标签:就是 .net https cin with pen span mes class
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\n", (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
标签:就是 .net https cin with pen span mes class
原文地址:https://www.cnblogs.com/widsom/p/9155012.html