1113 Integer Set Partition (25)

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
这题太水了， 排序求前n/2项和，再用总和相减就得到结果

 1 #include<iostream>
 2 #include<vector>
 3 #include<algorithm>
 4 using namespace std;
 5 int main(){
 6   int n, i, total=0;
 7   cin>>n;
 8   vector<int> v(n);
 9   for(i=0; i<n; i++) {scanf("%d", &v[i]); total+=v[i];}
10   sort(v.begin(), v.end());
11   int x=0;
12   for(i=0; i<n/2; i++) x += v[i];
13   printf("%d %d", n%2==0?0:1, total-2*x);
14   return 0;
15 }