山东省第七届ACM大学生程序设计竞赛 A Julyed

本文介绍了一个关于如何合理分配复习时间以准备CET-6考试中所需记忆的单词数量的问题。通过简单的数学计算，文章提供了一种方法来确定每天应该记忆的单词数，以避免临近考试时的压力过大。

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Julyed

Time Limit: 2000MS Memory limit: 65536K

题目描述

Julyed is preparing for her CET-6. She has N words to remember, but there is only M days left. If she can’t remember all these words, she won’t pass CET-6. If she can’t pass CET-6, she will be unhappy. But if she remember too many words in one day, she will be unhappy, too. If she is unhappy, tomriddly will be unhappy. So she will remember as little words as she can in one day. For the happiness of both tomriddly and Julyed, how many words at most will Julyed remember in one day?

输入

Multiple test cases. The first line is an integer T (T <= 10000), indicating the number of test cases.

Each test case is a line with two integers N, M (1 <= N, M <= 200), indicating the number of words to remember and the number of days left.

输出

One line per case, an integer indicates the answer.

示例输入

56 32 11 25 24 3

示例输出

提示

水题

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
int main()
{
int n;
scanf("%d",&n);
while(n--)
{
int a,b,c;
scanf("%d%d",&a,&b);
c=a/b;
if(a%b!=0)c++;
printf("%d\n",c);
}
return 0;

}