hdu 2662 Coin

Coin

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1020    Accepted Submission(s): 507

Problem Description

Moon has many coins, but only contains two value types which is 5 cents and 7 cents, Some day he find that he can get any value which greater than 23 cents using some of his coins. For instance, he can get 24 cents using two 5 cents coins and two 7 cents coins, he can get 25 cents using five 5 cents coins, he can get 26 cents using one 5 cents coins and three 7 cents coins and so on.

Now, give you many coins which just contains two value types just like Moon, and the two value types identified by two different prime number i and j. Can you caculate the integer n that any value greater than n can be created by some of the given coins.

 

 

Input

The first line contains an integer T, indicates the number of test cases.
For each test case, there are two different prime i and j separated by a single space.(2<=i<=1000000, 2<=j<=1000000)

 

 

Output

For each test case, output one line contains the number n adapt the problem description.

 

 

Sample Input

1

5 7

 

 

Sample Output

23

 

 

Source

HDU男生专场公开赛——赶在女生之前先过节（From WHU） 

 

题意：给你两种素数，每种素数个数不限，求哪个数字后所有的数字都可以通过这两种素数组合构成

 

做最短路的时候无意中看到的题，很简单的数学推理，代码短顺手敲了~

 

设所求为n，那么n+a、n+b可以用a、b线性表出，而n不可。
所以 n+a=x1*a+y1*b，n+b=x2*a+y2*b
所以 n=(x1-1)*a+y1*b n=x2*a+(y2-1)*b
因为n不能被线性表出，所以x1=0，y2=0
所以 n+a=y1*b，n+b=x2*a
所以 n+a=y1*b，n+a=(x2+1)*a-b
所以 (x2+1)*a-b是b的倍数
因为a、b互质，所以(x2+1)是b的倍数
因为求最小的n，所以选最小的x2值，所以取(x2+1)为b
所以 n+a=b*a-b，n=a*b-a-b
证毕

 

附上代码：

 

#include <iostream>
#include <cstdio>
#define ll __int64
using namespace std;
int main()
{
    int T,i,j;
    ll n,m;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%I64d%I64d",&n,&m);
        ll s=n*m-n-m;
        printf("%I64d\n",s);
    }
    return 0;
}