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 }

 

posted on 2014-08-19 08:48  Stomach_ache  阅读(179)  评论(0)  编辑  收藏  举报

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