PAT1113: Integer Set Partition

1113. Integer Set Partition (25)

时间限制

150 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

Given a set of N (> 1) positive integers, you are supposed to partition them into two disjoint sets A₁ and A₂ of n₁ and n₂numbers, 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 |n₁ - n₂| is minimized first, and then |S₁ - S₂| is maximized.

Input Specification:

Each input file contains one test case. For each case, the first line gives an integer N (2 <= N <= 10⁵), 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 2³¹.

Output Specification:

For each case, print in a line two numbers: |n₁ - n₂| and |S₁ - S₂|, 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

思路

水题--
1.先求min|n1-n2|
N为奇数时|n1-n2| = 1，反之|n1-n2| = 0;

2.由min|n1-n2| => max|S1-S2|
可以先把输入的数组num排序，然后令s1是num的前(N/2 - 1)个数的和,S2是num剩下数的和，此时|S1-S2|最小。

代码

#include<iostream>
#include<vector>
#include<algorithm>
#include<math.h>
using namespace std;
int main()
{
   int N;
   while(cin >> N)
   {
     int n = N % 2;
     vector<int> nums(N);
     for(int i = 0;i < N;i++)
     {
         cin >> nums[i];
     }
     sort(nums.begin(),nums.end());
     int s1 = 0,s2 = 0;
     int index = N/2 - 1;
     for(int i = 0;i <= index;i++)
     {
         s1 += nums[i];
     }
     for(++index;index < N;index++)
     {
         s2 += nums[index];
     }

     cout << n << " " << abs(s1-s2) << endl;
   }
}