# hdu 1024 Max Sum Plus Plus

xiaoxiao2021-02-28  5

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.

dp

#include <iostream> #include <algorithm> #include <cstdio> #include <cstring> using namespace std; int a; int dp;//表示前j个分成i-1分最大 int Max;//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; }