Source:

PAT A1113 Integer Set Partition (25 分)

Description:

Given a set of N (>) 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 ∣ is minimized first, and then ∣ is maximized.

Input Specification:

Each input file contains one test case. For each case, the first line gives an integer N (2), 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: ∣ and ∣, 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

Keys:

  • 简单模拟

Attention:

  • 408的一道真题,最快用O(N)规模完成,思路就是基于快排算法寻找中轴;但这里没卡时间就比较简单了-,-

Code:

 /*
Data: 2019-05-29 21:38:41
Problem: PAT_A1113#Integer Set Partition
AC: 08:30 题目大意:
给定N个整数的集合,把他们分为两个部分,要求两部分和的差最大且元素个数差最小
*/ #include<cstdio>
#include<algorithm>
using namespace std;
const int M=1e5+;
int s[M]; int main()
{
#ifdef ONLINE_JUDGE
#else
freopen("Test.txt", "r", stdin);
#endif int n,sum=;
scanf("%d", &n);
for(int i=; i<n; i++)
scanf("%d", &s[i]);
sort(s,s+n);
for(int i=; i<n/; i++)
sum+= (s[n--i]-s[i]);
if(n%==)
printf("0 ");
else{
printf("1 ");
sum += s[n/];
}
printf("%d\n", sum); return ;
}

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