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:

 1 #include <string>

 2 #include <vector>

 3 #include <algorithm>

 4 #include <numeric>

 5 #include <set>

 6 #include <map>

 7 #include <queue>

 8 #include <iostream>

 9 #include <sstream>

 10 #include <cstdio>

 11 #include <cmath>

 12 #include <ctime>

 13 #include <cstring>

 14 #include <cctype>

 15 #include <cassert>

 16 #include <limits>

 17 #include <bitset>

 18 #include <complex>

 19 #define rep(i,n) for(int (i)=0;(i)<(int)(n);++(i))

 20 #define rer(i,l,u) for(int (i)=(int)(l);(i)<=(int)(u);++(i))

 21 #define reu(i,l,u) for(int (i)=(int)(l);(i)<(int)(u);++(i))

 22 #define all(o) (o).begin(), (o).end()

 23 #define rall(o) (o).rbegin(), ((o).rend())

 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 const int maxn = 50005;  78 vector<int> ans;  79 char s1[maxn], s2[maxn];  80 

 81 void solve(vector<int> &res) {  82     //cerr << s1 << s2;

 83     int len1 = strlen(s1), len2 = strlen(s2);  84     for (int i = 0; i < len1; i++) s1[i] -= '0';  85     for (int i = 0; i < len2; i++) s2[i] -= '0';  86     vector<Complex> Lc(len1), Rc(len2);  87     for (int i = 0; i < len1; i++) Lc[i] = Complex(s1[len1-i-1], 0);  88     for (int i = 0; i < len2; i++) Rc[i] = Complex(s2[len2-i-1], 0);  89  convolution(Lc, Rc);  90     int n = len1 + len2 - 1;  91  res.resize(n);  92     rep(i, n) res[i] = Lc[i].real() + .5;  93 }  94 

 95 int main(void) {  96     while (~scanf("%s %s", s1, s2)) {  97  solve(ans);  98         ans.resize(ans.size() << 1);  99         for (int i = 0; i < ans.size(); i++) { 100             ans[i+1] += ans[i] / 10; 101             ans[i] %= 10; 102  } 103         int high = ans.size() - 1; 104         while (high > 0 && !ans[high]) high--; 105         for (int i = high; i >= 0; i--) putchar(ans[i] + '0'); 106         putchar('\n'); 107  } 108     return 0; 109 }