PAT_A1113#Integer Set Partition

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+10;
int s[M];

int main()
{
#ifdef    ONLINE_JUDGE
#else
    freopen("Test.txt", "r", stdin);
#endif

    int n,sum=0;
    scanf("%d", &n);
    for(int i=0; i<n; i++)
        scanf("%d", &s[i]);
    sort(s,s+n);
    for(int i=0; i<n/2; i++)
        sum+= (s[n-1-i]-s[i]);
    if(n%2==0)
        printf("0 ");
    else{
        printf("1 ");
        sum  += s[n/2];
    }
    printf("%d\n", sum);

    return 0;
}