UVA1555-- Garland(推导+二分)

 1 #include <cstdio>    
 2 int n;  
 3 double A, B, a[1005];  
 4 //h2的值  
 5 bool check(double x) {  
 6     a[2] = x;  
 7     for(int i = 3; i <= n; ++i) {  
 8         a[i] = 2 * a[i - 1] + 2 - a[i - 2];  
 9         if(a[i] < 0) return false;  
10     }  
11     B = a[n];  
12     return true;  
13 }  
14   
15 int main() {  
16     scanf("%d%lf", &n, &A);  
17     a[1] = A;  
18     double l = -1, r = 1e3 + 1;  
19     for(int i = 0; i < 100; ++i) {  
20         double mid = (l + r) / 2;  
21         if(check(mid)) r = mid;  
22         else l = mid;  
23     }  
24     printf("%.2f\n", B);  
25     return 0;  
26 }