PAT甲级1113 Integer Set Partition (25分)

最新推荐文章于 2024-01-19 11:40:52 发布

Seqlist_base

最新推荐文章于 2024-01-19 11:40:52 发布

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本文链接：https://blog.youkuaiyun.com/Seqlist_base/article/details/105244358

本文解析了一道408考试真题的算法题目，该题要求将一组正整数分为两个子集，使子集元素数量尽可能接近且元素总和之差最大化。文章提供了详细的解题思路和AC代码实现。

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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 S1and S2 denote the sums of all the numbers in A1and 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≤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: ∣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
这是一道408真题，目的是把原来的数据分成两个数组这两个数组元素数最接近且和相差最大。容易想到把小的元素放在一个数组里把大的元素放在另一个数组，且存放大元素的那个数组的大小永远等于另一个数组或大一个单位；
AC代码：

#include<algorithm>
#include<iostream>
#include<cstdio>
using namespace std;
const int maxn=100001;
int main(){
int n;
int a[maxn];
scanf("%d",&n);
for(int i=0;i<n;i++)
scanf("%d",&a[i]);
sort(a,a+n);
int num1=0,num2=0;
for(int i=0;i<n/2;i++)
    num1+=a[i];
for(int i=n/2;i<n;i++)
    num2+=a[i];
printf("%d %d",n%2,num2-num1);
system("pause");
return 0;
}