Big Number
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3386 Accepted Submission(s): 2305
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
1 #include <iostream> 2 #include <cstdio> 3 #include <cstdlib> 4 #include <cmath> 5 #include <cstring> 6 using namespace std; 7 char a[1000+10]; int b; 8 int main(void){ 9 #ifndef ONLINE_JUDGE 10 freopen("1212.in", "r", stdin); 11 #endif 12 while (~scanf("%s%d", a, &b)){ 13 int len = strlen(a); int x = a[len-1]-'0', po = 1; 14 x = x % b; int y; 15 for (int i = len - 2; i >= 0; --i){ 16 y = a[i] - '0'; po = (po * 10) % b; 17 x = (x + y*po) % b; 18 } 19 printf("%d\n", x); 20 } 21 return 0; 22 }
看了讨论版学会的，不过还是有一点儿小问题，学一下数论吧……