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Given two integers, a and b, you should check whether a is divisible by b or not. We know that an integer a is divisible by an integer b if and only if there exists an integer c such that a = b * c.

Input

Input starts with an integer T (≤ 525), denoting the number of test cases.

Each case starts with a line containing two integers a (-10²⁰⁰ ≤ a ≤ 10²⁰⁰) and b (|b| > 0, b fits into a 32 bit signed integer). Numbers will not contain leading zeroes.

Output

For each case, print the case number first. Then print ‘divisible‘ if a is divisible by b. Otherwise print ‘not divisible‘.

Sample Input

6

101 101

0 67

-101 101

7678123668327637674887634 101

11010000000000000000 256

-202202202202000202202202 -101

Sample Output

Case 1: divisible

Case 2: divisible

Case 3: divisible

Case 4: not divisible

Case 5: divisible

Case 6: divisible

就是判断前边的数字是否可以被后边的整除，，要按照字符串的格式处理

#include<iostream>
#include<string>
using namespace std;
int main(){
    int t,k=0;
    cin>>t;
    while(t--){
        k++;
        string a;
        int b;
        cin>>a>>b;
        long long ans=0,i=0;
        if(a[0]==‘-‘) i++;
        for(;i<a.size();i++){
            ans=(ans*10+a[i]-‘0‘)%b;
        }
        printf("Case %d: ",k);
        if(ans==0)puts("divisible");
        else puts("not divisible");
    }
    return 0;
}

大数相乘

标签：ber   sam   sign   cas   nta   number   output   visible   name   

原文地址：https://www.cnblogs.com/Accepting/p/11355292.html