HDU 1056 HangOver

HangOver

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1449    Accepted Submission(s): 471

Problem Description

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.



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.

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.

该题浙大，北大上都有，但在浙大-北大上提交过了的代码这里交不了，郁闷了好久，后来对输入加了限制，过了。

#include<iostream>
using namespace std;
int main(){
    int i;
    double aim,sum;
    while(cin>>aim&&aim!=0.00){
        if(aim<0.01||aim>5.20)continue;
        sum=0.0;
        i=1;
        while(sum<aim){
            i++;
            sum+=1.0/i;
        }
        printf("%d card(s)/n",i-1);
    }
    return 0;
}