POJ 1759 - Garland(二分搜索)

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题目:

http://poj.org/problem?id=1759

题意:

最左边灯的高度是a, 有n盏灯, 求出最右边灯的最小高度.

满足: 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

题意:

二分搜索, 对第一盏灯和第二盏灯的距离进行二分.

AC.

#include <iostream>#include <cstdio>using namespace std;const double esp = 1e-9;int n;double a, h[1005], b;bool can(double x){    h[1] = a-x;    for(int i = 2; i < n; ++i) {        h[i] = 2 * h[i-1] + 2 - h[i-2];        if(h[i] < 0) return false;    }    b = h[n-1];    return true;}void solve(){    double l = 0, r = a;    h[0] = a;    while(r - l > esp) {        double mid = (r + l) / 2;        if(can(mid)) l = mid;        else r = mid;    }    printf("%.2lf\n", b);}int main(){//freopen("in", "r", stdin);    while(~scanf("%d %lf", &n, &a)) {        solve();    }    return 0;}

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