LIGHTOJ-1214 - Large Division

绝地灬酷狼 2022-08-18 02:10 160阅读 0赞

1214 - Large Division

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   <td style="color:rgb(121,114,104); background:rgb(245,250,250)">Time Limit:&nbsp;<span style="color:rgb(180,95,4)">1 second(s)</span></td> 
   <td style="color:rgb(121,114,104); background:rgb(245,250,250)">Memory Limit:&nbsp;<span style="color:rgb(180,95,4)">32 MB</span></td> 
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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'**.

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   <td> <h1 style="color:green">Sample Input</h1> </td> 
   <td> <h1 style="color:green">Output for Sample Input</h1> </td> 
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   <td> <p>6</p> <p>101 101</p> <p>0 67</p> <p>-101 101</p> <p>7678123668327637674887634 101</p> <p>11010000000000000000 256</p> <p>-202202202202000202202202 -101</p> </td> 
   <td> <p>Case 1: divisible</p> <p>Case 2: divisible</p> <p>Case 3: divisible</p> <p>Case 4: not divisible</p> <p>Case 5: divisible</p> <p>Case 6: divisible</p> </td> 
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模拟除法过程！

#include<cstdio>
    #include<cstring>
    #include<algorithm>
    using namespace std;
    int main()
    {
    	char str[10001];
    	long long a,b,c,i,n,cot=1;
    	scanf("%lld",&n);
    	while(n--)
    	{
    		scanf("%s%lld",str,&a);
    		b=strlen(str);
    		if(str[0]!='-')
    		c=str[0]-'0';
    		else
    		c=str[1]-'0';
    		for(i=0;i<b-1;i++)
    		{
    			if(str[i]>='0'&&str[i]<='9')
    			{
    				c=c%a;
    				c=c*10+str[i+1]-'0';
    			}
    		}
    		c=c%a;
    		if(!c)
    		printf("Case %lld: divisible\n",cot++);
    		else
    		printf("Case %lld: not divisible\n",cot++);
    	}
    	return 0;
    }

#include<cstdio>
    #include<cstring>
    #include<algorithm>
    using namespace std;
    char str[910000];
    int main()
    {
    	long long a,b,i,j,m,n,x,y,len,cot=1;
    	scanf("%lld",&m);
    	while(m--)
    	{
    		scanf("%s%lld",str,&b);
    		b=abs(b);
    		i=0;
    		if(str[0]=='-')
    		i++;
    		a=str[i]-'0';
    		if(!a)
    		{
    			printf("Case %lld: divisible\n",cot++);
    			continue;
    		}
    		len=strlen(str);
    		i++;
    		while(a<b)
    		{
    		 	if(i==len)
    			break;
    			a=a*10+str[i++]-'0';
    		}
    		if(a<b)
    		{
    			printf("Case %lld: not divisible\n",cot++);
    			continue;
    		}
    		long long t=a%b;
    		while(i<len)
    		{
    			a=t*10+str[i++]-'0';
    			t=a%b;
    		}
    		if(t)
    		printf("Case %lld: not divisible\n",cot++);
    		else
    		printf("Case %lld: divisible\n",cot++);
    	}
    	return 0;
    }