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uoj#34. 多项式乘法

时间:2017-04-05 20:12:47      阅读:249      评论:0      收藏:0      [点我收藏+]

标签:namespace   ret   work   ring   lib   cst   amp   div   line   

题面:http://uoj.ac/problem/34

 

正解:$FFT$/$NTT$。

http://blog.xlightgod.com/%E3%80%90uoj34%E3%80%91%E5%A4%9A%E9%A1%B9%E5%BC%8F%E4%B9%98%E6%B3%95/

详见xlightgod学长博客。

 

FFT:

 1 //It is made by wfj_2048~
 2 #include <algorithm>
 3 #include <iostream>
 4 #include <complex>
 5 #include <cstring>
 6 #include <cstdlib>
 7 #include <cstdio>
 8 #include <vector>
 9 #include <cmath>
10 #include <queue>
11 #include <stack>
12 #include <map>
13 #include <set>
14 #define inf (1<<30)
15 #define pi acos(-1)
16 #define N (600010)
17 #define il inline
18 #define RG register
19 #define ll long long
20 #define C complex<double>
21 #define File(s) freopen(s".in","r",stdin),freopen(s".out","w",stdout)
22 
23 using namespace std;
24 
25 int rev[N],n,m,lg;
26 char sa[N],sb[N];
27 C a[N],b[N];
28 
29 il int gi(){
30     RG int x=0,q=1; RG char ch=getchar();
31     while ((ch<0 || ch>9) && ch!=-) ch=getchar();
32     if (ch==-) q=-1,ch=getchar();
33     while (ch>=0 && ch<=9) x=x*10+ch-48,ch=getchar();
34     return q*x;
35 }
36 
37 il void fft(C *a,int n,int f){
38     for (RG int i=0;i<n;++i) if (i<rev[i]) swap(a[i],a[rev[i]]);
39     for (RG int i=1;i<n;i<<=1){
40     C wn(cos(pi/i),sin(f*pi/i)),x,y;
41     for (RG int j=0;j<n;j+=(i<<1)){
42         C w(1,0);
43         for (RG int k=0;k<i;++k,w*=wn){
44         x=a[j+k],y=w*a[j+k+i];
45         a[j+k]=x+y,a[j+k+i]=x-y;
46         }
47     }
48     }
49 }
50 
51 il void work(){
52     n=gi()+1,m=gi()+1; for (RG int i=0;i<n;++i) a[i]=gi();
53     for (RG int i=0;i<m;++i) b[i]=gi(); m+=n; for (n=1;n<=m;n<<=1) lg++;
54     for (RG int i=1;i<=n;++i) rev[i]=(rev[i>>1]>>1)|((i&1)<<(lg-1));
55     fft(a,n,1),fft(b,n,1); for (RG int i=0;i<n;++i) a[i]*=b[i]; fft(a,n,-1);
56     for (RG int i=0;i<m-1;++i) printf("%d ",(int)(a[i].real()/n+0.5)); return;
57 }
58 
59 int main(){
60     File("fft");
61     work();
62     return 0;
63 }

 

 

NTT:

 1 //It is made by wfj_2048~
 2 #include <algorithm>
 3 #include <iostream>
 4 #include <complex>
 5 #include <cstring>
 6 #include <cstdlib>
 7 #include <cstdio>
 8 #include <vector>
 9 #include <cmath>
10 #include <queue>
11 #include <stack>
12 #include <map>
13 #include <set>
14 #define rhl (998244353)
15 #define inf (1<<30)
16 #define N (300010)
17 #define G (3)
18 #define il inline
19 #define RG register
20 #define ll long long
21 #define File(s) freopen(s".in","r",stdin),freopen(s".out","w",stdout)
22 
23 using namespace std;
24 
25 int rev[N],n,m,lg;
26 ll a[N],b[N];
27 
28 il int gi(){
29     RG int x=0,q=1; RG char ch=getchar();
30     while ((ch<0 || ch>9) && ch!=-) ch=getchar();
31     if (ch==-) q=-1,ch=getchar();
32     while (ch>=0 && ch<=9) x=x*10+ch-48,ch=getchar();
33     return q*x;
34 }
35 
36 il ll qpow(RG ll a,RG ll b){
37     RG ll ans=1;
38     while (b){
39     if (b&1) (ans*=a)%=rhl;
40     (a*=a)%=rhl,b>>=1;
41     }
42     return ans;
43 }
44 
45 il void NTT(ll *a,RG int n,RG int f){
46     for (RG int i=0;i<n;++i) if (i<rev[i]) swap(a[i],a[rev[i]]);
47     for (RG int i=1;i<n;i<<=1){
48     RG ll gn=qpow(G,(rhl-1)/(i<<1)),x,y;
49     for (RG int j=0;j<n;j+=(i<<1)){
50         RG ll g=1;
51         for (RG int k=0;k<i;++k,(g*=gn)%=rhl){
52         x=a[j+k],y=g*a[j+k+i]%rhl;
53         a[j+k]=(x+y)%rhl,a[j+k+i]=(x-y+rhl)%rhl;
54         }
55     }
56     }
57     if (f==1) return; reverse(a+1,a+n); RG ll inv=qpow(n,rhl-2);
58     for (RG int i=0;i<n;++i) (a[i]*=inv)%=rhl; return;
59 }
60 
61 il void work(){
62     n=gi()+1,m=gi()+1;
63     for (RG int i=0;i<n;++i) a[i]=gi();
64     for (RG int i=0;i<m;++i) b[i]=gi();
65     m+=n; for (n=1;n<=m;n<<=1) lg++;
66     for (RG int i=0;i<n;++i) rev[i]=(rev[i>>1]>>1)|((i&1)<<(lg-1));
67     NTT(a,n,1),NTT(b,n,1); for (RG int i=0;i<n;++i) (a[i]*=b[i])%=rhl;
68     NTT(a,n,-1); for (RG int i=0;i<m-1;++i) printf("%lld ",a[i]); return;
69 }
70 
71 int main(){
72     File("NTT");
73     work();
74     return 0;
75 }

 

uoj#34. 多项式乘法

标签:namespace   ret   work   ring   lib   cst   amp   div   line   

原文地址:http://www.cnblogs.com/wfj2048/p/6420629.html

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