【HDU1402】【FFT】A * B Problem Plus

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

Author

DOOM III

Recommend

DOOM III

 

【分析】

模板，不过我想说的是，这居然是06年的题目。。。太恐怖了。

 /*
 宋代苏轼
 《临江仙·夜饮东坡醒复醉》
 夜饮东坡醒复醉，归来仿佛三更。家童鼻息已雷鸣。敲门都不应，倚杖听江声。
 长恨此身非我有，何时忘却营营。夜阑风静縠纹平。小舟从此逝，江海寄余生。
 */
 #include <cstdio>
 #include <cstring>
 #include <algorithm>
 #include <cmath>
 #include <queue>
 #include <vector>
 #include <iostream>
 #include <string>
 #include <ctime>
 #define LOCAL
 const double Pi = acos(-1.0);
 const int MAXN =  + ;
 using namespace std;
 typedef long long ll;
 struct Num {
    double a , b;
     //构造函数
    Num(double x = ,double y = ) {a = x; b = y;}
     //复数的三种运算
    Num operator + (const Num &c) {return Num(a + c.a, b + c.b);}
    Num operator - (const Num &c) {return Num(a - c.a, b - c.b);}
    Num operator * (const Num &c) {return Num(a * c.a - b * c.b, a * c.b + b * c.a);}
 }x1[MAXN] , x2[MAXN];

 //注意loglen为log后的长度
 void change(Num *t, int len, int loglen){
     //蝶形变换后的序列编号
     for (int i = ; i < len; i++){
         int x = i, k = ;
         for (int j = ; j < loglen; j++, x >>= ) k = (k << ) | (x & );
         if (k < i) swap(t[k], t[i]);
     }
 }
 //基2-FFT
 void FFT(Num *x, int len, int loglen){
     change(x, len, loglen);
     int t = ;
     //t为长度
     for (int i = ; i < loglen; i++, (t <<= )){
         int l = , r = l + t;
         while (l < len){
             //初始化
             Num a, b, wo(cos(Pi / t), sin(Pi / t)), wn(, );
             for (int j = l; j < l + t; j++){
                 a = x[j];
                 b = x[j + t] * wn;
                 //蝶形计算
                 x[j] = a + b;
                 x[j + t] = a - b;
                 wn = wn * wo;
             }
             //注意是合并所以要走2t步
             l = r + t;
             r = l + t;
         }
     }
 }
 //离散傅里叶变换
 void DFT(Num *x, int len, int loglen){
     int t = (<<loglen);
     for (int i = ; i < loglen; i++){
         t >>= ;
         int l = , r = l + t;
         while (l < len){
             Num a, b, wn(, ), wo(cos(Pi / t), -sin(Pi / t));
             for (int j = l; j < l + t; j++){
                 a = x[j] + x[j + t];
                 b = (x[j] - x[j + t]) * wn;
                 x[j] = a;
                 x[j + t] = b;
                 wn = wn * wo;
             }
             l = r + t;
             r = l + t;
         }
     }
     change(x, len, loglen);
     for (int i= ; i < len; i++) x[i].a /= len;
 }
 int solve(char *a, char *b){
     int len1, len2, len, loglen;
     int t, over;
     len1 = strlen(a) << ;
     len2 = strlen(b) << ;
     len = ;
     loglen = ;
     while (len < len1) len <<= , loglen++;
     while (len < len2) len <<= , loglen++;
     //处理len1
     for (int i = ; i < len; i++){
         if (a[i]) x1[i].a = a[i] - '', x1[i].b = ;
         else x1[i].a = x1[i].b = ;
     }
     for (int i = ; i < len; i++){
         if (b[i]) x2[i].a = b[i] - '', x2[i].b = ;
         else x2[i].a = x2[i].b = ;
     }
     FFT(x1, len, loglen);
     FFT(x2, len, loglen);
     for (int i = ; i < len; i++) x1[i] = x1[i] * x2[i];

     DFT(x1, len, loglen);
     over = len = ;
     //转换成十进制的整数
     for (int i = ((len1 + len2) / ) - ; i >= ; i--){
         t = x1[i].a + over + 0.5;
         a[len++] = t % ;
         over = t / ;
     }
     while (over){
         a[len++] = over % ;
         over /= ;
     }
     return len;
 }
 //输出
 void print(char *str, int len){
      for(len--; len>= && !str[len];len--);
     if(len < ) putchar('');
     else for(;len>=;len--) putchar(str[len]+'');
     printf("\n");
 }
 char a[MAXN] , b[MAXN];

 int main() {

     //char a[MAXN], b[MAXN];
    while(scanf("%s%s", a, b) != EOF) {
           print(a, solve(a, b));
           memset(a, , sizeof(a));
           memset(b, , sizeof(b));
     }
     //printf("%.10lf\n", Pi);
    return ;
 }