题目链接

传送门

思路

首先我们对${\int }_0^{\infty }\frac{1}{\prod _{i=1}^n(a_i^2+x^2)}dx$进行裂项相消： $$\begin{array}{rlr}& \frac{1}{\prod _{i=1}^n(a_i^2+x^2)}& \\ =& \frac{1}{(a_1^2+x^2)(a_2^2+x^2)}\times \frac{1}{\prod _{i=3}^n(a_i^2+x^2)}& \\ =& \frac{1}{a_2^2-a_1^2}\times (\frac{1}{a_1^2+x^2}-\frac{1}{a_2^2+x^2})\times \frac{1}{\prod _{i=3}^n(a_i^2+x^2)}& \\ =& \frac{1}{a_2^2-a_1^2}\times (\frac{1}{a_1^2+x^2}\times \frac{1}{a_3^2+x^2}-\frac{1}{a_2^2+x^2}\times \frac{1}{a_3^2+x^2})\times \frac{1}{\prod _{i=4}^n(a_i^2+x^2)}& \\ =& \dots & \end{array}$$ 依次裂项相消，然后看系数的规律，可以手动推$n=2,3$的系数看规律，也可以计算，比赛的时候我$n=3$推到一半队友看到式子和我说这个他学过然后把系数告诉我就$A$了(队友$txdy$)。 每个$\frac{1}{a_i^2+x^2}$的系数为$\frac{1}{\prod _{j=1,j̸=i}^n(a_j^2-a_i^2)}$，因此最后题目要求的式子久变成了下式： $$\begin{array}{rlr}& \sum _{i=1}^n\frac{1}{\prod _{j=1,j̸=i}^n(a_j^2-a_i^2)}{\int }_0^{\infty }\frac{1}{a_i^2+x^2}dx& \\ =& \sum _{i=1}^n\frac{1}{\prod _{j=1,j̸=i}^n(a_j^2-a_i^2)\times a_i^2}{\int }_0^{\infty }\frac{1}{1+(\frac{x}{a_i}{)}^2}dx& \\ =& \sum _{i=1}^n\frac{1}{\prod _{j=1,j̸=i}^n(a_j^2-a_i^2)\times a_i}{\int }_0^{\infty }\frac{1}{1+(\frac{x}{a_i}{)}^2}d\frac{x}{a_i}& \end{array}$$ 积分符号里面的东西就是题目给的式子得到$\frac{\pi }{2}$，因此最后答案为 $$\begin{array}{rl}& \sum _{i=1}^n\frac{1}{2\times \prod _{j=1,j̸=i}^n(a_j^2-a_i^2)\times a_i}\end{array}$$

代码实现如下

#include <set> #include <map> #include <deque> #include <queue> #include <stack> #include <cmath> #include <ctime> #include <bitset> #include <cstdio> #include <string> #include <vector> #include <cstdlib> #include <cstring> #include <cassert> #include <iostream> #include <algorithm> #include <unordered_map> using namespace std; typedef long long LL; typedef pair<LL, LL> pLL; typedef pair<LL, int> pLi; typedef pair<int, LL> pil;; typedef pair<int, int> pii; typedef unsigned long long uLL; #define lson rt<<1 #define rson rt<<1|1 #define lowbit(x) x&(-x) #define name2str(name) (#name) #define bug printf("*********\n") #define debug(x) cout<<#x"=["<<x<<"]" <<endl #define FIN freopen("D://Code//in.txt","r",stdin) #define IO ios::sync_with_stdio(false),cin.tie(0) const double eps = 1e-8; const int mod = 1e9 + 7; const int maxn = 1e5 + 7; const double pi = acos(-1); const int inf = 0x3f3f3f3f; const LL INF = 0x3f3f3f3f3f3f3f3fLL; int n; int a[maxn], inv[maxn], cnt[maxn]; LL qpow(LL x, int n) { LL res = 1; while(n) { if(n & 1) res = res * x % mod; x = x * x % mod; n >>= 1; } return res; } int main() { int tmp = qpow(2, mod - 2); while(~scanf("%d", &n)) { for(int i = 1; i <= n; ++i) { scanf("%d", &a[i]); } for(int i = 1; i <= n; ++i) { cnt[i] = 1; for(int j = 1; j <= n; ++j) { if(i == j) continue; cnt[i] = 1LL * cnt[i] * ((1LL * a[j] * a[j] % mod - 1LL * a[i]* a[i] % mod) % mod + mod) % mod; } cnt[i] = qpow(cnt[i], mod - 2); cnt[i] = 1LL * cnt[i] * qpow(a[i], mod - 2) % mod; cnt[i] = 1LL * cnt[i] * tmp % mod; } LL ans = 0; for(int i = 1; i <= n; ++i) { ans = ((ans + cnt[i]) % mod + mod) % mod; } printf("%lld\n", ans); } return 0; }