POJ1003_Hangover(二分求上界)

Hangover

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 99245 Accepted: 48092

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

1.003.710.045.190.00

Sample Output

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

Source

Mid-Central USA 2001

解题报告

可以直接用二分求上界函数（求出大于等于n的元素地址）

lower_bound(num,num+cnt,n)

upper_bound()是求大于n的元素的地址

#include <iostream>#include <cstdio>#include <cstring>#include <cmath>#include <algorithm>using namespace std;double num[10000];int main(){    int cnt=0,i,j;    num[0]=0.00;    for(cnt=1;num[cnt-1]<5.20;cnt++)    {        num[cnt]=num[cnt-1]+1/(double)(cnt+1);    }    double n;    while(cin>>n,n)    {        int left=0,right=cnt-1;        while(left<right)        {            int mid=(left+right)/2;            if(num[mid]<n)                left=mid+1;            else right=mid;        }        cout<<right<<" card(s)"<<endl;        //cout<<lower_bound(num,num+cnt,n)-num<<" card(s)"<<endl;    }    return 0;}