D - Freefall

https://atcoder.jp/contests/abc279/tasks/abc279_d

思路

求凹函数的极小值

 https://www.cnblogs.com/luoyj/p/12408277.html#6

#include<bits/stdc++.h>
using namespace std;
const double eps = 1e-6;
int n;
double a[15];
double f(double x){       //计算函数值
    double s=0;
    for(int i=n;i>=0;i--) //注意函数求值的写法
        s = s*x + a[i];
    return s;
}
int main(){
    double L,R;
    scanf("%d%lf%lf",&n,&L,&R);
    for(int i=n;i>=0;i--) scanf("%lf",&a[i]);
    while(R-L > eps){  // for(int i = 0; i<100; i++){   //用for也行
        double k =(R-L)/3.0;
        double mid1 = L+k, mid2 = R-k;
        if(f(mid1) > f(mid2)) 
               R = mid2;
        else   L = mid1;
    }
    printf("%.5f\n",L);
    return 0;
}

Code

https://atcoder.jp/contests/abc279/submissions/36830825

void solve() {
    // double EPS = 0.0000001;
    cout << fixed << setprecision(10);
    double a, b;
    cin >> a >> b;
 
    auto cal = [&](int x) -> double {
        return x * b + a / sqrt(1+x);
    };
 
    // cout << fixed << setprecision(10);
    // int x = pow(a/2/b, 2/3) - 1;
    // cout << min({cal(x-1), cal(x), cal(x+1)}) << endl;
 
    int left = 0, right = INF;
    while(right - left > 2) {
        int mid1 = (left * 2 + right) / 3;
        int mid2 = (left + right * 2) / 3;
 
        double t1 = cal(mid1);
        double t2 = cal(mid2);
 
        if(t1 <= t2) {
            right = mid2;
        }else {
            left = mid1;
        }
    }
    
    double ans = a;
    repi(i, max(0LL, left-10), right+10) chmin(ans, cal(i));
    cout << ans << endl;
}
 
signed main() {
    cin.tie(nullptr);
    ios::sync_with_stdio(false);
    solve();
    return 0;
}