1113 Integer Set Partition （25 分）

1113 Integer Set Partition （25 分）

Given a set of N (>1) 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 ∣n​1​​−n​2​​∣ is minimized first, and then ∣S​1​​−S​2​​∣ is maximized.

Input Specification:

Each input file contains one test case. For each case, the first line gives an integer N (2≤N≤10​5​​), 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​​.

Output Specification:

For each case, print in a line two numbers: ∣n​1​​−n​2​​∣ and ∣S​1​​−S​2​​∣, 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

分析：真水题。。。先排序然后砍成两半就可以了

 /**
 * Copyright(c)
 * All rights reserved.
 * Author : Mered1th
 * Date : 2019-02-27-13.43.25
 * Description : A1113
 */
 #include<cstdio>
 #include<cstring>
 #include<iostream>
 #include<cmath>
 #include<algorithm>
 #include<string>
 #include<unordered_set>
 #include<map>
 #include<vector>
 #include<set>
 using namespace std;
 ;
 int a[maxn];
 int main(){
 #ifdef ONLINE_JUDGE
 #else
     freopen("1.txt", "r", stdin);
 #endif
     ;
     cin>>n;
     ;i<n;i++){
         scanf("%d",&a[i]);
         sum+=a[i];
     }
     sort(a,a+n);
     ,s=;
     ;i<m;i++){
         s+=a[i];
     }
     ==){
         printf("0 %d",sum-s-s);
     }
     else{
         printf("1 %d",sum-s-s);
     }
     ;
 }