题目链接:
https://vjudge.net/problem/UVA-562
题目大意:
给定n个硬币,要求将这些硬币平分以使两个人获得的钱尽量多,求两个人分到的钱最小差值
思路:
它所给出的n个钱币加起来sum,将sum/2当作体积,求出在sum/2下的最大值,sum-2*dp[sum/2]
1 #include<bits/stdc++.h> 2 using namespace std; 3 const int maxn = 105; 4 const int maxm = 1e5+10; 5 int a[maxn]; 6 int dp[maxm]; 7 int T, n, m; 8 int main() 9 { 10 cin >> T; 11 while(T--) 12 { 13 cin >> n; 14 int sum = 0; 15 memset(dp, 0, sizeof(dp)); 16 for(int i = 0; i < n; i++)cin >> a[i], sum += a[i]; 17 for(int i = 0; i < n; i++) 18 { 19 for(int j = sum / 2; j >= a[i]; j--) 20 dp[j] = max(dp[j], dp[j - a[i]] + a[i]); 21 } 22 cout<<(sum - 2 * dp[sum / 2])<<endl; 23 } 24 return 0; 25 }
UVA-562 Dividing coins---01背包+平分钱币
原文:https://www.cnblogs.com/fzl194/p/8783268.html