HDU 4609 3-idiots

Posted 2020-11-14 widsom

HDU - 4609

思路：
记录每个木棍长度出现的次数，然后就可以用用类似多项式的乘法（专业术语：卷积，因为是下标和为一特定值的积的和（x+y=k），相当于在笛卡尔坐标系中将这条直线卷起来，故得名卷积）的方法计算两个组合后每个长度的木棍的个数，然后用容斥减去多余的。

然后对它求个前缀和sum

假设两个组合最长木棍为l

然后再枚举n个木棍，假设当前的ai为最大的木棍，

另外两个木棍的和肯定大于ai，所以方案数加sum[l] - sum[ai]

减去另外两个都大于它的情况，也就是减去(n-i-1)*(n-i-2)/2

减去一大一小的情况，也就是减去i*(n-i-1)

减去一个为本身，一个为其他的情况，也就是减去n-1

最后拿方案数除以所有的情况n*(n-1)*(n-2)/6

代码：

#include<bits/stdc++.h>
using namespace std;
#define fi first
#define se second
#define pi acos(-1.0)
#define LL long long
#define mp make_pair
#define pb push_back
#define ls rt<<1, l, m
#define rs rt<<1|1, m+1, r
#define ULL unsigned LL
#define pll pair<LL, LL>
#define pii pair<int, int>
#define piii pair<int,pii>
#define mem(a, b) memset(a, b, sizeof(a))
#define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define fopen freopen("in.txt", "r", stdin);freopen("out.txt", "w", stout);
//head

const int N = 4e5 + 10;
int a[N], R[N];
LL num[N],sum[N];
struct Complex {
    double x, y;
    Complex(double _x=0, double _y=0) : x(_x),y(_y) {};
    Complex operator + (Complex &t) {return Complex(x+t.x, y+t.y);}
    Complex operator - (Complex &t) {return Complex(x-t.x, y-t.y);}
    Complex operator * (Complex &t) {return Complex(x*t.x - y*t.y, x*t.y + y*t.x);}
}X[N];
void fft(Complex *x, int n, int type) {
    for (int i = 0; i < n; i++) if(i < R[i]) swap(x[i], x[R[i]]);
    for (int i = 1; i < n; i<<=1) {
        Complex wn(cos(pi/i), type*sin(pi/i));
        for (int j = 0; j < n; j+=i<<1) {
            Complex w(1, 0);
            for (int k = 0; k < i; k++, w=w*wn) {
                Complex X = x[j+k];
                Complex Y = w*x[j+k+i];
                x[j+k] = X + Y;
                x[j+k+i] = X - Y;
            }
        }
    }
}
int main() {
    int T, n;
    scanf("%d", &T);
    while(T--) {
        mem(num, 0);
        scanf("%d", &n);
        for (int i = 0; i < n; i++) scanf("%d", &a[i]), num[a[i]]++;
        sort(a, a+n);
        int len = a[n-1] + 1;
        int l = 1, L = 0;
        for (int i = 0; i < len; i++) X[i] = Complex(num[i], 0);
        for ( ; l < len*2; l <<= 1) L++;
        for (int i = len; i < l; i++) X[i] = Complex(0, 0);
        for (int i = 0; i < l; i++) {
            R[i] = (R[i>>1]>>1)|((i&1)<<L-1);
        }
        fft(X, l, 1);
        for (int i = 0; i < l; i++) X[i] = X[i]*X[i];
        fft(X, l, -1);
        for (int i = 0; i < l; i++) num[i] = (LL)(X[i].x/l + 0.5);
        for (int i = 0; i < n; i++) num[a[i]+a[i]]--; 
        for (int i = 0; i < l; i++) num[i] /= 2;
        sum[0] = num[0];
        for (int i = 1; i < l; i++) sum[i] = sum[i-1] + num[i]; 
        LL ans = 0;
        for(int i = 0; i < n; i++) {
            ans += sum[l-1] - sum[a[i]];
            ans -= 1LL * (n-i-1) * i;
            ans -= 1LL * (n-i-1) * (n-i-2) / 2;
            ans -= n-1;
        } 
        LL t = 1LL*n*(n-1)*(n-2)/6;
        printf("%.7f
", (double)ans/t);
    }
    return 0;
}

PS:长度可以扩展一点，后面是0都没关系，只要长度是2的幂次就可以