大数对小数取余
Problem DescriptionAs we know,Big Number is always troublesome.But it's really important in our ACM. And today,your task is to write aprogram to calculate A mod B. InputThe input contains several testcases. Each test case consists of two positive integers A and B. The length ofA will not exceed 1000,and B will be smaller than 100000. Process to the endof file. OutputFor each test case,you have toouput the result of A mod B. Sample Input2 3 12 7 152455856554521 3250 Sample Output2 5 1521 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; 由此可得大数的每一位对小数求余最后结果再对小数求余就是大数对小数求余的结果 #include<stdio.h> #include<string.h> int main() { char str[1000]; int a,i,len,sum; while(scanf("%s%d",str,&a)!=EOF) { sum=0; len=strlen(str); for(i=0;i<len;i++) { sum=(sum*10+str[i]-'0')%a; } printf("%dn",sum); } return 0; } (编辑:李大同) 【声明】本站内容均来自网络,其相关言论仅代表作者个人观点,不代表本站立场。若无意侵犯到您的权利,请及时与联系站长删除相关内容! |