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$frac{1}{pi}int_{0}^{infty} frac{1}{prodlimits_{i=1}^{n}(a_i^2+x^2)}dx$模 10⁹ + 7。（可以证明这个积分一定是有理数）

分析

$$egin{align*}
&令c_i = frac{1}{prod_{j e i} (a_j^2 - a_i^2)} \
&则frac{1}{prodlimits_{i=1}^{n}(a_i^2+x^2)} = sumlimits_{i=1}^{n} frac{c_i}{a_i^2+x^2} \
&而int_{0}^{infty} frac{c_i}{a_i^2+x^2}dx = frac{c_i}{2a_i}pi \
&于是frac{1}{pi}int_{0}^{infty} frac{1}{prodlimits_{i=1}^{n}(a_i^2+x^2)}dx = sumlimits_{i=1}^{n} frac{c_i}{2a_i}
end{align*}$$

　　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] << " 
"[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 << "]" << "
";
    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, 1, BUF, stdin);
    return *bg++;
} 

inline int ri(){
    int x = 0, f = 1, c = gc();
    for(; c<48||c>57; f = c=='-'?-1:f, c=gc());
    for(; c>47&&c<58; x = x*10 + c - 48, 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-8;
const LL inf = 0x7fffffff;
const LL infLL = 0x7fffffffffffffffLL;
const LL mod = 1e9 + 7;
const int maxN = 1e3 + 7;
const LL ONE = 1;
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 = 1, y = 0;}
    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 == 1 ? (x % p + p) % p : -1;
}

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 = 0;
        For(i, 1, n) scanf("%lld", &a[i]);
        
        For(i, 1, n) {
            c[i] = 1;
            For(j, 1, 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], 2 * c[i])));
        }
        
        printf("%lld
", ans);
    }
    return 0;
}