• 3-idiots HDU


    给N个木棍问任选三个可组合出三角形的概率

    以前写的,整理整理存个档

    //#pragma comment(linker, "/STACK:1024000000,1024000000") 
    //#include<bits/stdc++.h>
    #include<stdio.h>
    #include<algorithm>
    #include<queue>
    #include<string.h>
    #include<iostream>
    #include<math.h>
    #include<set>
    #include<map>
    #include<vector>
    #include<iomanip>
    using namespace std;
    #define ll long long
    #define pb push_back
    #define FOR(a) for(int i=1;i<=a;i++)
    const double PI = acos(-1.0);
    
    struct complex
    {
    	double r, i;		//real and image
    	complex(double _r = 0, double _i = 0)
    	{
    		r = _r; i = _i;
    	}
    	complex operator +(const complex &b)
    	{
    		return complex(r + b.r, i + b.i);
    	}
    	complex operator -(const complex &b)
    	{
    		return complex(r - b.r, i - b.i);
    	}
    	complex operator *(const complex &b)
    	{
    		return complex(r*b.r - i*b.i, r*b.i + i*b.r);
    	}
    };
    void change(complex y[], int len)
    {
    	int i, j, k;
    	for (i = 1, j = len / 2; i < len - 1; i++)
    	{
    		if (i < j)swap(y[i], y[j]);
    		k = len / 2;
    		while (j >= k)
    		{
    			j -= k;
    			k /= 2;
    		}
    		if (j < k)j += k;
    	}
    }
    void fft(complex y[], int len, int on)
    {
    	change(y, len);
    	for (int h = 2; h <= len; h <<= 1)
    	{
    		complex wn(cos(-on * 2 * PI / h), sin(-on * 2 * PI / h));
    		for (int j = 0; j < len; j += h)
    		{
    			complex w(1, 0);
    			for (int k = j; k < j + h / 2; k++)
    			{
    				complex u = y[k];
    				complex t = w*y[k + h / 2];
    				y[k] = u + t;
    				y[k + h / 2] = u - t;
    				w = w*wn;
    			}
    		}
    	}
    	if (on == -1)
    		for (int i = 0; i < len; i++)
    			y[i].r /= len;
    }
    
    const int maxn = 4e5 + 6;
    complex x1[maxn];
    int arr[maxn >> 2];
    ll num[maxn]; ll sum[maxn];
    
    int main() {
    	int T, n;
    	scanf("%d", &T);
    	while (T--) {
    		memset(num, 0, sizeof num);
    		scanf("%d", &n);
    		for (int i = 0; i<n; i++) {
    			scanf("%d", &arr[i]);
    			num[arr[i]]++;
    		}
    		sort(arr, arr + n);
    		int len1 = arr[n - 1] + 1;                  //数域范围
    		int len = 1;
    		while (len<2 * len1)len <<= 1;             //扩充成最小2的幂次,即多项式项数
    		for (int i = 0; i<len1; i++)x1[i] = complex(num[i], 0);
    		for (int i = len1; i<len; i++)x1[i] = complex(0, 0);
    		fft(x1, len, 1);					//应用2n阶的fft计算出A(x)的点值表达
    		for (int i = 0; i<len; i++)x1[i] = x1[i] * x1[i];	//逐点相乘,O(n)
    		fft(x1, len, -1);					//对2n个点值应用fft,计算其逆dft
    		for (int i = 0; i<len; i++) {
    			num[i] = (long long)(x1[i].r + 0.5);
    		}
    		len = 2 * arr[n - 1];                       //从2的幂次缩回原来那么大
    /************************************************************************************************************/
    
    		//减掉相同组合
    		for (int i = 0; i<n; i++)num[arr[i] + arr[i]]--;
    		//选择无序,地位等价
    		for (int i = 1; i <= len; i++)num[i] /= 2;
    		sum[0] = 0;
    		for (int i = 1; i <= len; i++)sum[i] = sum[i - 1] + num[i];
    
    		ll cnt = 0;
    		for (int i = 0; i<n; i++) {
    			cnt += sum[len] - sum[arr[i]];
    			cnt -= (long long)(n - 1 - i)*i;
    			cnt -= n - 1;
    			cnt -= (long long)(n - 1 - i)*(n - i - 2) / 2;
    		}
    		ll tot = 1ll * n*(n - 1)*(n - 2) / 6;
    		printf("%.7lf
    ", (double)cnt / tot);
    	}
    }
    

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  • 原文地址:https://www.cnblogs.com/Drenight/p/8611212.html
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