HDU 1402 A * B Problem Plus ——（大数乘法，FFT）

Posted 2020-10-01 Storm_Spirit

　　因为刚学fft，想拿这题练练手，结果WA了个爽= =。

　　总结几点犯的错误：

　　1.要注意处理前导零的问题。

　　2.一定要注意数组大小的问题。（前一个fft的题因为没用到b数组，所以b就没管，这里使用了b数组，结果忘记给其大小乘以4倍了）

　　代码如下：

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 const double pi = atan(1.0)*4;
 4 const int N = 5e4 + 5;
 5 typedef long long ll;
 6 
 7 struct Complex {
 8     double x,y;
 9     Complex(double _x=0,double _y=0)
10         :x(_x),y(_y) {}
11     Complex operator + (Complex &tt) { return Complex(x+tt.x,y+tt.y); }
12     Complex operator - (Complex &tt) { return Complex(x-tt.x,y-tt.y); }
13     Complex operator * (Complex &tt) { return Complex(x*tt.x-y*tt.y,x*tt.y+y*tt.x); }
14 };
15 Complex a[N*4],b[N*4];
16 void fft(Complex *a, int n, int rev) {
17     // n是(大于等于相乘的两个数组长度)2的幂次 ； 比如长度是5 ,那么 n = 8    2^2 < 5          2^3 > 5
18     // 从0开始表示长度，对a进行操作
19     // rev==1进行DFT，==-1进行IDFT
20     for (int i = 1,j = 0; i < n; ++ i) {
21         for (int k = n>>1; k > (j^=k); k >>= 1);
22         if (i<j) std::swap(a[i],a[j]);
23     }
24     for (int m = 2; m <= n; m <<= 1) {
25         Complex wm(cos(2*pi*rev/m),sin(2*pi*rev/m));
26         for (int i = 0; i < n; i += m) {
27             Complex w(1.0,0.0);
28             for (int j = i; j < i+m/2; ++ j) {
29                 Complex t = w*a[j+m/2];
30                 a[j+m/2] = a[j] - t;
31                 a[j] = a[j] + t;
32                 w = w * wm;
33             }
34         }
35     }
36     if (rev==-1) {
37         for (int i = 0; i < n; ++ i) a[i].x /= n,a[i].y /= n;
38     }
39 }
40 
41 char s[N], t[N];
42 int c[N*2];
43 
44 int main(){
45     /*a[0] = Complex(0,0); //  a[0]: x的0次项。
46     a[1] = Complex(1,0);
47     a[2] = Complex(2,0);
48     a[3] = Complex(3,0);
49 
50     b[0] = Complex(3,0);
51     b[1] = Complex(2,0);
52     b[2] = Complex(1,0);
53     b[3] = Complex(0,0);
54     fft(a,8,1);
55     fft(b,8,1);
56     for(int i = 0 ; i < 8 ; i ++){
57         a[i] = a[i] * b[i];
58     }
59     fft(a,8,-1);
60     for(int i = 0 ; i < 8 ; i ++){
61         cout << i << "   " << a[i].x << endl;;
62     }*/
63     while(scanf("%s%s",s,t) == 2)
64     {
65         int n = strlen(s), m = strlen(t);
66         for(int i=0;i<n;i++) a[i] = Complex(s[n-i-1]-‘0‘, 0);
67         for(int i=0;i<m;i++) b[i] = Complex(t[m-i-1]-‘0‘, 0);
68         int len = n+m-1;
69         int LIM = 1;
70         while(LIM < len) LIM <<= 1;
71         for(int i=n;i<LIM;i++) a[i] = Complex(0, 0);
72         for(int i=m;i<LIM;i++) b[i] = Complex(0, 0);
73         fft(a, LIM, 1);
74         fft(b, LIM, 1);
75         for(int i=0;i<LIM;i++) a[i] = a[i] * b[i];
76         fft(a, LIM, -1);
77         memset(c, 0, sizeof c);
78         for(int i=0;i<len-1;i++)
79         {
80             c[i] += (int)(a[i].x + 0.1);
81             c[i+1] += c[i] / 10;
82             c[i] %= 10;
83         }
84         c[len-1] += (int)(a[len-1].x + 0.1);
85         if(c[len-1] > 9)
86         {
87             c[len] = c[len-1] / 10;
88             c[len-1] %= 10;
89             len++;
90         }
91         for(int i=len-1;i>0;i--) if(c[i] == 0) len--; else break; // 0 * 345 = 0 instead of 000
92         for(int i=0;i<len/2;i++) swap(c[i], c[len-i-1]);
93         for(int i=0;i<len;i++) printf("%d",c[i]);
94         puts("");
95     }
96     return 0;
97 }