CSL分蘋果(dp)
阿新 • • 發佈:2018-12-10
思路:dp[j]來表示當最多可容納 j 質量的質量,則dp方程:
#include<bits/stdc++.h> using namespace std; #define mem(a,b) memset(a,b,sizeof(a)) #define LL long long int main() { int n, sum, w[110], dp[10010]; while (cin >> n) { sum = 0; memset(dp, 0, sizeof(dp)); for (int i = 0; i < n; i++) { cin >> w[i]; sum += w[i]; } for (int i = 0; i < n; i++) for(int j = sum / 2; j >= w[i]; j--) dp[j] = max(dp[j], dp[j - w[i]] + w[i]); cout << dp[sum / 2] << " " << sum - dp[sum / 2] << endl; } return 0; }