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  • Light OJ 1214 Large Division (大数取模)

    题目:

    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 (-10^200 ≤ a ≤ 10^200) 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

    注:需要判断输入数字字符串a的首位为‘-’的情况和b为负数的情况。

    代码:

    #include <iostream>
    #include <cstring>
    using namespace std;
    char a[100000];
    int main()
    {
    	long long t,i,k,b,n,h;
    	cin>>t;
    	h=1;
    	while(t--)
    	{
    		cin>>a;
    		cin>>b;
    		if(b<0) b=-b;
    		n=strlen(a);
    		if(a[0]=='-')
    		{
    			for(i=0;i<n-1;i++)
    			{
    				a[i]=a[i+1];
    			}
    			n--;
    		}
    		k=0;
    		for(i=0;i<n;i++)
    		{
    			k=(k*10+a[i]-'0')%b;
    		}
    		cout<<"Case "<<h++<<": ";
    		if(!k) cout<<"divisible"<<endl;
    		else cout<<"not divisible"<<endl;
    	}
    	return 0;
    }
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  • 原文地址:https://www.cnblogs.com/cnlik/p/11851887.html
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