SPOJ Mixtures Interval DP

Harry Potter had n mixtures in front of him, arranged in a row. Each mixture has one of the different colors (colors having numbers from 0 to 99).

He wants to mix all these mixtures together. At each step, he's going to take the mixtures that stand next to each other and mix them together, and put the resulting Mixture in their place.

When mixing mixtures of colors A and B, the resulting mixture would have the color (a+b) mod 100.

Also, there'll is some smoke in the process. The amount of smoke generated when mixing-mixtures of colors A and B is a*b.

Find out what's the minimum amount of smoke that Harry can get when mixing all the mixtures together.

Input

There'll is a number of test cases in the input.

The first line of all test case would contain N, the number of mixtures, 1 <= n <= 100.

The second line would contain n integers between 0 and 99-the initial colors of the mixtures.

Output

For each test case, output the minimum amount of smoke.

Example

Input:218 19340Output:3422400

In the second test case, there is both possibilities:

• First Mix (smoke:2400), getting 0, then Mix 0 and (smoke:0); Total amount of smoke is 2400
• First mix and smoke:1200, getting, then Mix (smoke:3200); Total amount of smoke is 4400

The first scenario is a much better-to proceed.

Exercises

This is a stupid interval DP, there is no difficulty, pre-processing the interval and can

See the code:

#include <iostream> #include <cstring> #include <algorithm> #define MAX_N 105#define INF 0x3f3f3f3fusing namespace Std;int dp[max_n][max_n];int sum[max_n];int n;int a[max_n];int Main () {Cin.sync_with_stdio (f    Alse);                while (CIN >> N) {for (int i = 0, I <= N; i++) for (int j = 0; J <= N; j + +)        DP[I][J] = INF;        memset (sum, 0, sizeof (sum));            for (int i = 1; I <= n; i++) {cin >> a[i];            Dp[i][i] = 0;        Sum[i] = Sum[i-1] + a[i]; } for (int i = 1; I <= n; i++.) for (int j = 1; j + I <= N, j + +) for (int k = j; K & Lt j + i;                                                                k++) Dp[j][j + i] = min (dp[j][j + i], Dp[j][k] + dp[k + 1][j + i] +                                                                ((Sum[k]-sum[j-1])% 100) *        ((Sum[j + i]-sum[k]) (100)); cout << dp[1][n] << endl } return 0;}

　　

SPOJ Mixtures Interval DP