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
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.
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)
n adapt the problem description.
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
#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 ;
}
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