HDOJ 1212 Big Number【同余定理】

Big Number

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

Total Submission(s): 7156    Accepted Submission(s): 4938


[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

 

[align=left]Author[/align]
Ignatius.L
 

[align=left]Source[/align]
杭电ACM省赛集训队选拔赛之热身赛  

标准的同余定理!!

就是：(a*b)mod c = a mod c * b mod c

    (a+b)mod c = a mod c+ b mod c

写题的时候脑梗塞，一时间没想起来……虽然这题如此标准……

代码贴上，所有类似的题都是这个套路。

#include<stdio.h>
#include<string.h>
char a[111];
int main()
{
int b,i;
while(~scanf("%s%d",a,&b))
{
int l=strlen(a);
int sum=0;
for(i=0;i<l;i++)
sum=((sum*10)%b+(a[i]-'0')%b)%b;
printf("%d\n",sum);
}
return 0;
}