题意：k个马棚，n条马，黑马1， 白马0，每个马棚unhappy指数：黑马数*白马数，问最小的unhappy值是多少
分析：dp[i][j] 表示第i个马棚放j只马的最小unhappy值，状态转移方程：dp[i][j] = min (dp[i][j], dp[i-1][k-1] + cur * (j - k + 1 - cur)); 表示k到j匹马放在第i个马棚的最小unhappy值，dp[0][0] = 0。由于黑马数是1的和，前缀sum[i]表示前i匹马黑马的个数，白马就是总个数-黑马数。
收获：简单递推DP
代码：
/************************************************
* Author :Running_Time
* Created Time :2015-9-1 12:03:47
* File Name :URAL_1167.cpp
************************************************/ #include <cstdio>
#include <algorithm>
#include <iostream>
#include <sstream>
#include <cstring>
#include <cmath>
#include <string>
#include <vector>
#include <queue>
#include <deque>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <bitset>
#include <cstdlib>
#include <ctime>
using namespace std; #define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
typedef long long ll;
const int N = 5e2 + 10;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
int dp[N][N], sum[N], a[N]; int main(void) {
int n, m;
while (scanf ("%d%d", &n, &m) == 2) {
memset (sum, 0, sizeof sum);
for (int i=1; i<=n; ++i) scanf ("%d", &a[i]), sum[i] = sum[i-1] + a[i];
memset (dp, INF, sizeof (dp));
dp[0][0] = 0;
for (int i=1; i<=m; ++i) {
for (int j=1; j<=n; ++j) {
for (int k=1; k<=j; ++k) {
int cur = sum[j] - sum[k-1];
dp[i][j] = min (dp[i][j], dp[i-1][k-1] + cur * (j - k + 1 - cur));
}
}
}
printf ("%d\n", dp[m][n]);
} return 0;
}