题解：P12037 [USTCPC 2025] 数学分析
zhangxiaoyu008
·
·
题解
题目传送门
\mathtt{My \ Blog} —— P12037 [USTCPC 2025] 数学分析
由定积分性质
定义 f(n) = \displaystyle\int_{0}^1\frac{x^n}{x^2+1}dx
则 \displaystyle\int_{0}^1\frac{P(x)}{x^2+1}dx 可以转化为 \displaystyle\sum_{i=0}^n f(i)
然后用初中就学过的“分离系数法"。
\begin{aligned} f(n) & = \displaystyle\int_{0}^1\frac{x^n}{x^2+1}dx \\ & = \displaystyle\int_{0}^1\frac{x^n + x^{n-2}}{x^2+1}dx - \displaystyle\int_{0}^1\frac{x^{n-2}}{x^2+1}dx \\ & = \displaystyle\int_{0}^1x^{n-2}dx - \displaystyle\int_{0}^1\frac{x^{n-2}}{x^2+1}dx \end{aligned}
然后我们惊奇的发现 \displaystyle\int_{0}^1\frac{x^{n-2}}{x^2+1}dx 就是 f(n - 2) 。
定义 F(x, n) = \displaystyle\frac{x^{n - 1}}{n - 1}
\begin{aligned} f(n) & = F(1, n) - F(0, n) - f(n - 2) \\ & = \frac{1}{n - 1} - f(n - 2)\end{aligned}
f(0) = \displaystyle\int_{0}^1\frac{1}{x^2+1}dx = \big[\arctan x\big]_0^1 = \arctan 1
f(1) = \displaystyle\int_{0}^1\frac{x}{x^2+1}dx = \left[\frac{1}{2} \ln(x ^ 2 + 1)\right]_0^1 = \frac{\ln 2}{2}
Code
#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 10;
int n, x;
double ans, f[N];
int main()
{
scanf("%d", &n);
f[0] = atan(1);
f[1] = log(2) / 2;
for(int i = 0; i <= n; i ++) {
scanf("%d", &x);
if(i > 1) f[i] = 1.0 / (i - 1) - f[i - 2];
ans += x * f[i];
}
printf("%.10lf\n", ans);
return 0;
}