傳送門ios
首先咱們對\(\int_{0}^{\infty}\frac{1}{\prod\limits_{i=1}^{n}(a_i^2+x^2)}dx\)進行裂項相消:
\[ \begin{aligned} &\frac{1}{\prod\limits_{i=1}^{n}(a_i^2+x^2)}&\\ =&\frac{1}{(a_1^2+x^2)(a_2^2+x^2)}\times\frac{1}{\prod\limits_{i=3}^{n}(a_i^2+x^2)}&\\ =&\frac{1}{a_2^2-a_1^2}\times(\frac{1}{a_1^2+x^2}-\frac{1}{a_2^2+x^2})\times\frac{1}{\prod\limits_{i=3}^{n}(a_i^2+x^2)}&\\ =&\frac{1}{a_2^2-a_1^2}\times(\frac{1}{a_1^2+x^2}\times\frac{1}{a_3^2+x^2}-\frac{1}{a_2^2+x^2}\times\frac{1}{a_3^2+x^2})\times\frac{1}{\prod\limits_{i=4}^{n}(a_i^2+x^2)}&\\ =&\dots& \end{aligned} \]
依次裂項相消,而後看係數的規律,能夠手動推\(n=2,3\)的係數看規律,也能夠計算,比賽的時候我\(n=3\)推到一半隊友看到式子和我說這個他學過而後把係數告訴我就\(A\)了(隊友\(txdy\))。
每一個\(\frac{1}{a_i^2+x^2}\)的係數爲\(\frac{1}{\prod\limits_{j=1,j\not=i}^{n}(a_j^2-a_i^2)}\),所以最後題目要求的式子久變成了下式:
\[ \begin{aligned} &\sum\limits_{i=1}^{n}\frac{1}{\prod\limits_{j=1,j\not=i}^{n}(a_j^2-a_i^2)}\int_0^{\infty}\frac{1}{a_i^2+x^2}dx&\\ =&\sum\limits_{i=1}^{n}\frac{1}{\prod\limits_{j=1,j\not=i}^{n}(a_j^2-a_i^2)\times a_i^2}\int_0^{\infty}\frac{1}{1+(\frac{x}{a_i})^2}dx&\\ =&\sum\limits_{i=1}^{n}\frac{1}{\prod\limits_{j=1,j\not=i}^{n}(a_j^2-a_i^2)\times a_i}\int_0^{\infty}\frac{1}{1+(\frac{x}{a_i})^2}d\frac{x}{a_i}& \end{aligned} \]
積分符號裏面的東西就是題目給的式子獲得\(\frac{\pi}{2}\),所以最後答案爲
\[ \begin{aligned} &\sum\limits_{i=1}^{n}\frac{1}{2\times\prod\limits_{j=1,j\not=i}^{n}(a_j^2-a_i^2)\times a_i} \end{aligned} \]spa
#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; }