1113 Integer Set Partition (25 分)
标签：tee nta ica art rate pts scripts inf view
Given a set of N (>) positive integers, you are supposed to partition them into two disjoint sets A?1?? and A?2?? of n?1?? and n?2?? numbers, respectively. Let S?1?? and S?2?? denote the sums of all the numbers in A?1?? and A?2??, respectively. You are supposed to make the partition so that ∣ is minimized first, and then ∣ is maximized.
Each input file contains one test case. For each case, the first line gives an integer N (2), and then Npositive integers follow in the next line, separated by spaces. It is guaranteed that all the integers and their sum are less than 2?31??.
For each case, print in a line two numbers: ∣ and ∣, separated by exactly one space.
10
23 8 10 99 46 2333 46 1 666 555
0 3611
13
110 79 218 69 3721 100 29 135 2 6 13 5188 85
1 9359
1 #include <bits/stdc++.h> 2 3 using namespace std; 4 int n, an[200005]; 5 int main(){ 6 cin >> n; 7 for(int i = 0; i < n; i++){ 8 cin >> an[i]; 9 } 10 sort(an, an+n); 11 int ans = 0, cnt = 0; 12 for(int i = 0; i < n/2; i++){ 13 ans += an[i]; 14 } 15 for(int i = n/2; i < n; i++){ 16 cnt += an[i]; 17 } 18 cout << (n&1) <<" "<<cnt-ans<<endl; 19 return 0; 20 }
1113 Integer Set Partition (25 分)
标签：tee nta ica art rate pts scripts inf view
原文地址：https://www.cnblogs.com/zllwxm123/p/11334541.html
