HDU1042 A * B Problem Plus

Posted 2020-10-24 HWIM

 

A * B Problem Plus

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 24620    Accepted Submission(s): 6271

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

 

 

分析

一直不过，最后发现数组开小了qwq。

思路：把每个数分解成多项式

$n = a_0*10^0+a_1*10^1+...a_k*10^k$

然后多项式乘法，FFT模板题。

code 

 1 #include<cstdio>
 2 #include<algorithm>
 3 #include<cmath>
 4 #include<cstring>
 5 #include<iostream>
 6 
 7 using namespace std;
 8 
 9 const int N = 200100;
10 const double pi = acos(-1.0);
11 char a[N],b[N];
12 int ans[N]; //数组大小！
13 
14 struct Complex{
15     double x,y;
16     Complex() {x=0,y=0;}
17     Complex(double _x,double _y) {x = _x,y = _y;}
18 }A[N],B[N];
19 Complex operator + (Complex a,Complex b) {
20     return Complex(a.x+b.x,a.y+b.y);
21 }
22 Complex operator - (Complex a,Complex b) {
23     return Complex(a.x-b.x,a.y-b.y);
24 }
25 Complex operator * (Complex a,Complex b) {
26     return Complex(a.x*b.x-a.y*b.y,a.x*b.y+a.y*b.x);
27 }
28 void FFT(Complex *a,int n,int ty) {
29     for (int i=0,j=0; i<n; ++i) {
30         if (i < j) swap(a[i],a[j]);
31         for (int k=n>>1; (j^=k)<k; k>>=1); //妙啊！！！
32     }
33     for (int m=2; m<=n; m<<=1) {
34         Complex w1 = Complex(cos(2*pi/m),ty*sin(2*pi/m));
35         for (int i=0; i<n; i+=m) {
36             Complex w = Complex(1,0);
37             for (int k=0; k<(m>>1); ++k) {
38                 Complex t = w * a[i+k+(m>>1)];
39                 a[i+k+(m>>1)] = a[i+k] - t;
40                 a[i+k] = a[i+k] + t;
41                 w = w * w1;
42             }
43         }
44     }
45 }
46 int main () {
47     while (scanf("%s%s",a,b)!=EOF) {
48         int len1 = strlen(a),len2 = strlen(b);
49         int n = 1;
50         while (n < (len1+len2)) n <<= 1;
51         for (int i=0; i<n; ++i) {
52             if (i < len1) A[i] = Complex(a[len1-i-1]-‘0‘,0);
53             else A[i] = Complex(0,0);
54             if (i < len2) B[i] = Complex(b[len2-i-1]-‘0‘,0);
55             else B[i] = Complex(0,0);
56         }
57         FFT(A,n,1);
58         FFT(B,n,1);
59         for (int i=0; i<n; ++i) A[i] = A[i] * B[i];
60         FFT(A,n,-1);
61         for (int i=0; i<n; ++i) 
62             ans[i] = (int)(A[i].x/n+0.5);
63         for (int i=0; i<n-1; ++i) {
64             ans[i+1] += (ans[i]/10);
65             ans[i] %= 10;
66         }
67         bool fir = false;
68         for (int i=n-1; i>=0; --i) {
69             if (ans[i]) printf("%d",ans[i]),fir = true;
70             else if (fir || i==0) printf("0");
71         }
72         puts("");
73     }
74     return 0;
75 }