3-idiots

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 5646    Accepted Submission(s): 1960 Problem Description King OMeGa catched three men who had been streaking in the street. Looking as idiots though, the three men insisted that it was a kind of performance art, and begged the king to free them. Out of hatred to the real idiots, the king wanted to check if they were lying. The three men were sent to the king's forest, and each of them was asked to pick a branch one after another. If the three branches they bring back can form a triangle, their math ability would save them. Otherwise, they would be sent into jail. However, the three men were exactly idiots, and what they would do is only to pick the branches randomly. Certainly, they couldn't pick the same branch - but the one with the same length as another is available. Given the lengths of all branches in the forest, determine the probability that they would be saved.   Input An integer T(T≤100) will exist in the first line of input, indicating the number of test cases. Each test case begins with the number of branches N(3≤N≤10 ⁵). The following line contains N integers a_i (1≤a_i≤10 ⁵), which denotes the length of each branch, respectively.   Output Output the probability that their branches can form a triangle, in accuracy of 7 decimal places.   Sample Input 2 4 1 3 3 4 4 2 3 3 4   Sample Output 0.5000000 1.0000000   Source 2013 Multi-University Training Contest 1

题意：

给出n根棍子，让你求出可以组合成三角形的方法数

题解：

先任意选择两根棍子组合，这个用FFT即可

这里需要注意：

去重：选了两次本身，以及选了相同的组合两次

sort(a,a+n); len=2*a[n-1];///缩小区间 for(int i=0;i<n;i++)///去掉本身选两次的情况 num[a[i]+a[i]]--; for(int i=1;i<=len;i++)///选择的无序，除以2 num[i]/=2;

对于第三根棍子，我们先将棍子长度排好序：

然后从小到大进行选择：

每次选择的棍子当做一个三角形里面长度最短的，因为是枚举每一根棍子，所以可以覆盖所有的情况

同时，也要去掉错误的情况

LL cnt=0; for(int i=0;i<n;i++) { cnt+=sum[len]-sum[a[i]];///两两组合的前缀和之差 cnt-=(LL)(n-1-i)*i;///减掉一个取大，一个取小的 cnt-=(n-1);///减掉一个取本身，另外一个取其它 cnt-=(LL)i*(i-1)/2;///减掉小于它的取两个的组合 }

#include <stdio.h> #include <string.h> #include <iostream> #include <algorithm> #include <math.h> using namespace std; const double PI = acos(-1.0); //复数结构体 struct complex { double r,i; complex(double _r = 0.0,double _i = 0.0) { 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); } }; /* * 进行FFT和IFFT前的反转变换。 * 位置i和 （i二进制反转后位置）互换 * len必须去2的幂 */ 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]); //交换互为小标反转的元素，i<j保证交换一次 //i做正常的+1，j左反转类型的+1,始终保持i和j是反转的 k = len/2; while( j >= k) { j -= k; k /= 2; } if(j < k) j += k; } } /* * 做FFT * len必须为2^k形式， * on==1时是DFT，on==-1时是IDFT */ 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; } #define LL long long const int MAXN = 200010*2; complex x1[MAXN]; int a[MAXN/4]; LL sum[MAXN],num[MAXN]; int main() { int n,m,T; //freopen("in.txt","r",stdin); scanf("%d",&T); while(T--) { scanf("%d",&n); int len1=-1,len=1; memset(num,0,sizeof(num)); for(int i=0;i<n;i++){ scanf("%d",&a[i]); len1=len1>a[i]?len1:a[i]; num[a[i]]++; } len1++; while(len < len1*2) len<<=1; 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); //求DFT fft(x1,len,1); for(int i = 0;i < len;i++) x1[i] = x1[i]*x1[i]; fft(x1,len,-1); for(int i = 0;i < len;i++) num[i] = (LL)(x1[i].r+0.5); sort(a,a+n); len=2*a[n-1];///缩小区间 for(int i=0;i<n;i++)///去掉本身选两次的情况 num[a[i]+a[i]]--; for(int i=1;i<=len;i++)///选择的无序，除以2 num[i]/=2; sum[0]=0;///前缀和 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[a[i]]; cnt-=(LL)(n-1-i)*i;///减掉一个取大，一个取小的 cnt-=(n-1);///减掉一个取本身，另外一个取其它 cnt-=(LL)i*(i-1)/2;///减掉小于它的取两个的组合 } LL tot=(LL)n*(n-1)*(n-2)/6; printf("%.7lf\n",(double)cnt/tot); } return 0; }