POJ1759-Garland

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探讨了新年花环灯谜题的背景与数学模型，采用二分查找法确定最右侧灯泡的最低悬挂高度，确保所有灯泡不接触地面。通过C++实现算法并提供完整代码。

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Garland

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 2168		Accepted: 932

Description

The New Year garland consists of N lamps attached to a common wire that hangs down on the ends to which outermost lamps are affixed. The wire sags under the weight of lamp in a particular way: each lamp is hanging at the height that is 1 millimeter lower than the average height of the two adjacent lamps.

The leftmost lamp in hanging at the height of A millimeters above the ground. You have to determine the lowest height B of the rightmost lamp so that no lamp in the garland lies on the ground though some of them may touch the ground.

You shall neglect the lamp's size in this problem. By numbering the lamps with integers from 1 to N and denoting the ith lamp height in millimeters as Hi we derive the following equations:

H1 = A
Hi = (H _i-1 + H _i+1)/2 - 1, for all 1 < i < N
HN = B
Hi >= 0, for all 1 <= i <= N

The sample garland with 8 lamps that is shown on the picture has A = 15 and B = 9.75.

Input

The input file consists of a single line with two numbers N and A separated by a space. N (3 <= N <= 1000) is an integer representing the number of lamps in the garland, A (10 <= A <= 1000) is a real number representing the height of the leftmost lamp above the ground in millimeters.

Output

Write to the output file the single real number B accurate to two digits to the right of the decimal point representing the lowest possible height of the rightmost lamp.

Sample Input

692 532.81

Sample Output

446113.34

Source

Northeastern Europe 2000

题意：n个灯泡离地H_i，满足H1 = A ，Hi = (Hi-1 + Hi+1)/2 – 1，HN = B ，求最小B

解题思路：二分第二个灯泡的高度

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <queue>
#include <cmath>
#include <map>
#include <bitset>
#include <set>
#include <vector>
#include <functional>

using namespace std;

#define LL long long
const int INF = 0x3f3f3f3f;

double a,b,x[1005],ans;
int n;

bool check(double k)
{
    x[1]=a,x[2]=k;
    for(int i=3;i<=n;i++)
    {
        x[i]=(x[i-1]+1)*2-x[i-2];
        if(x[i]<0) return 0;
    }
    ans=x[n];
    return 1;
}

int main()
{
	while(~scanf("%d%lf",&n,&a))
    {
        double l=0,r=a;
        while(fabs(r-l)>=1e-8)
        {
            double mid=(l+r)/2;
            if(check(mid)) {b=mid;r=mid;}
            else l=mid;
        }
        printf("%.2f\n",ans);
    }
	return 0;
}