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本文介绍了一种使用C/C++实现基本算术运算中除法和取模操作的方法，通过字符串处理和逐位计算，解决了大整数除法和取模的问题。文章提供了两种实现思路，并附带了详细的代码示例。

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Description

The Problem

The first project for the poor student was to make a calculator that can just perform the basic arithmetic operations.

But like many other university students he doesn’t like to do any project by himself. He just wants to collect programs from here and there. As you are a friend of him, he asks you to write the program. But, you are also intelligent enough to tackle this kind of people. You agreed to write only the (integer) division and mod (% in C/C++) operations for him.

Input

Input is a sequence of lines. Each line will contain an input number. One or more spaces. A sign (division or mod). Again spaces. And another input number. Both the input numbers are non-negative integer. The first one may be arbitrarily long. The second number n will be in the range (0 < n < 231).

Output

A line for each input, each containing an integer. See the sample input and output. Output should not contain any extra space.

Sample Input

110 / 100

99 % 10

2147483647 / 2147483647

2147483646 % 2147483647

Sample Output

2147483646

HINT

#include<iostream>
#include<string.h>
using namespace std;
int main()
{
	char str[1000],a[100],b[100],c;
	int t,i,j,x,y;
    int	 num,s,d[1000],e,m;
	while(gets(str))
	{
		t=0;
		for(i=0;str[i];i++)
		{
			if(str[i]=='/'||str[i]=='%')
			{
				c=str[i];
				a[t-1]='\0';
				break;
			}
			else
				a[t++]=str[i];
		}
		a[t]='\0';
		t=0;
		for(j=i+2;str[j];j++)
			b[t++]=str[j];
		b[t]='\0';
		x=strlen(a);
		y=strlen(b);
		s=0;
		if(strcmp(a,b)<0&&x<y)
		{
			if(c=='/')
				cout<<"0"<<endl;
			else if(c=='%')
			{
				cout<<a<<endl;
			}
		}
		else if(strcmp(a,b)==0)
		{
			if(c=='/')
				cout<<"1"<<endl;
			else if(c=='%')
			{
				cout<<"0"<<endl;
			}
		}	
		else
		{
			for(i=0;i<y;i++)
				s=s*10+(b[i]-'0');
			num=0;
			if(c=='/')
			{
				e=0;
				for(i=0;i<x;i++)
				{
					num=num*10+(a[i]-'0');//从高位向低位逐渐的除（例子123/5,1/5=0,* 1%5=1 * ,(1*10+2)/5=2,* 12%5=2 *  ,(2*10+3)/5=4）所以为024
					m=num/s;
					d[e++]=m;
					num%=s;
				}
				for(i=0;i<e;i++)
				{
					if(d[i]!=0)
						break;
				}
				for(j=i;j<e;j++)
					cout<<d[j];
					//printf("%I64d",d[j]);
				cout<<endl;
			}
			else if(c=='%')
			{
				for(i=0;i<x;i++)
				{
					num=num*10+(a[i]-'0');//从高位向低位逐渐的取余（例子123%5, 1%5=1 ,(1*10+2)%5=2,(2*10+3)%5=3）所以为余数3
					num%=s;
				}
				cout<<num<<endl;
				//printf("%I64d\n",num);
			}
			
		//	cout<<s<<endl;
		}
		//cout<<a<<endl<<x<<endl;
		//cout<<b<<endl<<y<<endl;		
	}
	return 0;
}

另一种方法更简洁

#include<stdio.h>
#include<string.h>
int main()
{
	int b,t,len,i,j,k;
	char a[1000],flag;
	while(scanf("%s %c %d",a,&flag,&b)!=EOF)
	{
		len=strlen(a);
		if(flag=='/')
			t=0; //标志作用
		else if(flag=='%')
			t=1;
		k=0;
		j=0;
		for(i=0;i<len;i++)
		{
			k=k*10+(a[i]-'0');
			if((k/b)!=0||(j!=0))
			{
				if(t==0)
					printf("%d",k/b);
				k%=b;
				j=1;//标志除了第一次整除的零不输出外，其余均输出，例如100/5输出的是20而不是020
			}
		}
		if(j==0&&t==0) printf("0");//整除的特殊情况例如2/5的值是零
		if(t!=0) printf("%d",k);
		printf("\n");
	}
	return 0;
}

转载于:https://www.cnblogs.com/NYNU-ACM/p/4237322.html