一些代码动态区间

#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>#include<map>#include<set>#include<bitset>#include<cmath>#include<string>#include<cstdlib>#include<vector>#defineXfirst#defineYsecond#definempmake_pair#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>#include<map>#include<set>#include<bitset>#include<cmath>#include<string>#include<cstdlib>#include<vector>#defineXfirst#defineYsecond#definempmake_pair#definesqr(x)((x)*#defineRep(i,n)for(inti=0;i<(n);#defineforeach(it,n)for(typeof(n.begin())it=n.begin();it!=n.end();it++)usingnamespacestd;typedeflonglongLL;typedefpair<int,int>pii;#defineesp1e-8#defineN5010intn,v[N],vector<pii>que;intdp[N];voidcheck(int&x,inty)if(x==-1||x>y)x=}intinit()if(n==1)returnintl=0,r=n-1,l_num=1,r_num=1;LLl_val=v[0],r_val=v[n-1];whilelrif(l_val<r_val)if(l+1==r)returnl_num+r_num;l_val+=v[++l];l_num}elseif(l_val>r_val)if(l+1==r)returnl_num+r_num;r_val+=v[--r];r_num}elseque.push_back(mp(l_num,ifl1rreturn0;//ifl2rreturn1;//中间有一个数，最后操作要算上合并一个数的代价l_num=r_num=1;l_valv[++l];//进行下一组关节点的查找r_val=v[--r];}}return}intmain()while(scanf("%d",&n),n>Rep(i,n)scanf("%d",Rep(i,n)scanf("%d",&a[i+1]);intmid_element=init();intnn=que.size();memset(dp,-1,sizeofdp);dp[0]=0;for(inti=0;i<nn;i++){intnum1=0,num2=0;forfor(intj=i;j<nn;j++)num1+=que[j].X,num2+=que[j].Y;check(dp[j+1],dp[i]+a[num1]+a[num2]);}}intans=-for(inti=nn;i>=0;i--)check(ansdp[ia[mid_element]);//if(i)mid_element+=que[i-1].X+que[i-}printf("%d\n",}return}eam120可以进行合并，合并得到的能量是前一个数字的头数*后一个数字的头数*后一个数字的尾把每次合并得到的能量加起来，问最后怎么合并会使能量和最大，输出能量。dp，dp[i][j]ij(1)(2,3,...,N)1N一切为二，前半段聚合成一个珠子，后半段聚合成一个珠子，再把聚合后forij之f(1,N)的值，也即题目的答案e[N+i]=e[i];for循环的循环条件#include<iostream>#include<cstdio>#include<cstring>#include<iomanip>#include<algorithm>usingnamespacestd;intN;intintintmain{memset(f,0,sizeof(f));inti,j,k,max=0;scanf("%d",&N);for(i=1;i<=N;{scanf("%d",e[Ni]e[i];//}for(j=2;j<=2*N-1;for(i=j-1;i>=1&&j-i<N;i--)for(k=i;k<j;k++){if(f[i][k]f[k1][j]e[i]*e[k1]*e[j1]>f[i][j])//k把ij分成f[i][j]=f[i][k]+f[k+1][j]+e[i]*e[k+1]*e[j+}for(i=1;i<=N;{if(f[i][iN1]>max)//遍历每个起点到终点找最大max=f[i][i+N-1];}printf("%d\n",max);return0;}Codes159D-Palindromepairs给你一串字符串，问你找出不重复的两段回文子串，问能#define_CRT_SECURE_NO_WARNINGS#include<iostream>#include<cstdio>#include<cstdlib>#include<cstring>#include<cmath>#include<climits>#include<string>#include<vector>#include<set>#include<queue>#include<stack>#include<map>#include<iomanip>#include<algorithm>usingnamespacestd;typedeflonglongconstdoublepi=acos(-1.0);constintmod= constdoubleeps=1e-12;llmaxx,minn;llmaxi,mini;intcases=intl[2010],intmain{inti,j,k;stringa;intlen;while(cin>>{memset(l,0,memset(r,0,sizeof(r));len=a.length();for(i0;ilen;i++)//{for(j=i,k=i;j>=0&&k<len&&a