题目：
Garland
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 2778 | Accepted: 1170 |
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.
题意：有N个灯，第一个灯的高度是a，第2个到第（n-1）个灯的高度值都满足：Hi =（H i-1 + H i + 1）/ 2-1 的的关系，求第N个灯的最低高度，使这n个灯的高度都 >= 0。
思路：这题是不能二分答案的，因为这样无论如何都无法进行递推，找到最低的位置。
我们将题目给的式子稍加变动得：h（i+1） = 2*h（i）- h（i-1）+ 2 。（1式子）则发现由前两个灯的高度可以推出第3个灯的高度。
又因为：h（i+2） = 2*h（i+1）- h（i）+ 2 ， 把1式子带入 得： h（i+2） = 3*h（i）- 2*h（i-1）+ 6。（2式子）
又因为：h（i+3） = 2*h（i+2）- h（i+1）+ 2 ，把1式子2式子代入 得：h（i+3） = 4*h（i）- 3*h（i-1）+ 12。（3式子）
又因为：h（i+4） = 2*h（i+3）- h（i+2）+ 2 ，把2式子3式子代入 得：h（i+4） = 5*h（i）- 4*h（i-1）+ 20。（4式子）
又因为……
然后我们得到了h（i+n） = （n+1）*h（i）- n*h（i-1）+ n*(n+1)。 令i-1 = 1，则i==2.（同时使i+n == N）我们可以发现，第N个灯的高度(也就是答案)与第2个灯的高度成正相关。
于是我们就可以二分第2个灯的高度，然后递推判断是否可行，和求第n个灯的高度了。
代码：
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<queue>
#include<map>
#include<stack>
#include<set>
#define sd(x) scanf("%d",&x)
#define ss(x) scanf("%s",x)
#define sc(x) scanf("%c",&x)
#define sf(x) scanf("%f",&x)
#define slf(x) scanf("%lf",&x)
#define slld(x) scanf("%lld",&x)
#define me(x,b) memset(x,b,sizeof(x))
#define pd(d) printf("%d\n",d);
#define plld(d) printf("%lld\n",d);
#define eps 1.0E-8
// #define Reast1nPeacetypedef long long ll;using namespace std;const int INF = 0x3f3f3f3f;double a;
int n;
double ans;
double t[1010];bool judge(double x){t[1] = a;t[2] = x;for(int i = 3 ; i<=n ; i++){t[i] = 2*t[i-1] - t[i-2] + 2;if(t[i]<0){return false;}}ans = t[n];return true;
}int main(){
#ifdef Reast1nPeacefreopen("in.txt", "r", stdin);freopen("out.txt", "w", stdout);
#endifios::sync_with_stdio(false);cin>>n>>a;double l = 0 ; double r = INF;for(int i = 0 ; i<100 ; i++){double mid = (l+r)/2;if(judge(mid)){r = mid;}else{l = mid;}}printf("%.2lf\n",ans);return 0;
}