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 themone 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
-
3 4 3 10 3 100 5
- 样例输出
-
5 14 124
- 来源
- [rooot]原创
- 上传者
- rooot
- 题意:
- 输入一个数T,代表测试样例的个数,然后每个测试样例输入两个数n,k;n代表刚开始有n根香烟,k代表每k根烟头能够形成1根香烟,最终让求他能够吸多少根香烟!
思路:
因为每一根新烟都会产生1根烟头(就算是他又新拼接的新烟也是如此),所以要用刚开始的n根新烟加上n根烟头所能产生的新烟的个数,这n根烟头产生的新烟重新被吸之后还要产生烟头,所以要重新加到烟头的总数上面,然后再进行组装新烟,具体看代码:
//这一道题的题意比较难理解,理解题意后就非常简单了! #include <stdio.h> #include <string.h> int main() { int T; scanf("%d",&T); while(T--) { int n,k; scanf("%d%d",&n,&k);//n代表开始的时候有n根烟!k代表有k根烟头的话就能够形成一根新烟! int t=0; int N=n; while(n) { n-=k; if(n<0)//应该是<不是<=; break; t++; n++; } printf("%d\n",N+t); } return 0; } //通过特殊样例得到正确结果:输入:9 3;输出13;
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时间: 2024-10-04 14:15:26