeelse}}}return}Sample)[)(Sample66406usingSample66406usingnamespacestd;chars[110];boolboolmatch(inta,int{if(s[a]=='('&&s[b]==')')returnif(s[a]=='['&&s[b]==']')returntrue;returnfalse;}void{inti,j,k;{{}{}}for{{}}}int{{{}}}int{{}return}大和所在的子区间（L，R。0~9#intmaxsum(inta[],intleft,int{intintintsum1=0;intlefts=0;intsum1=0;intlefts=0;}intsum2=0;intrights=0;if(rights>sum2)sum2=rights;}if(localsum<leftsum)localsum=leftsum;}return}int{intintn,temp,i,j,k,max;}return}intmain(){intintn,temp,i,j,k,max;{{inttemp=0;}}}return}HDU100356-154}}}return}HDU100356-154-706-11-6-Sample144Case71publicclassMain{publicstaticvoidargs)Scannerscan=newm=a[]=newint[100001];i,j,m,n;(j=1;j<=m;j++)intsum=0,maxsum,startn=scan.nextInt();=0,=0,k=for(i}maxsum0;i<n;i++)=for(i0;i<n;i++)sum+=if(sum>{maxsum=sum;end=i;start=}ifk}<=i0)+}if(j!={System.out.println("Case}ifk}<=i0)+}if(j!={System.out.println("Case"+j+""+":");.println(maxsum+""+(start+1)+""+(end+1));}elseSystem.out.println("Case"+j+""+":");.println(maxsum+""+(start+1)+""+(end+}}}}A-ParkAnintegerT(T≤20)willexistinthefirstlineofinput,indicatingthenumberoftestcases.EachtestcasebeginswithtwointegersNandM(1≤N,M≤105),whichrespectivelydenotesthenumberofnodesandqueries.ThefollowingThenodesMlines,eachwithanlabeledfrom1toN.Sample1431424Foreachquery,outputtheminimumwalkingdistance,oneperSampleOutput4SampleOutput4到（因为是一棵树S-Tk，则我们只需要沿着路径一直走就2*xxnum直径的求法：两遍BFS(orDFS)于是原问题可以在O(Euv如果u在最长路上，那么v一定是最长路的一端。可以用反证法：假设v不是最长路的一端，则存在另一点v’使得(u→v’len(u→v’)len(u→v)。但这与条件“v是距u最远的点”矛盾。如果u不在最长路上，则u到其距最远点v的路与最长路一定有一交点c，且(c→v)与最长路的后半段重合(why?)，即v一定是最长路的一端一部分l1，一部分l2l2对称，取l1l2都一样）usingnamespaceconstintstructintconstintstructintvoidaddedge(intcu,int}voidDFS(intfor(inti=head[u];i!=-1;i=edge[i].nxt){intv=edge[i].to;}}}intt;intu,v;i=1;i<n;i++){scanf("%d%d",&u,&v);addedge(u,v);}intfor(inti=1;i<=n;i++)}for(inti=1;i<=n;i++)intfor(inti=1;i<=n;i++)int}}return}3Test4TestTest这算是八皇后问题的变种了