题目大意：给出一个$N$次函数，保证在范围$[l,r]$内存在一点x，使得$[l,x]$上单调增，$[x,r]$上单调减。试求出$x$的值。

题解：求导，就变成了求零点，二分答案即可

卡点：无

 

C++ Code：

#pragma GCC optimize($1 orz)#pragma GCC optimize(3)#pragma GCC optimize("Ofast")#pragma GCC optimize("inline")#pragma GCC optimize("-fgcse")#pragma GCC optimize("-fgcse-lm")#pragma GCC optimize("-fipa-sra")#pragma GCC optimize("-ftree-pre")#pragma GCC optimize("-ftree-vrp")#pragma GCC optimize("-fpeephole2")#pragma GCC optimize("-ffast-math")#pragma GCC optimize("-fsched-spec")#pragma GCC optimize("unroll-loops")#pragma GCC optimize("-falign-jumps")#pragma GCC optimize("-falign-loops")#pragma GCC optimize("-falign-labels")#pragma GCC optimize("-fdevirtualize")#pragma GCC optimize("-fcaller-saves")#pragma GCC optimize("-fcrossjumping")#pragma GCC optimize("-fthread-jumps")#pragma GCC optimize("-funroll-loops")#pragma GCC optimize("-fwhole-program")#pragma GCC optimize("-freorder-blocks")#pragma GCC optimize("-fschedule-insns")#pragma GCC optimize("inline-functions")#pragma GCC optimize("-ftree-tail-merge")#pragma GCC optimize("-fschedule-insns2")#pragma GCC optimize("-fstrict-aliasing")#pragma GCC optimize("-fstrict-overflow")#pragma GCC optimize("-falign-functions")#pragma GCC optimize("-fcse-skip-blocks")#pragma GCC optimize("-fcse-follow-jumps")#pragma GCC optimize("-fsched-interblock")#pragma GCC optimize("-fpartial-inlining")#pragma GCC optimize("no-stack-protector")#pragma GCC optimize("-freorder-functions")#pragma GCC optimize("-findirect-inlining")#pragma GCC optimize("-fhoist-adjacent-loads")#pragma GCC optimize("-frerun-cse-after-loop")#pragma GCC optimize("inline-small-functions")#pragma GCC optimize("-finline-small-functions")#pragma GCC optimize("-ftree-switch-conversion")#pragma GCC optimize("-foptimize-sibling-calls")#pragma GCC optimize("-fexpensive-optimizations")#pragma GCC optimize("-funsafe-loop-optimizations")#pragma GCC optimize("inline-functions-called-once")#pragma GCC optimize("-fdelete-null-pointer-checks")#include 
     
      #define maxn 20using namespace std;const double eps = 1e-5;int n;double s[maxn];double l, r, p, mid;bool check(double mid) {    double tmp = 1, ans = 0;    for (int i = 0; i < n; i++) ans += s[i] * tmp, tmp *= mid;    return ans > 0;}int main(){    scanf("%d%lf%lf", &n, &l, &r);    for (int i = n; i; i--) {        scanf("%lf", &s[i - 1]);        s[i - 1] *= i;    }    while (r - l > eps) {        mid = (l + r) / 2;        if (check(mid)) l = mid;        else r = mid;    }    printf("%.5lf\n", l);}