2019年牛客多校第一场B题 Integration 数学

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

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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"=["<
#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;
}