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hdu1576

A/B

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2430    Accepted Submission(s): 1760


Problem Description
要求(A/B)%9973,但由于A很大,我们只给出n(n=A%9973)(我们给定的A必能被B整除,且gcd(B,9973) = 1)。
 

 

Input
数据的第一行是一个T,表示有T组数据。
每组数据有两个数n(0 <= n < 9973)和B(1 <= B <= 10^9)。
 

 

Output
对应每组数据输出(A/B)%9973。
 

 

Sample Input
21000 5387 123456789
 

 

Sample Output
79226060
 

 

Author
xhd
 

 

Source
HDU 2007-1 Programming Contest
 

 

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#include<iostream>#include<cstdio>#include<cstring>#include<string>#include<cmath>#include<algorithm>#include<queue>#include<cstdlib>#include<vector>#include<set>using namespace std;#define MOD 9973int t,n,b,ans;void egcd(int a,int b,int &x,int &y){      if(b==0)      {            x=1;            y=0;            return ;      }      else      {            egcd(b,a%b,x,y);            int t=x;            x=y;            y=t-a/b*y;      }}int main(){      int x,y;      scanf("%d",&t);      while(t--)      {            scanf("%d%d",&n,&b);            egcd(b,MOD,x,y);            x=n*x;            ans=(x%MOD+MOD)%MOD;            printf("%d\n",ans);      }      return 0;}

  

hdu1576