luogu3382模板三分法

Posted 2020-11-20 headboy2002

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。

看代码即可。

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

const int MAX_N = 15;
const double EPS = 0.0000001;
double A[MAX_N];
int N;

double F(double x)
{
	double ans = A[1];
	for(int i = 2; i <= N; i++)
		ans = ans * x + A[i];
	return ans;
}

int main()
{
	double l, r;
	scanf("%d%lf%lf", &N, &l, &r);
	N++;
	for(int i = 1; i <= N; i++)
		scanf("%lf", A + i);
	while(r - l > EPS)
	{
		double mid1 = (r - l) / 3 + l, mid2 = (r - l) / 3 * 2 + l;
		if (F(mid1) > F(mid2))
			r = mid2;
		else
			l = mid1;
	}
	printf("%.5f
", l);
	return 0;
}