Hdu 4609 (FFT)
3-idiots
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2165 Accepted Submission(s): 740Problem DescriptionKing 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.
InputAn 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≤105).
The following line contains N integers a_i (1≤a_i≤105), which denotes the length of each branch, respectively.
OutputOutput the probability that their branches can form a triangle, in accuracy of 7 decimal places.
Sample Input2 4 1 3 3 4 4 2 3 3 4
Sample Output0.5000000 1.0000000
给出ｎ个数，从中选出三个数，问能组成三角形的概率。
首先记录每个数出现的次数。然后和自己做多项式乘法，得到两个数组合起来可以拼成的数的情况。
当然，这里应该去掉自身和自身组合，也要考虑组合的无序性。
然后将所有数排序，假设a[i]作为三角形中最大的边。代码里有相关注释。
Accepted Code:
1 #define _CRT_SECURE_NO_WARNINGS
2 #include <string>
3 #include <vector>
4 #include <algorithm>
5 #include <numeric>
6 #include <set>
7 #include <map>
8 #include <queue>
9 #include <iostream>
10 #include <sstream>
11 #include <cstdio>
12 #include <cmath>
13 #include <ctime>
14 #include <cstring>
15 #include <cctype>
16 #include <cassert>
17 #include <limits>
18 #include <bitset>
19 #include <complex>
20 #define rep(i,n) for(int (i)=0;(i)<(int)(n);++(i))
21 #define rer(i,l,u) for(int (i)=(int)(l);(i)<=(int)(u);++(i))
22 #define reu(i,l,u) for(int (i)=(int)(l);(i)<(int)(u);++(i))
23 #define all(o) (o).begin(), (o).end()
24 #define pb(x) push_back(x)
25 #define mp(x,y) make_pair((x),(y))
26 #define mset(m,v) memset(m,v,sizeof(m))
27 #define INF 0x3f3f3f3f
28 #define INFL 0x3f3f3f3f3f3f3f3fLL
29 using namespace std;
30 typedef vector<int> vi; typedef pair<int,int> pii; typedef vector<pair<int,int> > vpii;
31 typedef long long ll; typedef vector<long long> vl; typedef pair<long long,long long> pll; typedef vector<pair<long long,long long> > vpll;
32 typedef vector<string> vs; typedef long double ld;
33 template<typename T, typename U> inline void amin(T &x, U y) { if(y < x) x = y; }
34 template<typename T, typename U> inline void amax(T &x, U y) { if(x < y) x = y; }
35
36 typedef long double Num; //??????long double?????
37 const Num PI = 3.141592653589793238462643383279L;
38 typedef complex<Num> Complex;
39 //n?????
40 //a?????
41 void fft_main(int n, Num theta, Complex a[]) {
42 for(int m = n; m >= 2; m >>= 1) {
43 int mh = m >> 1;
44 Complex thetaI = Complex(0, theta);
45 rep(i, mh) {
46 Complex w = exp((Num)i*thetaI);
47 for(int j = i; j < n; j += m) {
48 int k = j + mh;
49 Complex x = a[j] - a[k];
50 a[j] += a[k];
51 a[k] = w * x;
52 }
53 }
54 theta *= 2;
55 }
56 int i = 0;
57 reu(j, 1, n-1) {
58 for(int k = n >> 1; k > (i ^= k); k >>= 1) ;
59 if(j < i) swap(a[i], a[j]);
60 }
61 }
62
63 void fft(int n, Complex a[]) { fft_main(n, 2 * PI / n, a); }
64 void inverse_fft(int n, Complex a[]) { fft_main(n, -2 * PI / n, a); }
65
66 void convolution(vector<Complex> &v, vector<Complex> &w) {
67 int n = 1, vwn = v.size() + w.size() - 1;
68 while(n < vwn) n <<= 1;
69 v.resize(n), w.resize(n);
70 fft(n, &v[0]);
71 fft(n, &w[0]);
72 rep(i, n) v[i] *= w[i];
73 inverse_fft(n, &v[0]);
74 rep(i, n) v[i] /= n;
75 }
76
77 // solve problem....
78 const int MAXN = 100002;
79 int a[MAXN];
80 long long num[MAXN], sum[MAXN<<2];
81
82 void read(int &res) {
83 res = 0;
84 char c = ' ';
85 while (c < '0' || c > '9') c = getchar();
86 while (c >= '0' && c <= '9') res = res * 10 + c - '0', c = getchar();
87 }
88
89 vector<long long> calc(const vector<long long> &A) {
90 vector<Complex> Lc(all(A)), Rc(all(A));
91 convolution(Lc, Rc);
92 long long n = A.size() + A.size() - 1;
93 vector<long long> res(n);
94 rep(i, n) res[i] = (long long)(Lc[i].real() + .5);
95 // rep(i, n) cerr << res[i] << ", "; cerr << endl;
96 return res;
97 }
98
99 int main() {
100 int T; read(T);
101 while (T--) {
102 int N; read(N);
103 //a[i]出现次数
104 memset(num, 0, sizeof(num));
105 rep(i, N) {
106 read(a[i]);
107 num[a[i]]++;
108 }
109 sort(a, a + N);
110 int len = a[N-1] * 2;
111 vector<long long> A(num, num + a[N-1] + 1);
112 vector<long long> ans = calc(A);
113 //重复使用
114 rep(i, N) {
115 ans[a[i]+a[i]]--;
116 }
117 //无序
118 rep(i, len) {
119 ans[i+1] /= 2;
120 }
121 //前缀和
122 sum[0] = 0;
123 rep(i, len) {
124 sum[i+1] = sum[i] + ans[i+1];
125 }
126 long long cnt = 0;
127 //假设a[i] 为最大的边
128 rep(i, N) {
129 cnt += sum[len] - sum[a[i]];
130 //a[i]本身
131 cnt -= N - 1;
132 //一大一小
133 cnt -= (long long)i * (N - i - 1);
134 //两大
135 cnt -= (long long)(N-i-1) * (N-i-2) / 2;
136 }
137 //总的方法数 = C(N, 3)
138 long long tot = (long long)N * (N-1) * (N-2) / 6;
139 printf("%.7f\n", (cnt+0.0) / tot);
140 }
141 return 0;
142 }
posted on 2014-08-19 10:42 Stomach_ache 阅读(204) 评论(0) 编辑 收藏 举报
