大數求餘
阿新 • • 發佈:2018-12-23
題目描述
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.
輸入
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.
輸出
For each test case, you have to ouput the result of A mod B.
樣例輸入
2 3
12 7
152455856554521 3250
樣例輸出
2
5
1521
分析:
此題重點考察大數取餘。
我們會發現一個規律。
123456 mod 7
即:
https://blog.csdn.net/qq_32779119/article/details/79513480
#include"stdio.h" #include"string.h" int main() { char a[1001]; int B; int mod; int i,j,l; while(~scanf("%s %d",a,&B)) { mod=0; l=strlen(a); for(i=0;i<l;i++) { mod=(mod*10+(a[i]-'0'))%B; } printf("%d\n",mod); } }