NYoj 156 Hangover

Hangover
时间限制：1000 ms | 内存限制：65535 KB
难度：1
描述
How far can you make a stack of cards overhang a table? If you have one card, you can create a maximum overhang of half a card length. (We’re assuming that the cards must be perpendicular to the table.) With two cards you can make the top card overhang the bottom one by half a card length, and the bottom one overhang the table by a third of a card length, for a total maximum overhang of 1/2 + 1/3 = 5/6 card lengths. In general you can make n cards overhang by 1/2 + 1/3 + 1/4 + … + 1/(n + 1) card lengths, where the top card overhangs the second by 1/2, the second overhangs tha third by 1/3, the third overhangs the fourth by 1/4, etc., and the bottom card overhangs the table by 1/(n + 1). This is illustrated in the figure below.

输入
The input consists of one or more test cases, followed by a line containing the number 0.00 that signals the end of the input. Each test case is a single line containing a positive floating-point number c whose value is at least 0.01 and at most 5.20; c will contain exactly three digits.
输出
For each test case, output the minimum number of cards necessary to achieve an overhang of at least c card lengths. Use the exact output format shown in the examples.
样例输入
1.00
3.71
0.04
5.19
0.00
样例输出
3 card(s)
61 card(s)
1 card(s)
273 card(s)

这个题是难度不大。但有好几个要注意的点。首先是判定出界的条件我觉得很巧。可能接触的还是太少。>=时出界。还有就是我写的时候一直出错，最后才注意到，我定义的sum是double，但接收的数字却是%f，这两个比较是没法比较的。不知道是编译器的原因还是没学好，这一地方要一一的对应。算是知道了新的东西。

#include"cstdio"#include"cstring"#include"cmath"#include"algorithm"using namespace std;  int main (){    int t;    int i, j, k, m, x, y, p;    double n;    while(scanf("%lf",&n)!=EOF)    {        if(n==0)        return 0;        double sum=0;        for(i=2;;i++)        {            sum=sum+1.0/i;            if(sum>=n)            {                printf("%d card(s)\n",i-1);                break;            }        }    }    return 0;}