• 1113 Integer Set Partition (25 分)


    1113 Integer Set Partition (25 分)
     

    Given a set of N (>) positive integers, you are supposed to partition them into two disjoint sets A1​​ and A2​​ of n1​​ and n2​​ numbers, respectively. Let S1​​ and S2​​ denote the sums of all the numbers in A1​​ and A2​​, respectively. You are supposed to make the partition so that ∣ is minimized first, and then ∣ is maximized.

    Input Specification:

    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 231​​.

    Output Specification:

    For each case, print in a line two numbers: ∣ and ∣, separated by exactly one space.

    Sample Input 1:

    10
    23 8 10 99 46 2333 46 1 666 555
    

    Sample Output 1:

    0 3611
    

    Sample Input 2:

    13
    110 79 218 69 3721 100 29 135 2 6 13 5188 85
    

    Sample Output 2:

    1 9359
    
     
    排个序,判个奇偶
     
     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 }
     
     
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  • 原文地址:https://www.cnblogs.com/zllwxm123/p/11334541.html
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