HDOJ 1212 Big Number（大数同余）

Big Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 7269    Accepted Submission(s): 5022

[align=left]Problem Description[/align]
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.

 

[align=left]Input[/align]
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.

 

[align=left]Output[/align]
For each test case, you have to ouput the result of A mod B.

 

[align=left]Sample Input[/align]

2 3
12 7
152455856554521 3250

 

[align=left]Sample Output[/align]

2
5
1521

这一题用了以下两个同余定理公式：

              (A + B) mod M = ( A mod M + B mod M ) mod M

                                                                    

              (A * B) mod M = ((A mod M) *( B mod M)) mod M  

#include<stdio.h>
#include<string.h>
int main()
{
int n,len,i,ans;
char str[1010];
while(scanf("%s%d",str,&n)!=EOF)
{
len=strlen(str);
ans=0;
for(i=0;i<len;i++)
{
ans=ans*10+(str[i]-'0');
ans=ans%n;
}
printf("%d\n",ans);
}
return 0;
}