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 <cstdio> #define maxn 20 using 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 ", l); }