PAT A1113 Integer Set Partition （25 分）——排序题

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 N positive 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

 #include <stdio.h>
 #include <algorithm>
 using namespace std;
 const int maxn=;
 int seq[maxn];
 int total[maxn];
 int main(){
   int n;
   int sum=;
   scanf("%d",&n);
   for(int i=;i<n;i++){
     scanf("%d",&seq[i]);
   }
   sort(seq,seq+n);
   for(int i=;i<n;i++){
     sum+=seq[i];
     total[i]=sum;
   }
   printf("%d %d",n%,total[n-]-*total[n/-]);
 }

注意点：计算差的时候由于总和算了前面部分，要多减一次前半部分。感觉直接算和然后相减也不会超时