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Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 8311 Accepted Submission(s): 5726
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
核心代码
for (int i =0; i < len; ++i){ ans = (ans*10 + big_num[i] - '0') % num; }
#include#include int main() { int num; char big_num[1024] = { 0}; while(scanf("%s%d",&big_num ,&num) == 2) { int ans = 0; int len = strlen(big_num); for (int i =0; i < len; ++i){ ans = (ans*10 + big_num[i] - '0') % num; } printf("%d\n",ans); } return 0; }
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