NYOJ94cigarettes

最新推荐文章于 2024-03-09 17:40:10 发布

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最新推荐文章于 2024-03-09 17:40:10 发布

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原文链接：http://www.cnblogs.com/zhaojiedi1992/archive/2012/08/02/zhaojiedi_2012_08

本文介绍了一个关于计算在保留烟蒂后剩余香烟数量的算法，通过输入香烟总数和可以重新卷烟的烟蒂数，输出最大可能的香烟数量。使用循环和整数除法来实现计算。

cigarettes

时间限制：3000 ms | 内存限制：65535 KB

难度：2

描述

Tom has many cigarettes. We hypothesized that he has n cigarettes and smokes them

one by one keeping all the butts. Out of k > 1 butts he can roll a new cigarette.
Now，do you know how many cigarettes can Tom has?

输入

First input is a single line,it's n and stands for there are n testdata.then there are n lines ,each line contains two integer numbers giving the values of n and k.

输出

For each line of input, output one integer number on a separate line giving the maximum number of cigarettes that Peter can have.

样例输入

 
#include<stdio.h>
int main()
{
    int n,a,b;
    int t1,t2,t3,k;
    scanf("%d",&n);
    while(n--)
    {
       scanf("%d%d",&a,&b);
        t2=0;
        k=a;
      while(a/b>0)
      {
          t1=a/b;
          t3=a%b;
          t2+=t1;
          //printf("t2==%d",t2);//
          a=t1+t3;
      } 
      printf("%d\n",k+t2);
    }
   return 0;
}