poj 1003 Hangover

Hangover

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.

Input

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.

Output

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.

Sample Input

1.00
3.71
0.04
5.19
0.00

Sample Output

3 card(s)
61 card(s)
1 card(s)
273 card(s)

宿醉

描述

你能讓一堆卡片懸在桌子上方多遠?如果你有一張卡片，你可以創(chuàng)造出半張卡片長度的最大懸垂。(我們假設紙牌必須垂直于桌子。)用兩張卡片，你可以使頂部的卡片懸在底部的卡片長度的一半，底部的卡片懸在桌子的卡片長度的三分之一，總的最大懸在1/2 + 1/3 = 5/6卡片長度。一般來說，你可以讓n張卡片以1/2 + 1/3 + 1/4 +…+ 1/(n + 1)牌長，上面的牌比第二張高出1/2，第二張比第三張高出1/3，第三張比第四張高出1/4，等等，下面的牌比桌子高出1/(n + 1)，如下圖所示。

輸入

輸入由一個或多個測試用例組成，后面跟著一行包含數(shù)字0.00，表示輸入的結(jié)束。每個測試用例都是包含一個浮點數(shù)c的單行，其值至少為0.01，最多為5.20;C正好包含三個數(shù)字。

輸出

對于每個測試用例，輸出所需的最小卡數(shù)，以達到至少c卡長度的懸垂。使用示例中所示的輸出格式。

Sample Input

1.00
3.71
0.04
5.19
0.00

Sample Output

3 card(s)
61 card(s)
1 card(s)
273 card(s)


題意：求最小的n讓1+1/2+1/3+...+1/n大于給的一個實數(shù)。
思路：找一個數(shù)sum記錄結(jié)果，寫一個循環(huán)不斷累加 sum+=1.0/n，
直到sum大于c，輸出n-2.

代碼：
#include<stdio.h>
int main()
{
    double c,sum;
    int n,card;
    while(scanf("%lf",&c)!=EOF&&c)
    {
        if((int)(c*100)==0)
	    break;
        sum=0;
        n=2;
        while(sum<=c)//循環(huán)模擬。直到超過
	{
            sum+=1.0/n;
            n++;
        }
        printf("%d card(s)\n",n-2);
    }
    return 0;
}