Write at the top:
A few days ago, I got two important ways to communicate with my classmates:
1. Do not write ready-made functions, such as qsort, which will become a deduction point (unless the title requires you to write);
2. Give priority to the boundary value, and then consider the general situation, so that it is not easy to make mistakes; (this question)
1. Topic information
This problem requires the calculation of A/B, where A is a positive integer with no more than 1000 digits and B is a 1-digit positive integer. You need to output the quotient Q and the remainder R such that A = B * Q + R holds.
Input format:
The input gives A and B sequentially in 1 line, separated by 1 space.
Output format:
Output Q and R sequentially in 1 line, separated by 1 space.
2. Basic ideas
Because it is a 1000-digit number, only an array can solve this problem, then the first method to consider is to divide every two bits (i, i+1) by B, output the quotient, and assign the remainder to i+1, and perform next operation.
Third, the specific implementation code and the problems encountered
#include<stdio.h>
#include<string.h>
intmain ()
{
char number[1000];
int n,i,sum=0;
unsigned long m;
scanf("%s %d",number,&n);
m = strlen (number); /*m represents the number of digits in the input number*/
if(m == 1)
printf("%d %d\n",(number[0]-48)/n,(number[0]-48)%n);
for(i=0,number[0] -= 48;i<m-1;i++) /*没减干净，继续*/
{
number[i+1] -= 48;
sum=(number[i]) * 10 + (number[i+1]);/*前两位的和,商存入Q，余数存入number[i+1]*/
if(i==m-2)
printf("%d %d\n",sum/n,sum-(sum/n)*n);
else
{
printf("%d",sum/n);
number[i+1] = sum % n;
}
}
}
评测结果
| 时间 | 结果 | 得分 | 题目 | 语言 | 用时(ms) | 内存(kB) | 用户 |
|---|---|---|---|---|---|---|---|
| 5月08日 15:19 | 答案正确 | 20 | 1017 | C (gcc 4.7.2) | 2 | 264 | chauncyyoung |
测试点
| 测试点 | 结果 | 用时(ms) | 内存(kB) | 得分/满分 |
|---|---|---|---|---|
| 0 | 答案正确 | 1 | 264 | 12/12 |
| 1 | 答案正确 | 2 | 264 | 2/2 |
| 2 | 答案正确 | 2 | 264 | 2/2 |
| 3 | 答案正确 | 2 | 256 | 2/2 |
| 4 | 答案正确 | 2 | 264 | 2/2 |