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hdu 1212 Big Number(数论)


Big Number


Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3256    Accepted Submission(s): 2219


Problem Description


As we know, Big Number is always troublesome. But it's really important in our ACM. And today, your task is to write a program to calculate A mod B.

To make the problem easier, I promise that B will be smaller than 100000.

Is it too hard? No, I work it out in 10 minutes, and my program contains less than 25 lines.



Input


The input contains several test cases. Each test case consists of two positive integers A and B. The length of A will not exceed 1000, and B will be smaller than 100000. Process to the end of file.




Output


For each test case, you have to ouput the result of A mod B.



Sample Input

2 3
12 7
152455856554521 3250




Sample Output


2
5
1521



Author


Ignatius.L




Source


​​杭电ACM省赛集训队选拔赛之热身赛 ​​




Recommend


Eddy

 

思路:一个数模上另一个数等于10进制数展开的模相加的模。

        如 532 mod 7 =(500%7+30%7+2%7)%7;

       当然还有a*b mod c=(a mod c+b mod c)mod c;

       如35 mod 3=((5%3)*(7%3))%3

知道这些这题就很容易了。

 

#include<iostream>
#include<cstring>
using namespace std;
const int mm=100020;
char s[mm];
int ss;
int main()
{
while(cin>>s>>ss)
{
int len=strlen(s);
int ans=0;
for(int i=0;i<len;i++)
{
ans=ans*10+s[i]-'0';
ans%=ss;
}
cout<<ans<<"\n";
}
}

 

 

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