B Integration
题意：
给定$a_1,a_2,...,a_n$, 计算 $$\frac{1}{π}\int_{0}^{\infty}\frac{1}{\prod\limits_{i=1}^{n}(a_i^2+x^2)}dx$$ 在mod（1E9+7）意义下的答案。
思路：
裂项化乘为和的方法
可以得到
$$\frac{1}{2}\sum_{i=1}^n \quad \frac{1}{\prod_{j=1,j \ne i}^n \quad a_j^2 - a_i^2} \quad \frac{1}{a_i}$$
参考：https://www.cnblogs.com/Dillonh/p/11209476.html
#include <iostream> #include <vector> #include <queue> using namespace std; #define pb push_back #define fi first #define se second #define debug(x) cerr<<#x << " := " << x << endl; #define bug cerr<<"-----------------------"<<endl; #define FOR(a, b, c) for(int a = b; a <= c; ++ a) typedef long long ll; typedef long double ld; typedef pair<int, int> pii; typedef pair<ll, ll> pll; template<class T> void _R(T &x) { cin >> x; } void _R(int &x) { scanf("%d", &x); } void _R(ll &x) { scanf("%lld", &x); } void _R(double &x) { scanf("%lf", &x); } void _R(char &x) { scanf(" %c", &x); } void _R(char *x) { scanf("%s", x); } void R() {} template<class T, class... U> void R(T &head, U &... tail) { _R(head); R(tail...); } template<typename T> inline T read(T&x){ x=0;int f=0;char ch=getchar(); while (ch<'0'||ch>'9') f|=(ch=='-'),ch=getchar(); while (ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar(); return x=f?-x:x; } const int inf = 0x3f3f3f3f; const int mod = 1e9+7; /**********showtime************/ const int maxn = 1e3+9; int a[maxn]; ll ksm(ll a, ll b) { ll res = 1; while(b > 0) { if(b & 1) res = res * a % mod; a = a * a % mod; b = b >> 1; } return res; } int main(){ int n; while(~scanf("%d", &n)){ for(int i=1; i<=n; i++) scanf("%d", &a[i]); ll sum = 0; for(int i=1; i<=n; i++) { ll tmp = 1; for(int j=1; j<=n; j++) { if(i == j) continue; ll s = (1ll*a[j] * a[j]%mod - 1ll*a[i]*a[i]%mod + mod)%mod; tmp = tmp * ksm(s, mod-2) % mod; } tmp = tmp * ksm(a[i], mod-2) % mod; sum = (sum + tmp )% mod; } sum = sum * ksm(2, mod-2) % mod; printf("%lld\n", sum); } return 0; }