hdu 1792 A New Change Problem(数论)
2016-06-17 21:10
387 查看
A New Change Problem
Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1236 Accepted Submission(s): 737
Problem Description
Now given two kinds of coins A and B,which satisfy that GCD(A,B)=1.Here you can assume that there are enough coins for both kinds.Please calculate the maximal value that you cannot pay and the total number that you cannot pay.
Input
The input will consist of a series of pairs of integers A and B, separated by a space, one pair of integers per line.
Output
For each pair of input integers A and B you should output the the maximal value that you cannot pay and the total number that you cannot pay, and with one line of output for each line in input.
Sample Input
2 3
3 4
Sample Output
1 1
5 3
题意:求a和b的最大不能组合数和不能组合数的个数
思路:a和b的最大不能组合数为(a-1)*b-a,不能组合数的个数为(a-1)*(b-1)/2
代码:
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
int main()
{
int n,m;
while(~scanf("%d %d",&m,&n))
{
printf("%d %d\n",(m-1)*n-m,(n-1)*(m-1)/2);
}
return 0;
}
相关文章推荐
- struts1完整处理响应示例
- 大数据框架 Hadoop 和 Spark 的异同
- 【06-17】开发中遇到的异常
- BZOJ2194: 快速傅立叶之二
- Java - PAT - 1036. 跟奥巴马一起编程(15)
- Q392判断是否为平衡二叉树(后序遍历实现)
- 调整数组顺序使奇数位于偶数前面
- oracle命令识记
- sql server 查询出错(校验和失败)
- 关于web.xml不同版本之间的区别
- BZOJ 2179FFT快速傅立叶
- IOC_Bean作用域
- tikz-相对坐标
- shell脚本之――字符串截取
- 手贱装了win10之后用hotmail的邮箱登录了小娜
- 使用ESP8266的AT指令调用网络数据接口
- 获取android系统应用
- VirtualBox——User Manual
- Iterator接口
- Dalvik VM和JVM