ZOJ - 1880 Tug of War

题意：求在两边人数不相差超过1个的情况下，实力尽量相等的情况

思路：从实力和的一半开始类背包操作

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int MAXN = 45010;const int MAXM = 110;int a[MAXM];int dp[MAXN][MAXM];int n;int main(){int t,flag = 0;while (scanf("%d", &n) != EOF){int r = n/2;if (n & 1)r++;int sum = 0;for (int i = 0; i < n; i++){scanf("%d", &a[i]);sum += a[i];}memset(dp,0,sizeof(dp));dp[0][0] = 1;for (int i = 0; i < n; i++)for (int j = sum/2; j >= a[i]; j--)for (int k = r; k >= 1; k--)if (dp[j-a[i]][k-1])dp[j][k] = 1;int ans = 0;for (int i = sum/2; i >= 0; i--){if (n%2 == 0){if (dp[i][r]){ans = i;break;}}if (n%2 == 1){if (dp[i][r] || dp[i][r-1]){ans = i;break;}}}printf("%d %d\n", ans, sum-ans);}return 0;}