Codeforces 389A (最大公约数)
2016-04-23 16:47
351 查看
Fox and Number Game
Submit Status
Description
Fox Ciel is playing a game with numbers now.
Ciel has n positive integers: x1, x2, ..., xn. She can do the following operation as many times as needed: select two different indexes i and jsuch that xi > xj hold, and then apply assignment xi = xi - xj. The goal is to make the sum of all numbers as small as possible.
Please help Ciel to find this minimal sum.
Input
The first line contains an integer n (2 ≤ n ≤ 100). Then the second line contains n integers: x1, x2, ..., xn (1 ≤ xi ≤ 100).
Output
Output a single integer — the required minimal sum.
Sample Input
Input
Output
Input
Output
Input
Output
Input
Output
Hint
In the first example the optimal way is to do the assignment: x2 = x2 - x1.
In the second example the optimal sequence of operations is: x3 = x3 - x2, x2 = x2 - x1.
Source
Codeforces Round #228 (Div. 2)
题解:将题目转化一下,就是求n个数的最大公约数再乘以n。
Time Limit: 1000MS | Memory Limit: 262144KB | 64bit IO Format: %I64d & %I64u |
Description
Fox Ciel is playing a game with numbers now.
Ciel has n positive integers: x1, x2, ..., xn. She can do the following operation as many times as needed: select two different indexes i and jsuch that xi > xj hold, and then apply assignment xi = xi - xj. The goal is to make the sum of all numbers as small as possible.
Please help Ciel to find this minimal sum.
Input
The first line contains an integer n (2 ≤ n ≤ 100). Then the second line contains n integers: x1, x2, ..., xn (1 ≤ xi ≤ 100).
Output
Output a single integer — the required minimal sum.
Sample Input
Input
2 1 2
Output
2
Input
3 2 4 6
Output
6
Input
2 12 18
Output
12
Input
5 45 12 27 30 18
Output
15
Hint
In the first example the optimal way is to do the assignment: x2 = x2 - x1.
In the second example the optimal sequence of operations is: x3 = x3 - x2, x2 = x2 - x1.
Source
Codeforces Round #228 (Div. 2)
题解:将题目转化一下,就是求n个数的最大公约数再乘以n。
#include <iostream> #include <stdlib.h> #include <stdio.h> #include <algorithm> #include <string.h> #include <math.h> using namespace std; typedef long long ll; ll gcd(ll a,ll b) { if (b==0) return a; else return gcd(b,a%b); } int main() { int i,n,ans,a[105]; while (cin>>n) { cin>>a[0]; ans=a[0]; for (i=1;i<n;i++) { cin>>a[i]; ans=gcd(ans,a[i]); } cout<<ans*n<<endl; } return 0; }
相关文章推荐
- 4.23 GDOI赛前模拟 总结
- LeetCode(69)-Reverse String
- LeetCode(69)-Reverse String
- LeetCode(69)-Reverse String
- (2)在VMware虚拟机上装CentOS6_Linux系统
- Java 注解全面解析
- 儒家
- JavaScript中的Date类型详解与moment简介
- 深入浅出java虚拟机系列:(二)GC&垃圾收集算法
- salt-api安装、配置、使用
- Android面试题相关
- android 5.0 自动接听电话
- 60款顶级大数据开源工具
- Java当中产生随机数
- [改善Java代码]提防包装类型的null值
- symfony静态文件解决方案(未完待续)
- hibernate---性能优化, 1+N问题
- [leetcode ]239. Sliding Window Maximum
- 如何把AChartEngine加载到view视图?
- 40条Android开发优化建议