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HDU1212 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): 4594 Accepted Submission(s): 3175
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.
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
用到同余定理：（a+b）%c=(a%c+b%c)%c=（a+b%c）%c; 附：（a*b）%c=(a%c*b%c)%c;
#include <stdio.h>
#define maxn 1002
char str[maxn];
int main()
{
int m, ans, i;
while(scanf("%s%d", str, &m) != EOF){
ans = 0;
for(i = 0; str[i]; ++i){
ans = (ans * 10 + (str[i] - '0') % m) %m;
}
printf("%d\n", ans);
}
return 0;
}HDU1212 Big Number 【同余定理】
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