hdu 1792 A New Change Problem（数论）

硬币组合问题解析

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本文针对两种硬币A和B，给出了求解最大不可组合数值及不可组合数值总数的算法。利用数学公式直接计算出结果，提供了一种简洁高效的解决方法。

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;
}