USACO 3.3.5

博弈问题，可以使用dp解决.
f[i][j] =s[i][j] - min{f[i+1][j], f[i][j-1]}
s[i][j]表示从i加到j的和,枚举l=j-i.

Executing...
   Test1: TEST OK [0.011 secs, 3108 KB]
   Test2: TEST OK [0.011 secs, 3108 KB]
   Test3: TEST OK [0.011 secs, 3108 KB]
   Test4: TEST OK [0.011 secs, 3108 KB]
   Test5: TEST OK [0.011 secs, 3108 KB]
   Test6: TEST OK [0.011 secs, 3108 KB]
   Test7: TEST OK [0.000 secs, 3108 KB]
   Test8: TEST OK [0.000 secs, 3108 KB]
   Test9: TEST OK [0.000 secs, 3108 KB]
   Test10: TEST OK [0.000 secs, 3108 KB]
   Test11: TEST OK [0.011 secs, 3108 KB]
   Test12: TEST OK [0.011 secs, 3108 KB]
   Test13: TEST OK [0.000 secs, 3108 KB]
   Test14: TEST OK [0.000 secs, 3108 KB]
   Test15: TEST OK [0.011 secs, 3108 KB]
   Test16: TEST OK [0.000 secs, 3108 KB]

All tests OK.

YOUR PROGRAM('game1') WORKED FIRST TIME! That's fantastic -- and a rare thing.Please accept these special automated congratulations.

[Code]

#include<stdio.h>#include<iostream>using namespace std;const int MAXN = 100+10;int N[MAXN], f[MAXN][MAXN] = {0}, s[MAXN][MAXN] = {0}, sum[MAXN] = {0};int min(int x, int y) {return (x < y) ? x : y;}int main(){ FILE *fin, *fout; fin = fopen("game1.in", "r"); fout = fopen("game1.out", "w"); int n=0, i, j, l; fscanf(fin, "%d", &n); for (i = 1; i <= n; i++) fscanf(fin, "%d", &N[i]); for (i = 1; i <= n; i++) { s[i][i] = N[i]; for (j = i+1; j <= n; j++) s[i][j] = s[i][j-1] + N[j]; } for (i = 1; i <= n; i++) f[i][i] = N[i]; for (l = 1; l < n; l++) for (i = 1; i <= n-l; i++) f[i][i+l] = s[i][i+l] - min(f[i+1][i+l], f[i][i+l-1]); fprintf(fout, "%d %d/n", f[1][n], s[1][n]-f[1][n]);}