一些代码-数学问题

UVAOJ10061nb0。Log10(a*b)=log10(a)+log(b)0N!分解为素数的积，记录下<=B的素数因子的个数（B的原因代码中解释了N!的因子中（BN!B的因子用来B时结束B0.usingnamespacestd;intsu[ intdigit(intnn,int{intdoublesum=0;}intzero(intnn,int{inti,j;intintnum;{{{}}}{{{elsegotoend;}}}return}int{intnn,bb;inti;intzz,dd;{cout<<zz<<""<<dd<<endl;}return}:超长方体(1+2+…+n)^4-超立方体usingnamespacestd;intmain(){longlonglonglongs2[101],s3[101],s4[101];longlongr2[101],r3[101],r4[101];{r2[i]=(i*(i+1)/2)*(i*(i+1)/2)-r3[i]=(i*(i+1)/2)*(i*(i+1)/2)*(i*(i+1)/2)-}{cout<<s2[n]<<""<<r2[n]<<""<<r3[n]<<""<<s4[n]<<"}return}UVAOJ10014a0,an+1,c1,c2,…cn,ai=(a(i-1)+a(i+1))/2-ci,a1。先全部累加：a1+an=a0+a(n+1)–2Σci(1<=i<=n)按照上面得出的公式再累加:(n+1)a1n*a0a(n+12ΣΣcj(1<=j<=i#include<stdio.h>#include<string.h>intmain(){intnCases;intnN;doublea0,an1;doublea1;doubleci[3000+10];doublec;doublesum;scanf("%d",&nCases);while(nCases--){scanf("%d",&nN);scanf("%lf%lf",&a0,&an1);sum=0.0;memset(ci,0,sizeof(ci));for(intj=1;j<=nN;++j){scanf("%lf",ci[j]ci[j1]c;//ci[j]代表数列ci中第1-j的sum+=}a1=(nN*a0+an1-2*sum)/(nN+printf("%.2f",if(nCases){}}return}HDU486120141k，pk1^i+2^i+...+(p-1)^imodp（1i<=k)，现在两人轮流取球，最后球的值总和大的人赢，问先手是否能赢#include<cstdio>#include#include<cstdio>#include<cstring>#include<cmath>longlongk,intmain()while(~scanf("%lld%lld",&k,&p)){if(k/(p-1)%2)printf("YES\n");elseprintf("NO\n");}return}≡1（moda是整数，pa,pa的(p-1)p的余1。φ函数的值通式：φ(x)=x(1-1/p1)(1-1/p2)(1-1/p3)(1-1/p4)…..(1-1/pn),p1,p2……pnx的所有质因数，x0的整数。φ(1)=1（1互质的数(1)1本身。(12=2*2*3那么φ（12）=12*（1-1/2）*(1-1/3)=4nnφ(n)nn素的正整数的个数，称为n的函数值，这里函数φ：N→N，n→φ(n)称为函数。m,n互质，φ(mn)=φ(m)φ(n)。n为奇数时，φ(2n)=φ(n),PrimitiveRoot假设一个数gPg^imodP的结果两两不同,且有1<g<P0<i<P,g可以称为是P的一个原根,g^(P-11modP)当且仅当指数P-1的时候成立.(P是素数).简单来说，g^imodpg^jmodp（p为素数sum次操作的话，其实有很多牌是经过了多次翻11的总数的奇偶性是固定的。数公式必定有除法C（n,m）=n!/(m!*(n-m)!)但是我们知道费定理a^(p-1)=1%p那么a^(p-1)/a=1/a%p得到a^(p-2)=#include<iostream>#include<stdio.h>#include<string.h>#include#include<iostream>#include<stdio.h>#include<string.h>#include<algorithm>usingnamespacestd;#definemod#define #definemaxn100000+5LLf[maxn];void{inti;f[0]=1;for(i=1;i<maxn;i++)f[i]=(f[i-1]*i)%mod;}LLquickmod(LLa,LL{LLans=1;{{ans=b--}a=}return}int{intn,m,i,j,k,l,r,x,ll,rr;{l=r=for(i=0;i<n;{if(l>=xlll-x;//当最小的1个数大于x，把x1elseif(r>=xlll%2)==(x%2))?0:1;//当l<x<=r，由于无论怎么翻，其奇偶性必定相等，所以看l的奇偶性与x是否相同，相同那么知道最小必定变为0，否则变为1elsellx-r;//当x>r,那么在把1全部变为0的同时，还有x-r0变为if(r+x<=mrrr+x;//当r+x<=m的情况下，全部变为elseif(l+x<=mrrl+x)%2m%2)?m:m-1);//r+x>m但是elserr2*m-(l+x);//在l+x>m的情况下，等于我首先把m1变为了0，那么我还要翻（l+x-m）张，所以最终得到m-(l+x-m)1l=ll,r=}LLsum=for(i=l;i<=r;i+=2)//使用 }return}HDU49192014 n最大是先乘不部拿出1就是外面加上个kk异或0=k所以拿出外面也k。JAVA方便，上面的递推式用简单的递归就行了。importimportjava.util.*;importjava.io.*;importpublicclassMainpublicstaticBigIntegertwo=BigInteger.valueOf(2);publicstaticBigIntegerfour=BigInteger.valueOf(4);publicstaticBigIntegersix=BigInteger.valueOf(6);publicstaticHashMap<BigInteger,BigInteger>map=newHashMap<BigInteger,publicstaticBigIntegerF(BigIntegern){if(map.containsKey(n))returnBigIntegerBigIntegertmp1=n.divide(two);BigIntegertmp2=n.mod(two);BigIntegertmp3=tmp1.subtract(BigInteger.ONE);if( pareTo(BigInteger.ONE)==0)tmp1=F(tmp1).multiply(four).add(tmp1.multiply(six));tmp1=map.put(n,tmp1);returntmp1;}publicstaticvoidmain(String[]args){Scannercin=newScanner(System.in);map.put(BigInteger.ZERO,BigInteger.ZERO);map.put(BigInteger.ONE,BigInteger.ZERO);while(cin.hasNext()){BigIntegern=cin.nextBigInteger();}}}HDU2058找一个子序列的和便是Ma+1a+2...数学题，等差数列的运用。Sna1+an)*n2a1a1n-1*d)*n/2#include<stdio.h>#include<math.h>int{intn,m,#include<stdio.h>#include<math.h>int{intn,m,a,while(scanf("%d%d",&n,&m)&&(n||{len=(int)sqrt(2*m);while(len--){a=m/(len+1)-len/if((2*a+len)*(len+1)/2==m)printf("[%d,%d]\n",a,}}return}HDU49232014AB数列(0<=bi<=1)A0...01...1bi=ai0i<jbi<=bjbibjbibj的值都可以得到更优的结果。所以，bi=bj。bi值一定相同。这样，我们可以得到一个中间序列，x1x1,x2,...,x2,xm,...xm.x1xmxixixibi相等时取得最优值。因此，我们可以把这两段合并，求出B序列一定是这样：y1y1ym,ym。所以我们可以用一个单调栈来维bi值。#include<iostream>#include<cstdio>#include<cstring>#include<cstdlib>#include<cmath>#include<algorithm>#include<string>#include<queue>#include<stack>#include<vector>#include<map>#include<set>#include#include<iostream>#include<cstdio>#include<cstring>#include<cstdlib>#include<cmath>#include<algorithm>#include<string>#include<queue>#include<stack>#include<vector>#include<map>#include<set>#include<functional>#include<time.h>usingnamespacetypedefpair<double,int>constintINF=constintMAXN=(int)intinta[MAXN];doubleb[MAXN];PDIsk[MAXN//inttail//栈顶指voidvoidsolve()tail0//for(inti=0;i<n;i++)sk[tailmake_pair(1.0*a[i1//whiletail>1&&sk[tail-1].first<=sk[tail-2].first//如果栈不满足单调intcnt=sk[tail-1].second+sk[tail-2].second;doubletmp=sk[tail-2]=make_pair(tmp,cnt);}}//for(inti=0,j=0;i<tail;for(intk=0;k<sk[i].second;k++)b[j++]=sk[i].first;doubleans=for(inti=0;i<n;ans+=(a[i]-b[i])*(a[i]-printf("%f\n",}intmain(){intT;scanf("%d",while(T--)scanf("%d",for(inti=0;i<n;i++)scanf("%d",&a[i]);}return}HDU49372014nbase3，4，5，6组成的话，basenn，求有多少个幸运进制。无穷多个的话输出-1，单个位置9用相应的字符表示。3100001e124位，直接枚举每一位计算N即可#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>#include<vector>#include<utility>#include<stack>#include<queue>#include<map>#include<deque>#include<cmath>#definemax(x,y)#definemin(x,y)usingnamespace#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>#include<vector>#include<utility>#include<stack>#include<queue>#include<map>#include<deque>#include<cmath>#definemax(x,y)#definemin(x,y)usingnamespacelonglongx,N;intans;inta[6];map<int,bool>mapp;boolcheck(longlongx)//{{if(x>=3&&x<=6)return1;elsereturn0;}for(inti=3i<=min(6,N-1i++)//6N-1if((x-i)%N==0&&check((x-i)/N))//3-6能被三整出，returnreturn}voiddfs(intp)//100003，4，5，6枚举这三位二分进制算{if(p==3||{longlongl=1e4,r=1e12;//最大最小边界if(p==4r=1e6;//如果四位那么最大边界可以缩小{longlongm=(l+r)/2;longlongtmp=0;for(inti=1;i<=p-1;i++)if(tmp>x)r=m;elsel=m;}longlongfor(inti=1;i<=p-1;tmp=tmp*l+a[i];//dfs出来的结果来算xif(tmp==x&&l!=1e4&&mapp[l])//10000map中之前没有这进制{}if(l!=r&&r!=1e4)//r{for(inti=1;i<=p-1;i++)if(tmp==x&&{}}}if(p==4)for(inti=3i<=6i++)//3-6{}}int{inttt;for(intfor(intt=1;t<=tt;{printf("Case#%d:",t);{}for(N=4;N<=10000;N++)//4到10000直 进}return}HDU49522014xkkxxk的倍x。(1<=xk<=10^10)#include#include<cstdio>longlongx,s;int;intmain(){while(1){if(x==0&&s==0)break;for(inti=1;i<s;i++){if(x<i+1)}printf("Case#%d:%I64d\n",++}}SPOJ10232问在自然数中，第n3intcntfor(inti=2;i<=n;{ifni0)//{while(n%i==0{n/=}}}returncnt>=#include<iostream>#include<cstdio>usingnamespacestd;intans[1111]={…};intmain(){intn,m;{}}livearchive59861，表示他们之中有人赢了后面某几人。3346n(n-1)/2。usingnamespacestd;intints[51],p[51];intt,i,n,res;}}}return}c/d0<c<d<busingnamespacestd;intan[100]; (inta,int{intr=0;{}return}intfenjie(inta,intb,intan[])//{intn=0;intt;{}returnn;}int{intn,A,B,C,D;{for(int{intd= {}intlen=fenjie(A,B,an);for(intj=len-2;j>=0;j--{intt=an[j]*D+C;}}}return}扩展算法是用来在已知a,b求解一组x，y使得ax+by= (a,b)=d(解一定存在，intex(inta,intb,int&x,int&y){if(b==0){x=1;y=0;return}intr=ex(b,a%b,x,intt=x;x=y;y=t-a/b*y;returnr;}1（0的话表示可以约分，直接输出约分后的分数。分母尽可能大。a*d-b*c=1b*c-由扩展a*x+b*y=(a,b);所以d=x,c=-y或者c=y,d=-int64int64int64int64 int64 (int64a,int64b,int64&x,int64{int64if(b==0){x=1;y=0;returna;else{ans=extended_ (b,a%b,x,y);t=x;x=y;y=t-(a/b)*y;}returnans;}int{{ d1=(x+b)%b;c1=(-c2=(y+a)%a;d2=(-if(d1<d2)printf("%I64d/%I64d\n",c2,d2);elseprintf("%I64d/%I64d\n",c1,d1);}return}CodeforcesRound#266a*b6个直接，从6*n开始找出这个数能拆成的两个数的积的可能性，然后和原先的ab进行比#include<iostream>usingnamespacestd;typedeflonglongll;intmain(){lln,a,b;lli,j;while(cin>>n>>a>>{n*=if(a*b>={cout<<a*b<<cout<<a<<""<<b<<endl;}while{for(i=min(a,b);i*i<=n;i++)//拆当前{if(n%i=={j=n/if(i>=min(a,b)&&j>=max(a,{cout<<n<<if(a>b)cout<<j<<""<<i<<endl;elsecout<<i<<""<<j<<endl;gotoend;}}}}}return}Dnk(xix平均)^2xi^2x平均^22*xi*x平均所以方差=∑xi^2+n*x平均-2*∑xi*x平均事先计算∑xi^2和∑xi，便可以以o(n)#define#define_CRT_SECUR