HDU 1402 A * B Problem Plus 快速傅里叶变换 FFT

Posted 2020-11-10 鲸头鹳

http://acm.hdu.edu.cn/showproblem.php?pid=1402

快速傅里叶变换优化的高精度乘法。

https://blog.csdn.net/ggn_2015/article/details/68922404 这个写的很详细了。

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<algorithm>
 4 #include<cmath>
 5 #include<iostream>
 6 #include<vector>
 7 #include<complex>
 8 using namespace std;
 9 #define LL long long
10 const int maxn=150010;
11 typedef complex< double >cd;
12 char s1[maxn]={},s2[maxn]={};
13 int ans[maxn]={};
14 int rev[maxn]={}; int s,bit;
15 cd a[maxn]={},b[maxn]={};
16 double Pi;
17 inline void getrev(){ for(int i=0;i<s;i++)rev[i]=(rev[i>>1]>>1)|((i&1)<<(bit-1)); }
18 inline void fft(cd *c,int n,int dft){
19     for(int i=1;i<=s;i++)if(rev[i]>i)swap(c[i],c[rev[i]]);
20     for(int step=1;step<n;step<<=1){
21         cd shu=exp(cd(0,dft*Pi/step));
22         for(int i=0;i<n;i+=step<<1){
23             cd z=cd(1,0); 
24             for(int j=i;j<i+step;j++){
25                 cd x=c[j];cd y=c[j+step]*z;
26                 c[j]=x+y; c[j+step]=x-y;
27                 z*=shu;
28             }
29         }
30     }
31     if(dft==-1)for(int i=0;i<n;i++)c[i]/=n;
32 }
33 int main(){
34     cd cle(0,0);Pi=2.0*acos(0.0);
35     while(~scanf("%s%s",s1,s2)){
36         memset(rev,0,sizeof(rev));
37         int l1=strlen(s1),l2=strlen(s2),n=l1+l2-1;
38         bit=1;s=2; for(;s<n;++bit)s<<=1;
39         getrev();
40         for(int i=1;i<=s;i++){a[i]=cle;b[i]=cle;}
41         for(int i=0;i<l1;i++)a[i]=(double)(s1[l1-i-1]-‘0‘);
42         for(int i=0;i<l2;i++)b[i]=(double)(s2[l2-i-1]-‘0‘);
43         fft(a,s,1);fft(b,s,1);
44         for(int i=0;i<s;i++)a[i]*=b[i];
45         fft(a,s,-1);
46         memset(ans,0,sizeof(ans));
47         for(int i=0;i<s;i++){
48             ans[i]+=(int)(a[i].real()+0.5);
49             ans[i+1]+=ans[i]/10;
50             ans[i]%=10;
51         }
52         int i;
53         for(i=l1+l2;!ans[i]&&i>0;--i);
54         if(i==0)printf("%d\n",ans[0]);
55         else{
56             for(;i>=0;--i)printf("%d",ans[i]);
57             printf("\n");
58         }
59     }
60     return 0;
61 }

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