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\n", l);
return 0;
}
原文:https://www.cnblogs.com/headboy2002/p/9218907.html