HDU 1212 Big Number

Big Number

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

Total Submission(s): 7133    Accepted Submission(s): 4922


[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

 

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题意：大数取余

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int main()
{
char a[1100];
int b[1100],c,d,i,j,k,l;
while(scanf("%s",a)!=EOF)
{
scanf("%d",&c);
int len=strlen(a);
for(i=0;i<len;i++)
b[i]=a[i]-'0';
int sum=b[0];
for(i=1;i<len;i++)
{
sum=sum*10+b[i];
sum=sum%c;
}
printf("%d\n",sum);
}
return 0;
}