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, S
3, S
4 ... S
x, ... S
n (1 ≤ x ≤ n ≤ 1,000,000, -32768 ≤ S
x ≤ 32767). We define a function sum(i, j) = S
i + ... + 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(i
1, j
1) + sum(i
2, j
2) + sum(i
3, j
3) + ... + sum(i
m, j
m) maximal (i
x ≤ i
y ≤ 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(i
x, j
x)(1 ≤ x ≤ m) instead. ^_^
Input Each test case will begin with two integers m and n, followed by n integers S
1, S
2, S
3 ... S
n.
Process to the end of file.
Output Output the maximal summation described above in one line.
Sample Input
1 3 1 2 3
2 6 -1 4 -2 3 -2 3 Sample Output
6
8
Hint
Huge input, scanf and dynamic programming is recommended.
题意
从m个数中选n段，使累加和的累积和最大
思路
dp
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
using namespace std;
int a[1100000];
int dp[1010000];//表示前j个分成i-1分最大
int Max[1000100];//表示前j个分成i分最大
int main()
{
int n,m,maxx;
while(cin>>n>>m)
{
for(int i=1;i<=m;i++)
{
scanf("%d",&a[i]);
}
memset(dp,0,sizeof(dp));
memset(Max,0,sizeof(Max));
for(int i=1;i<=n;i++)
{
maxx = -0x3f3f3f3f;//刚上来必须新分一组
for(int j=i;j<=m;j++)
{
dp[j]=max(dp[j-1]+a[j],Max[j-1]+a[j]);//判断分成i组，第j个是新分一组还是分到前一组
Max[j-1] = maxx;//前j-1结尾分成i组最大值
maxx = max(maxx,dp[j]);//前j个分成i组最大值
}
}
cout<<maxx<<endl;
}
return 0;
}
