【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。

  1 /*

  2 宋代苏轼

  3 《临江仙·夜饮东坡醒复醉》

  4 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。

  5 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。  

  6 */

  7 #include <cstdio>

  8 #include <cstring>

  9 #include <algorithm>

 10 #include <cmath>

 11 #include <queue>

 12 #include <vector>

 13 #include <iostream>

 14 #include <string>

 15 #include <ctime>

 16 #define LOCAL

 17 const double Pi = acos(-1.0);

 18 const int MAXN = 200000 + 10;

 19 using namespace std;

 20 typedef long long ll;

 21 struct Num {

 22    double a , b;

 23     //构造函数

 24    Num(double x = 0,double y = 0) {a = x; b = y;}

 25     //复数的三种运算

 26    Num operator + (const Num &c) {return Num(a + c.a, b + c.b);}

 27    Num operator - (const Num &c) {return Num(a - c.a, b - c.b);}

 28    Num operator * (const Num &c) {return Num(a * c.a - b * c.b, a * c.b + b * c.a);}

 29 }x1[MAXN] , x2[MAXN];

 30 

 31 //注意loglen为log后的长度

 32 void change(Num *t, int len, int loglen){

 33     //蝶形变换后的序列编号

 34     for (int i = 0; i < len; i++){

 35         int x = i, k = 0;

 36         for (int j = 0; j < loglen; j++, x >>= 1) k = (k << 1) | (x & 1);

 37         if (k < i) swap(t[k], t[i]);

 38     }

 39 }

 40 //基2-FFT

 41 void FFT(Num *x, int len, int loglen){

 42     change(x, len, loglen);

 43     int t = 1;

 44     //t为长度

 45     for (int i = 0; i < loglen; i++, (t <<= 1)){

 46         int l = 0, r = l + t;

 47         while (l < len){

 48             //初始化

 49             Num a, b, wo(cos(Pi / t), sin(Pi / t)), wn(1, 0);

 50             for (int j = l; j < l + t; j++){

 51                 a = x[j];

 52                 b = x[j + t] * wn;

 53                 //蝶形计算

 54                 x[j] = a + b;

 55                 x[j + t] = a - b;

 56                 wn = wn * wo;

 57             }

 58             //注意是合并所以要走2t步

 59             l = r + t;

 60             r = l + t;

 61         }

 62     }

 63 }

 64 //离散傅里叶变换

 65 void DFT(Num *x, int len, int loglen){

 66     int t = (1<<loglen);

 67     for (int i = 0; i < loglen; i++){

 68         t >>= 1;

 69         int l = 0, r = l + t;

 70         while (l < len){

 71             Num a, b, wn(1, 0), wo(cos(Pi / t), -sin(Pi / t));

 72             for (int j = l; j < l + t; j++){

 73                 a = x[j] + x[j + t];

 74                 b = (x[j] - x[j + t]) * wn;

 75                 x[j] = a;

 76                 x[j + t] = b;

 77                 wn = wn * wo;

 78             }

 79             l = r + t;

 80             r = l + t;

 81         }    

 82     }

 83     change(x, len, loglen);

 84     for (int i= 0; i < len; i++) x[i].a /= len;

 85 }

 86 int solve(char *a, char *b){

 87     int len1, len2, len, loglen;

 88     int t, over;

 89     len1 = strlen(a) << 1;

 90     len2 = strlen(b) << 1;

 91     len = 1;

 92     loglen = 0;

 93     while (len < len1) len <<= 1, loglen++;

 94     while (len < len2) len <<= 1, loglen++;

 95     //处理len1

 96     for (int i = 0; i < len; i++){

 97         if (a[i]) x1[i].a = a[i] - '0', x1[i].b = 0;

 98         else x1[i].a = x1[i].b = 0;

 99     }

100     for (int i = 0; i < len; i++){

101         if (b[i]) x2[i].a = b[i] - '0', x2[i].b = 0;

102         else x2[i].a = x2[i].b = 0;

103     }

104     FFT(x1, len, loglen);

105     FFT(x2, len, loglen);

106     for (int i = 0; i < len; i++) x1[i] = x1[i] * x2[i];

107 

108     DFT(x1, len, loglen);

109     over = len = 0;

110     //转换成十进制的整数

111     for (int i = ((len1 + len2) / 2) - 2; i >= 0; i--){

112         t = x1[i].a + over + 0.5;

113         a[len++] = t % 10;

114         over = t / 10;

115     }

116     while (over){

117         a[len++] = over % 10;

118         over /= 10;

119     }

120     return len;

121 }

122 //输出

123 void print(char *str, int len){

124      for(len--; len>=0 && !str[len];len--);

125     if(len < 0) putchar('0');

126     else for(;len>=0;len--) putchar(str[len]+'0');

127     printf("\n");

128 }

129 char a[MAXN] , b[MAXN];

130 

131 int main() {

132     

133     //char a[MAXN], b[MAXN];

134    while(scanf("%s%s", a, b) != EOF) {

135           print(a, solve(a, b));

136           memset(a, 0, sizeof(a));

137           memset(b, 0, sizeof(b));

138     }

139     //printf("%.10lf\n", Pi);

140    return 0;

141 }

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