最大m字段和
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Max Sum Plus Plus
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 30233 Accepted Submission(s): 10634
Problem Description
Now I think you have got an AC in Ignatius.L's "Max Sum" problem. To be a brave ACMer, we always challenge ourselves to more difficult problems. Now you are faced with a more difficult problem.
Given a consecutive number sequence S1, S2, S3, S4 ... Sx, ... Sn (1 ≤ x ≤ n ≤ 1,000,000, -32768 ≤ Sx ≤ 32767). We define a function sum(i, j) = Si + ... + Sj (1 ≤ i ≤ j ≤ n).
Now given an integer m (m > 0), your task is to find m pairs of i and j which make sum(i1, j1) + sum(i2, j2) + sum(i3, j3) + ... + sum(im, jm) maximal (ix ≤ iy ≤ jx or ix ≤ jy ≤ jx is not allowed).
But I`m lazy, I don't want to write a special-judge module, so you don't have to output m pairs of i and j, just output the maximal summation of sum(ix, jx)(1 ≤ x ≤ m) instead. ^_^
Given a consecutive number sequence S1, S2, S3, S4 ... Sx, ... Sn (1 ≤ x ≤ n ≤ 1,000,000, -32768 ≤ Sx ≤ 32767). We define a function sum(i, j) = Si + ... + Sj (1 ≤ i ≤ j ≤ n).
Now given an integer m (m > 0), your task is to find m pairs of i and j which make sum(i1, j1) + sum(i2, j2) + sum(i3, j3) + ... + sum(im, jm) maximal (ix ≤ iy ≤ jx or ix ≤ jy ≤ jx is not allowed).
But I`m lazy, I don't want to write a special-judge module, so you don't have to output m pairs of i and j, just output the maximal summation of sum(ix, jx)(1 ≤ x ≤ m) instead. ^_^
Input
Each test case will begin with two integers m and n, followed by n integers S1, S2, S3 ... Sn.
Process to the end of file.
Process to the end of file.
Output
Output the maximal summation described above in one line.
Sample Input
1 3 1 2 32 6 -1 4 -2 3 -2 3
Sample Output
68
dp[i][j]表示前j项中i个字段和的最大值
dp[i][j]=max(dp[i][j-1]+dp[i-1][t])+num[j]
#include<iostream>//最大m字段和问题#include<stdio.h>#include<string.h>using namespace std;int val[1000001],ans[1000001],lans[1000001];int max(int a,int b){ return a>b?a:b; }int main(){ int n,m,i,j; //cout<<max(7,8); while(cin>>n>>m) { memset(ans,0,sizeof(ans)); memset(lans,0,sizeof(lans)); for(i=1;i<=m;i++) { scanf("%d",&val[i]); } long maxx; for(i=1;i<=n;i++) { maxx=-99*1000000; for(j=i;j<=m;j++) { ans[j]=max(ans[j-1],lans[j-1])+val[j];//当前的最大值是前一项的最大值和上一个状态较大的值加上目前的值 lans[j-1]=maxx ;//修改前一项的值,不能和前一行互换 if(ans[j]>maxx ) maxx=ans[j]; } } cout<<maxx<<endl; } return 0;}
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