Max Sum Plus Plus

1 3 1 2 3
2 6 -1 4 -2 3 -2 3

6
8

Hint
Huge input, scanf and dynamic programming is recommended.

/*
** dp[i][j]表示以第i个数字结尾且选定并分成j份能得到的最大值。转移方程为
** dp[i][j] = max(dp[i-1][j], max(dp[1...i-1][j-1])) + arr[i];
** 假设开二维数组的话内存会超，所以得用滚动数组省空间。preMax[j]保存
** 上一轮得到的dp[1...i][j]中的最大值，ans每次读取当前dp数组最大值
** 用以更新preMax数组，最后一轮循环后ans保存的就是答案。
*/

#include <stdio.h>
#include <string.h>

#define maxn 1000010
#define inf 0x7fffffff

int dp[maxn], preMax[maxn], arr[maxn];

int max(int a, int b) {
	return a > b ? a : b;
}

int main() {
	int n, m, i, j, ans;
	while(scanf("%d%d", &n, &m) == 2) {
		for(i = 1; i <= m; ++i) {
			scanf("%d", &arr[i]);
			preMax[i] = dp[i] = 0;
		}
		preMax[0] = dp[0] = 0;
		for(j = 1; j <= n; ++j) { // 分成j份
			ans = -inf;
			for(i = j; i <= m; ++i) {
				dp[i] = max(dp[i-1], preMax[i-1]) + arr[i];
				preMax[i-1] = ans;
				ans = max(ans, dp[i]);
			}
		}
		printf("%d\n", ans);
	}
	return 0;
}