hdu 4609 3-idiots

FFT乘法

fft在超大整数乘法中的应用

貌似能把时间从o(n^2) 降低到 o(n logn)

#include <stdio.h>   #include <string.h>#include <math.h>#include <algorithm>#ifdef __GNUC__#endif // __GNUC__using namespace std;typedef long long ll;const int MAXN = 400100;const double PI = acos(-1.0);struct mcomplex{double r, i;mcomplex(double rr=0.0, double ii = 0.0):r(rr),i(ii){}mcomplex operator + (mcomplex &a){return mcomplex(r+a.r, i+a.i);}mcomplex operator - (mcomplex &a){return mcomplex(r-a.r, i-a.i);}mcomplex operator * (mcomplex &a){return mcomplex(r*a.r-i*a.i, r*a.i+i*a.r);}};int a[MAXN>>2];mcomplex xl[MAXN];ll num[MAXN], sum[MAXN];void change(mcomplex 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(mcomplex y[], int len, int on){change(y, len);for (int h =2; h <= len; h<<=1){mcomplex wn(cos(-on*2*PI/h), sin(-on*2*PI/h));for (int j = 0; j< len; j+=h){mcomplex w(1, 0);for (int k = j; k< j+h/2; ++k){mcomplex u = y[k];mcomplex 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;}int main(){#ifndef ONLINE_JUDGE    freopen("in.txt", "r", stdin);#endif // ONLINE_JUDGEint t, n, i;scanf("%d", &t);while (t--){scanf("%d", &n);memset(num, 0, sizeof num);for (i = 0; i< n; ++i){scanf("%d", &a[i]);num[a[i]]++;}sort(a, a+n);int crest = a[n-1] + 1;int len = 1;while ( len < 2*crest ) len <<= 1;for ( i = 0; i< crest; ++i)xl[i] = mcomplex(num[i], 0.0);for (i = crest ; i< len; ++i)xl[i] = mcomplex(0, 0);fft(xl, len, 1);for (i = 0; i< len; ++i)xl[i] = xl[i]*xl[i];fft(xl, len, -1);for (i = 0; i< len; ++i)num[i] = (ll)(xl[i].r + 0.5);///.....................................len = 2*a[n-1];/// 与自身组合for (i = 0; i< n; ++i)--num[a[i]<<1];/// 顺序for (i = 1; i<= len; ++i)num[i] >>= 1;sum[0] = 0;for (i = 1; i<= len; ++i)sum[i] = sum[i-1] + num[i];ll cnt = 0;for (i = 0; i< n; ++i){cnt += sum[len] - sum[a[i]];/// 假设a[i]是最长的棒子，枚举长度大于a[i]的组合数量 cnt -= (ll)(n-1-i)*i; /// 一根大于，一根小于cnt -= n-1;/// 另外两根中不能包括自身cnt -= (ll)(n-1-i)*(n-i-2)/2;/// 两根都大于}ll tot = (ll) n*(n-1)*(n-2)/6;printf("%.7lf\n", (double)cnt/tot);}    return 0;}