PAT (Advanced Level)1113. Integer Set Partition (25)

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

愿你一生平坦考试全遇水题

愿你通宵熬夜归来仍是浓密黑发

//1113. Integer Set Partition(25)
#include <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;
int main() {
	int n, a[100000],sumin=0,sumax=0;
	cin >> n;
	for (int i = 0; i < n; i++) {
		cin >> a[i];
	}
	int mid = n / 2;
	sort(a, a + n);
	for (int i = 0; i < mid; i++)
		sumin += a[i];
	for (int i = mid; i < n; i++)
		sumax += a[i];
	printf("%d ", n % 2);
	cout << sumax - sumin;
	return 0;
}