题目链接
题解
对于每一项展开
的到(atk+frac{dnf}{b}a + dnf + frac{atk}{a} b)
令$T = frac{a}{b} $
原式$=atk+Tdnf + dnf + frac{atk}{T} $
这就是那个单峰的对勾函数，
把单峰函数复合为求最值，发现也是个单峰函数(下凸壳)
三分就好了
或者维护一个最大值得下凸壳
代码
#include<cstdio>
#include<algorithm>
inline int read() {
int x = 0,f = 1;
char c = getchar();
while(c < '0' || c > '9') c = getchar();
while(c <= '9' && c >= '0') x = x * 10 + c - '0',c = getchar();
return x * f;
}
#define db double
const int maxn = 5000007;
db atk[maxn],dnf[maxn];
int n;
db calc(db T) {
double ret = 0.0;
for(int i = 1;i <= n;++ i) {
db tmp = dnf[i] + atk[i] + dnf[i] / T + atk[i] * T;
ret = std::max(tmp,ret);
}
return ret;
}
int main() {
n = read();
for(int i = 1;i <= n;++ i)
scanf("%lf%lf",atk + i,dnf + i);
db l = 0,r = 10.0;
for(;r - l >= (1e-12);) {
db m1 = l + (r - l) / 3.0,m2 = r - (r - l) / 3.0;
db c1 = calc(m1),c2 = calc(m2);
if(c1 > c2) l = m1;
else r = m2;
}
printf("%.4lf
",std::min(calc(r),calc(l))) ;
return 0;
}