Va 562-Dividing coins (01 backpack)

The nominal value of n coins should be divided into two heaps (the average number is not required), and the absolute value of the sum of the two heaps of coins should be minimized, the absolute value of the minimum output value. Solution: Consider half of the sum of half as the capacity of the backpack, and then perform the 01 backpack processing, and then take the reachable value cur as close as possible to half, two times of half-cur is the required value ans. Note that when sum is an odd number, ans must add 1.

#include <stdio.h>  #include <string.h>  const int N = 50005;  const int M = 105;    int n, sum, dp[N], coin[M], Max;    int solve() {      memset(dp, 0, sizeof(dp));      dp[0]= 1;      int half = sum / 2;        for (int i = 0; i < n; i++) {      for (int j = half - 1; j >= 0; j--)          if (dp[j])  dp[j + coin[i]] = 1;      }        for (int i = 0; i <= half; i++)      if (dp[half - i])   return i * 2;  }    int main() {      int cas;      scanf("%d", &cas);      while (cas--) {      sum = 0;      scanf("%d", &n);      memset(coin, 0, sizeof(coin));      for (int i = 0; i < n; i++) {          scanf("%d", &coin[i]);          sum += coin[i];      }      printf("%d\n", solve() + sum % 2);      }      return 0;  }