1017 A divided by B (20 minutes)
This problem requires calculation of A / B, where A is not more than 1000 bit positive integer, B is a positive integer. You need to output quotient Q and a remainder R, such that A = B × Q + R established.
Input format:
input sequence is given in row A and B, separated by an intermediate space.
Output format:
sequentially outputs Q and R in a row, separated by an intermediate space.
Sample input:
123456789050987654321 7
Sample output:
17636684150141093474 3
The title three pit
pit 1: 1000 even with a positive integer unsigned long long long is not enough, can only be solved with a string
Pit 2: When the business for more than the first and is 0 (test point 0 and 2), separate output (not the first place to the output is also 0)
pit 3: when a business is and is 0 (test point 1), normal output (output to 0) to
#include <stdio.h>
int main()
{
char a[1001],c[1001];
int b,n=1,yu=0;
scanf("%s %d",&a,&b);
c[0]=(a[0]-'0')/b+'0';
yu=(a[0]-'0')%b;
while(a[n]!='\0')
{
c[n]=(a[n]-'0'+yu*10)/b+'0';
yu=(a[n]-'0'+yu*10)%b;
n++;
}
c[n]='\0';
if(c[0]=='0' && c[1]!='\0') //注意商为0或商首位为0的情况
for(int i=1;i<n;i++)
printf("%c",c[i]);
else
printf("%s",c);
printf(" %d",yu);
return 0;
}
/*
本题三个坑
坑1：1000位的正整数即使使用unsigned long long 也不够长，只能用字符串解决
坑2：当商为多位并且第一位为0时（测试点0和2），分开输出（不能把首位的0也输出来了）
坑3：当商为一位且为0时（测试点1），正常输出（把0输出）即可
*/
