Hdu 1402 （FFT）

题目链接

A * B Problem Plus

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 12490    Accepted Submission(s): 2206

Problem Description

Calculate A * B.

Input

Each line will contain two integers A and B. Process to end of file.

Note: the length of each integer will not exceed 50000.

Output

For each case, output A * B in one line.

Sample Input

1
2
1000
2

Sample Output

2
2000

初学FFT，根本不懂怎么实现的。直接套模板

Accepted Code:

 #include <string>
 #include <vector>
 #include <algorithm>
 #include <numeric>
 #include <set>
 #include <map>
 #include <queue>
 #include <iostream>
 #include <sstream>
 #include <cstdio>
 #include <cmath>
 #include <ctime>
 #include <cstring>
 #include <cctype>
 #include <cassert>
 #include <limits>
 #include <bitset>
 #include <complex>
 #define rep(i,n) for(int (i)=0;(i)<(int)(n);++(i))
 #define rer(i,l,u) for(int (i)=(int)(l);(i)<=(int)(u);++(i))
 #define reu(i,l,u) for(int (i)=(int)(l);(i)<(int)(u);++(i))
 #define all(o) (o).begin(), (o).end()
 #define rall(o) (o).rbegin(), ((o).rend())
 #define pb(x) push_back(x)
 #define mp(x,y) make_pair((x),(y))
 #define mset(m,v) memset(m,v,sizeof(m))
 #define INF 0x3f3f3f3f
 #define INFL 0x3f3f3f3f3f3f3f3fLL
 using namespace std;
 typedef vector<int> vi; typedef pair<int,int> pii; typedef vector<pair<int,int> > vpii;
 typedef long long ll; typedef vector<long long> vl; typedef pair<long long,long long> pll; typedef vector<pair<long long,long long> > vpll;
 typedef vector<string> vs; typedef long double ld;
 template<typename T, typename U> inline void amin(T &x, U y) { if(y < x) x = y; }
 template<typename T, typename U> inline void amax(T &x, U y) { if(x < y) x = y; }

 typedef long double Num;    //??????long double?????
 const Num PI = .141592653589793238462643383279L;
 typedef complex<Num> Complex;
 //n?????
 //a?????
 void fft_main(int n, Num theta, Complex a[]) {
     for(int m = n; m >= ; m >>= ) {
         int mh = m >> ;
         Complex thetaI = Complex(, theta);
         rep(i, mh) {
             Complex w = exp((Num)i*thetaI);
             for(int j = i; j < n; j += m) {
                 int k = j + mh;
                 Complex x = a[j] - a[k];
                 a[j] += a[k];
                 a[k] = w * x;
             }
         }
         theta *= ;
     }
     int i = ;
     reu(j, , n-) {
     for(int k = n >> ; k > (i ^= k); k >>= ) ;
         if(j < i) swap(a[i], a[j]);
     }
 }

 void fft(int n, Complex a[]) { fft_main(n,  * PI / n, a); }
 void inverse_fft(int n, Complex a[]) { fft_main(n, - * PI / n, a); }

 void convolution(vector<Complex> &v, vector<Complex> &w) {
     int n = , vwn = v.size() + w.size() - ;
     while(n < vwn) n <<= ;
     v.resize(n), w.resize(n);
     fft(n, &v[]);
     fft(n, &w[]);
     rep(i, n) v[i] *= w[i];
     inverse_fft(n, &v[]);
     rep(i, n) v[i] /= n;
 }

 const int maxn = ;
 vector<int> ans;
 char s1[maxn], s2[maxn];

 void solve(vector<int> &res) {
     //cerr << s1 << s2;
     int len1 = strlen(s1), len2 = strlen(s2);
     for (int i = ; i < len1; i++) s1[i] -= '';
     for (int i = ; i < len2; i++) s2[i] -= '';
     vector<Complex> Lc(len1), Rc(len2);
     for (int i = ; i < len1; i++) Lc[i] = Complex(s1[len1-i-], );
     for (int i = ; i < len2; i++) Rc[i] = Complex(s2[len2-i-], );
     convolution(Lc, Rc);
     int n = len1 + len2 - ;
     res.resize(n);
     rep(i, n) res[i] = Lc[i].real() + .;
 }

 int main(void) {
     while (~scanf("%s %s", s1, s2)) {
         solve(ans);
         ans.resize(ans.size() << );
         for (int i = ; i < ans.size(); i++) {
             ans[i+] += ans[i] / ;
             ans[i] %= ;
         }
         int high = ans.size() - ;
         while (high >  && !ans[high]) high--;
         for (int i = high; i >= ; i--) putchar(ans[i] + '');
         putchar('\n');
     }
     return ;
 }