Big Number 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 Author Ignatius.L Source 杭電ACM省賽集訓隊選拔賽之熱身賽 Recommend Eddy 在做題之前，先了解這樣一些結論： A*B % C = (A%C * B%C)%C (A+B)%C = (A%C + B%C)%C 如 532 mod 7 =（500%7+30%7+2%7)%7; 當然還有a*b mod c=(a mod c+b mod c)mod c; 如35 mod 3=((5%3)*(7%3))%3 有了這一些結論，題目就好做了！ 代碼如下： [cpp] #include<iostream> #include<string.h> using namespace std; const int MAX=100010; int main() { char str[MAX]; int s,len,sum; while(scanf("%s%d",str,&s)!=EOF) { len=strlen(str); sum=0; for(int i=0;i<len;i++) sum=(sum*10+(str[i]-'0')%s)%s; cout<<sum<<endl; } return 0; }
