码迷,mamicode.com
首页 > 编程语言 > 详细

PAT A1113 Integer Set Partition (25 分)——排序题

时间:2019-02-26 16:58:05      阅读:269      评论:0      收藏:0      [点我收藏+]

标签:each   numbers   lang   exactly   8 8   fonts   rate   pat   one   

Given a set of N (>1) 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 n?1??n?2??∣ is minimized first, and then S?1??S?2??∣ is maximized.

Input Specification:

Each input file contains one test case. For each case, the first line gives an integer N (2N10?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: n?1??n?2??∣ and S?1??S?2??∣, 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
 
技术图片
 1 #include <stdio.h>
 2 #include <algorithm>
 3 using namespace std;
 4 const int maxn=100010;
 5 int seq[maxn];
 6 int total[maxn];
 7 int main(){
 8   int n;
 9   int sum=0;
10   scanf("%d",&n);
11   for(int i=0;i<n;i++){
12     scanf("%d",&seq[i]);
13   }
14   sort(seq,seq+n);
15   for(int i=0;i<n;i++){
16     sum+=seq[i];
17     total[i]=sum;
18   }
19   printf("%d %d",n%2,total[n-1]-2*total[n/2-1]);
20 }
View Code

注意点:计算差的时候由于总和算了前面部分,要多减一次前半部分。感觉直接算和然后相减也不会超时

PAT A1113 Integer Set Partition (25 分)——排序题

标签:each   numbers   lang   exactly   8 8   fonts   rate   pat   one   

原文地址:https://www.cnblogs.com/tccbj/p/10438206.html

(0)
(0)
   
举报
评论 一句话评论(0
登录后才能评论!
© 2014 mamicode.com 版权所有  联系我们:gaon5@hotmail.com
迷上了代码!