POJ-1003

#include<iostream>
using namespace std;

int recursionFunc(int cardNum, float len){
	if(1.0/(cardNum+1)>=len){
		return cardNum;
	}else{
		return recursionFunc(cardNum+1, len-(1.0/(cardNum+1)));
	}
}

int main(int argc, char *argv[]){
	float input;
	while(1){
		cin>>input;
		if(input==0.0){
			break;
		}

		cout<<recursionFunc(1,input)<<" card(s)"<<endl;
	}

	return 0;
}

版权声明:本文为博主原创文章,未经博主允许不得转载。

时间: 2024-11-08 22:30:13

POJ-1003的相关文章

poj 1003:Hangover(水题,数学模拟)

Hangover Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 99450   Accepted: 48213 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 as

poj 1003 Hangover

#include <iostream> using namespace std; int main() { double len; while(cin >> len && len) { double sum = 0.0; double i = 1.0; int n = 2; while(sum < len) { sum += i/n; ++n; } cout << n-2 << " card(s)" <<

[POJ] #1003# 487-3279 : 桶排序/字典树(Trie树)/快速排序

一. 题目 487-3279 Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 274040   Accepted: 48891 Description Businesses like to have memorable telephone numbers. One way to make a telephone number memorable is to have it spell a memorable word or

POJ 1003 Hangover&amp;&amp;NYOJ 156 Hangover【数学题】

计算1+1/2+1/3+++1/n Hangover Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 104558   Accepted: 50926 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 car

[POJ] #1003# Hangover : 浮点数运算

一. 题目 Hangover Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 116593   Accepted: 56886 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. (W

Hangover POJ - 1003

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 overha

POJ 1003 Max Sum

Max Sum Problem Description Given a sequence a[1],a[2],a[3]......a[n], your job is to calculate the max sum of a sub-sequence. For example, given (6,-1,5,4,-7), the max sum in this sequence is 6 + (-1) + 5 + 4 = 14. Input The first line of the input

POJ题目Java代码(一)

POJ 1001 Exponentiation import java.math.BigDecimal; import java.util.Scanner; public class Poj1001 { public static void main(String[] args) { Scanner sc = new Scanner(System.in); while(sc.hasNext()){ BigDecimal bigDecimal = new BigDecimal(sc.next())

Hangover

POJ - 1003 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

HDU - 1003 - Max Sum &amp;&amp; POJ - 1050 - To the Max (经典DP问题)

题目传送:HDU - 1003 思路:最大子序列和 dp[i]= a[i]   (dp[i-1]<0) dp[i]= dp[i-1]+a[i]   (dp[i-1]>=0) AC代码: #include <cstdio> #include <cstring> #include <iostream> #include <algorithm> #include <cmath> #include <queue> #include