2019 牛客多校第一场 B Integration

题目链接：https://ac.nowcoder.com/acm/contest/881/B

题目大意

　　给定 n 个不同的正整数 a_i，求$\frac{1}{\pi}\int_{0}^{\infty} \frac{1}{\prod\limits_{i=1}^{n}(a_i^2+x^2)}dx$模 10⁹ + 7。（可以证明这个积分一定是有理数）

分析

$$\begin{align*}
&令c_i = \frac{1}{\prod_{j \ne i} (a_j^2 - a_i^2)} \\
&则\frac{1}{\prod\limits_{i=1}^{n}(a_i^2+x^2)} = \sum\limits_{i=1}^{n} \frac{c_i}{a_i^2+x^2} \\
&而\int_{0}^{\infty} \frac{c_i}{a_i^2+x^2}dx = \frac{c_i}{2a_i}\pi \\
&于是\frac{1}{\pi}\int_{0}^{\infty} \frac{1}{\prod\limits_{i=1}^{n}(a_i^2+x^2)}dx = \sum\limits_{i=1}^{n} \frac{c_i}{2a_i}
\end{align*}$$

　　补：关于裂项，也就是$c_i$怎得得出来，以两项为例。

　　以 $x$ 代替 $x^2$，$-y_i$ 代替 $a_i^2$。

　　则$\frac{1}{(x - y_1)(x - y_2)} = \frac{1}{y_1 - y_2} * (\frac{1}{x - y_1} - \frac{1}{x - y_2}) = \frac{1}{y_1 - y_2} * \frac{1}{x - y_1} + \frac{1}{y_2 - y_1} * \frac{1}{x - y_2}$。

　　三项以至于更多项同理，可以找出规律。

　　熟练以后建议当作公式来记住。

　　你所以为的顿悟，常常只是别人的基本功。

代码如下

 #include <bits/stdc++.h>
 using namespace std;

 #define INIT() ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
 #define Rep(i,n) for (int i = 0; i < (n); ++i)
 #define For(i,s,t) for (int i = (s); i <= (t); ++i)
 #define rFor(i,t,s) for (int i = (t); i >= (s); --i)
 #define ForLL(i, s, t) for (LL i = LL(s); i <= LL(t); ++i)
 #define rForLL(i, t, s) for (LL i = LL(t); i >= LL(s); --i)
 #define foreach(i,c) for (__typeof(c.begin()) i = c.begin(); i != c.end(); ++i)
 #define rforeach(i,c) for (__typeof(c.rbegin()) i = c.rbegin(); i != c.rend(); ++i)

 #define pr(x) cout << #x << " = " << x << "  "
 #define prln(x) cout << #x << " = " << x << endl

 #define LOWBIT(x) ((x)&(-x))

 #define ALL(x) x.begin(),x.end()
 #define INS(x) inserter(x,x.begin())
 #define UNIQUE(x) x.erase(unique(x.begin(), x.end()), x.end())
 #define REMOVE(x, c) x.erase(remove(x.begin(), x.end(), c), x.end()); // ?? x ?????? c
 #define TOLOWER(x) transform(x.begin(), x.end(), x.begin(),::tolower);
 #define TOUPPER(x) transform(x.begin(), x.end(), x.begin(),::toupper);

 #define ms0(a) memset(a,0,sizeof(a))
 #define msI(a) memset(a,inf,sizeof(a))
 #define msM(a) memset(a,-1,sizeof(a))

 #define MP make_pair
 #define PB push_back
 #define ft first
 #define sd second

 template<typename T1, typename T2>
 istream &operator>>(istream &in, pair<T1, T2> &p) {
     in >> p.first >> p.second;
     return in;
 }

 template<typename T>
 istream &operator>>(istream &in, vector<T> &v) {
     for (auto &x: v)
         in >> x;
     return in;
 }

 template<typename T>
 ostream &operator<<(ostream &out, vector<T> &v) {
     Rep(i, v.size()) out << v[i] << " \n"[i == v.size()];
     return out;
 }

 template<typename T1, typename T2>
 ostream &operator<<(ostream &out, const std::pair<T1, T2> &p) {
     out << "[" << p.first << ", " << p.second << "]" << "\n";
     return out;
 }

 inline int gc(){
     static const int BUF = 1e7;
     static char buf[BUF], *bg = buf + BUF, *ed = bg;

     if(bg == ed) fread(bg = buf, , BUF, stdin);
     return *bg++;
 } 

 inline int ri(){
     int x = , f = , c = gc();
     for(; c<||c>; f = c=='-'?-:f, c=gc());
     for(; c>&&c<; x = x* + c - , c=gc());
     return x*f;
 }

 template<class T>
 inline string toString(T x) {
     ostringstream sout;
     sout << x;
     return sout.str();
 }

 inline int toInt(string s) {
     int v;
     istringstream sin(s);
     sin >> v;
     return v;
 }

 //min <= aim <= max
 template<typename T>
 inline bool BETWEEN(const T aim, const T min, const T max) {
     return min <= aim && aim <= max;
 }

 typedef long long LL;
 typedef unsigned long long uLL;
 typedef pair< double, double > PDD;
 typedef pair< int, int > PII;
 typedef pair< int, PII > PIPII;
 typedef pair< string, int > PSI;
 typedef pair< int, PSI > PIPSI;
 typedef set< int > SI;
 typedef set< PII > SPII;
 typedef vector< int > VI;
 typedef vector< double > VD;
 typedef vector< VI > VVI;
 typedef vector< SI > VSI;
 typedef vector< PII > VPII;
 typedef map< int, int > MII;
 typedef map< int, string > MIS;
 typedef map< int, PII > MIPII;
 typedef map< PII, int > MPIII;
 typedef map< string, int > MSI;
 typedef map< string, string > MSS;
 typedef map< PII, string > MPIIS;
 typedef map< PII, PII > MPIIPII;
 typedef multimap< int, int > MMII;
 typedef multimap< string, int > MMSI;
 //typedef unordered_map< int, int > uMII;
 typedef pair< LL, LL > PLL;
 typedef vector< LL > VL;
 typedef vector< VL > VVL;
 typedef priority_queue< int > PQIMax;
 typedef priority_queue< int, VI, greater< int > > PQIMin;
 const double EPS = 1e-;
 const LL inf = 0x7fffffff;
 const LL infLL = 0x7fffffffffffffffLL;
 const LL mod = 1e9 + ;
 const int maxN = 1e3 + ;
 const LL ONE = ;
 const LL evenBits = 0xaaaaaaaaaaaaaaaa;
 const LL oddBits = 0x5555555555555555;

 //ax + by = gcd(a, b) = d
 // 扩展欧几里德算法
 /**
  *    a*x + b*y = 1
  *    如果ab互质，有解
  *    x就是a关于b的逆元
  *    y就是b关于a的逆元
  *
  *    证明：
  *        a*x % b + b*y % b = 1 % b
  *        a*x % b = 1 % b
  *        a*x = 1 (mod b)
  */
 inline void ex_gcd(LL a, LL b, LL &x, LL &y, LL &d){
     if (!b) {d = a, x = , y = ;}
     else{
         ex_gcd(b, a % b, y, x, d);
         y -= x * (a / b);
     }
 }

 // 求a关于p的逆元，如果不存在，返回-1
 // a与p互质，逆元才存在
 inline LL inv_mod(LL a, LL p = mod){
     LL d, x, y;
     ex_gcd(a, p, x, y, d);
     return d ==  ? (x % p + p) % p : -;
 }

 LL add_mod(LL a, LL b) {
     return (a + b) % mod;
 }

 LL mul_mod(LL a, LL b) {
     return (a * b) % mod;
 }

 LL sub_mod(LL a, LL b) {
     return (a - b + mod) % mod;
 }

 LL n, a[maxN], c[maxN], ans;

 int main(){
     //freopen("MyOutput.txt","w",stdout);
     //freopen("input.txt","r",stdin);
     //INIT();
     while(~scanf("%lld", &n)) {
         ans = ;
         For(i, , n) scanf("%lld", &a[i]);

         For(i, , n) {
             c[i] = ;
             For(j, , n) {
                 if(i == j) continue;
                 c[i] = mul_mod(c[i], sub_mod(mul_mod(a[j], a[j]), mul_mod(a[i], a[i])));
             }
             ans = add_mod(ans, inv_mod(mul_mod(a[i],  * c[i])));
         }

         printf("%lld\n", ans);
     }
     return ;
 }