HDU
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Max Sum Plus Plus
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 S 1, S 2, S3, S 4 ... S x, ... Sn (1 ≤ x ≤ n ≤ 1,000,000, -32768 ≤ S x ≤ 32767). We define a function sum(i, j) = Si + ... + S j (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, j 1) + sum(i 2, j 2) + sum(i 3, j 3) + ... + sum(im, j m) maximal (i x ≤ iy ≤ j x or i x ≤ j y ≤ j x 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, j x)(1 ≤ x ≤ m) instead. ^_^
Process to the end of file.
1 3 1 2 32 6 -1 4 -2 3 -2 3
68
Huge input, scanf and dynamic programming is recommended.
题意:
求最大m段子序列和
(可以直接用模板)
代码:
#include<stdio.h> #include<algorithm> #include<string.h> #define N 1000000 #define INF 0x3f3f3f3fint a[N+5]; int now[N+5]; int pre[N+5]; using namespace std; //最大m段和 (数组下标从一开始) int MaxSum(int m,int n,int *a){int maxx;memset(now,0,sizeof(now)); memset(pre,0,sizeof(pre)); for(int i=1;i<=m;i++) { maxx=-INF; for(int j=i;j<=n;j++) { now[j]=max(now[j-1]+a[j],pre[j-1]+a[j]); pre[j-1]=maxx; if(now[j]>maxx) { maxx=now[j]; } } } return maxx;}int main() { int n,m,maxx,i,j; while(scanf("%d%d",&m,&n)!=EOF) { for(i=1;i<=n;i++) scanf("%d",&a[i]); printf("%d\n",MaxSum(m,n,a)); } return 0; }
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