E - Pairs Gym - 100935E

题意：已知五个数的相互相加的和，这十个数均不同，求这五个数的值。
In the secret book of ACM, it’s said: “Glory for those who write short ICPC problems. May they live long, and never get Wrong Answers” . Everyone likes problems with short statements. Right? Let’s have five positive numbers: X1,X2,X3,X4,X5. We can form 10 distinct pairs of these five numbers. Given the sum of each one of the pairs, you are asked to find out the original five numbers.
Input
The first line will be the number of test cases T. Each test case is described in one line which contains 10 numbers, these are the sum of the two numbers in each pair. Notice that the input has no particular order, for example: the first number doesn’t have to be equal to {X1+ X2}. All numbers are positive integers below 100,000,000.
Output
For each test case, print one line which contains the number of the test case, and the five numbers X1,X2,X3,X4,X5 in ascending order, see the samples and follow the output format. There always exists a unique solution.

Input
2
15 9 7 15 6 12 13 16 21 14
12 18 13 10 17 20 21 15 16 14

Output
Case 1: 2 4 5 10 11
Case 2: 4 6 8 9 12
题解：这十个数最小的是a+b,最大的是d+e,易得
c = sum/4-(a+b)-(d+e); 剩下的易推。

#include<bits/stdc++.h>
using namespace std;int main()
{int t;cin >> t;for(int k = 1; k <= t; k ++){int a, b, c, d, e;int sum[11];int num = 0;for(int i = 0; i < 10; i ++){cin >> sum[i];num+=sum[i];}sort(sum, sum+10);c = num/4-sum[0]-sum[9];a = sum[1] - c;b = sum[0] - a;e = sum[8] - c;d = sum[9] - e;printf("Case %d: ", k);cout << a << ' ' << b << ' ' << c <<' ' << d <<' ' << e <<endl;}return 0;
}

E - Pairs Gym - 100935E相关推荐

I - Por Costel and the Pairs Gym - 100923I _思维啊——可惜我现在还没
We don't know how Por Costel the pig arrived at FMI's dance party. All we know is that he did. The d ...
Codeforces Gym 100463A Crossings 逆序数
Crossings Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/gym/100463 Description ...
Codeforces gym 2013-2014 Samara SAU ACM ICPC Quarterfinal Qualification Contest
题目地址:http://codeforces.com/gym/100247 2014.10.27版 26号训练的题,今天发现A题有更高效的解法,用到了前缀最大值和后缀最大值 A. The Power ...
gym.wrappers介绍
原地址:https://blog.csdn.net/u013745804/article/details/78403353 gym wrappers用于将环境进行打包,我们可以使用这个模块来记录自己的 ...
Codeforces Gym 100114 H. Milestones 离线树状数组
H. Milestones Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/gym/100114 Descripti ...
Pairs Forming LCM LightOJ - 1236
Pairs Forming LCM LightOJ - 1236 题意问共有多少组数的最大公约数是n 分析组合数学 ,唯一分解定理参考代码 int Prime[670000]; const in ...
强化学习（三） - Gym库介绍和使用，Markov决策程序实例，动态规划决策实例
强化学习(三) - Gym库介绍和使用,Markov决策程序实例,动态规划决策实例 1. 引言在这个部分补充之前马尔科夫决策和动态规划部分的代码.在以后的内容我会把相关代码都附到相关内容的后面.本部 ...
Gym - 102082G
Gym - 102082G https://vjudge.net/problem/2198225/origin 对于数列中任意一个数,要么从最左边到它不递减,要么从最右边到到它不递减,为了满足这个条件 ...
373. Find K Pairs with Smallest Sums （java，优先队列）
题目: You are given two integer arrays nums1 and nums2 sorted in ascending order and an integer k. Def ...

E - Pairs Gym - 100935E

E - Pairs Gym - 100935E

E - Pairs Gym - 100935E相关推荐

最新文章

热门文章