UVa 1555 - Garland

链接

传送门

题意

有$n$座灯塔,第一个高度为$a$,第$n$个高度为$b$,其余的高度满足公式$h[i]=(h[i-1]+h[i+1])/2+1$,给出$a$,求最小的$b$。

思路

二分第二个灯塔的高度,然后可以计算出所有灯塔的高度,取$b$的最小值。

代码

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#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
const int inf = 0x3f3f3f3f;
const double eps = 1e-4;
int n;
double a, b;
bool judge(double h) {
double cur = h, pre = a;
for (int i = 2; i < n; i++) {
h = cur;
cur = 2 * cur + 2 - pre;
if (cur < 0) return false;
pre = h;
}
b = cur;
return true;
}
int main() {
while (~scanf("%d%lf", &n, &a)) {
double l = 0, r = a, pre = -inf;
while (l < r) {
double m = (l + r) / 2;
if (judge(m)) {
if (fabs(pre - b) < eps) {
break;
}
pre = b;
r = m;
}
else {
l = m;
}
}
printf("%.2lf\n", b);
}
return 0;
}