洛谷P1043数字游戏

题目

区间DP，将$maxn[i][j][k]$表示为i到j区间内分为k个区间所得到的最大值，$minn$表示最小值。

然后可以得到状态转移方程：

$$maxn[i][j][k]= max(maxn[i][j][k],maxn[i][l][k-1]*(l到j的\%10后的和))$$。 然后判断一下边界就能得到最终答案。 ```c++ #include <bits/stdc++.h> using namespace std; int n, m, data[1001010]; int sum[111], ha[111][111];//每一段区间的和 int maxn[111][111][111], minn[111][111][111]; int ans1, ans2 = 2147483647; int main() { scanf("%d%d", &n, &m); for (int i = 1; i <= n; i++) { scanf("%d", &data[i]); data[i + n] = data[i]; } n *= 2; for (int i = 1; i <= n; i++) sum[i] = sum[i - 1] + data[i]; memset(minn, 12, sizeof(minn)); for (int i = 1; i <= n; i++) for (int j = i + 1; j <= n; j++) ha[i][j] = (sum[j] - sum[i] + 100000) % 10, maxn[i][j][1] = minn[i][j][1] = ha[i][j]; for (int k = 2; k <= m; k++) for (int i = 1; i <= n; i++) for (int j = i + k - 1; j <= n; j++) { minn[i][j][k] = 214748367; for (int l = i + k - 2; l < j; l++) { maxn[i][j][k] = max(maxn[i][l][k - 1] * ha[l][j], maxn[i][j][k]); minn[i][j][k] = min(minn[i][l][k - 1] * ha[l][j], minn[i][j][k]); } } for (int i = 1; i <= n / 2; i++) ans1 = max(maxn[i][i + n / 2][m], ans1), ans2 = min(minn[i][i + n / 2][m], ans2); printf("%d\n%d", ans2, ans1); return 0; } ```$$