1 /* 2 Welcome Hacking 3 Wish You High Rating 4 */ 5 #include<iostream> 6 #include<cstdio> 7 #include<cstring> 8 #include<ctime> 9 #include<cstdlib> 10 #include<algorithm> 11 #include<cmath> 12 #include<string> 13 #include<complex> 14 using namespace std; 15 int read(){ 16 int xx=0,ff=1;char ch=getchar(); 17 while(ch>'9'||ch<'0'){if(ch=='-')ff=-1;ch=getchar();} 18 while(ch>='0'&&ch<='9'){xx=(xx<<3)+(xx<<1)+ch-'0';ch=getchar();} 19 return xx*ff; 20 } 21 const int maxn=(1<<18)+1; 22 const double PI=acos(-1.0); 23 typedef complex<double> C; 24 int N,M,L,R[maxn]; 25 C a[maxn],b[maxn]; 26 void FFT(C a[],int arg){ 27 for(int i=0;i<N;i++) 28 if(i<R[i]) 29 swap(a[i],a[R[i]]); 30 for(int i=1;i<N;i<<=1){ 31 C wn(cos(PI/i),arg*sin(PI/i)); 32 for(int p=i<<1,j=0;j<N;j+=p){ 33 C w(1,0); 34 for(int k=0;k<i;k++,w*=wn){ 35 C x=a[j+k],y=w*a[j+k+i]; 36 a[j+k]=x+y,a[j+k+i]=x-y; 37 } 38 } 39 } 40 } 41 int main(){ 42 //freopen("in","r",stdin); 43 N=read(),M=read(); 44 for(int i=0;i<=N;i++) 45 a[i]=read(); 46 for(int i=0;i<=M;i++) 47 b[i]=read(); 48 M+=N; 49 for(N=1;N<=M;N<<=1) 50 L++; 51 for(int i=0;i<N;i++) 52 R[i]=(R[i>>1]>>1)|((i&1)<<(L-1)); 53 FFT(a,1);FFT(b,1); 54 for(int i=0;i<=N;i++) 55 a[i]*=b[i]; 56 FFT(a,-1); 57 for(int i=0;i<=M;i++) 58 printf("%d ",(int)(a[i].real()/N+0.5)); 59 return 0; 60 }
FFT还有蝶形数组那部分不懂，emmmm，先这样吧，抄一个黄学长的模板，体验acmer用模板切题的快感。
hdu1402
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.
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
直接上模板改一改，注意清除前导0和进位（错了一次，逃）
1 /* 2 Welcome Hacking 3 Wish You High Rating 4 */ 5 #include<iostream> 6 #include<cstdio> 7 #include<cstring> 8 #include<ctime> 9 #include<cstdlib> 10 #include<algorithm> 11 #include<cmath> 12 #include<string> 13 #include<complex> 14 using namespace std; 15 int read(){ 16 int xx=0,ff=1;char ch=getchar(); 17 while(ch>'9'||ch<'0'){if(ch=='-')ff=-1;ch=getchar();} 18 while(ch>='0'&&ch<='9'){xx=(xx<<3)+(xx<<1)+ch-'0';ch=getchar();} 19 return xx*ff; 20 } 21 const int maxn=(1<<18)+1; 22 const double PI=acos(-1.0); 23 typedef complex<double> C; 24 int N,M,L,R[maxn]; 25 C a[maxn],b[maxn]; 26 void FFT(C a[],int arg){ 27 for(int i=0;i<N;i++) 28 if(i<R[i]) 29 swap(a[i],a[R[i]]); 30 for(int i=1;i<N;i<<=1){ 31 C wn(cos(PI/i),arg*sin(PI/i)); 32 for(int p=i<<1,j=0;j<N;j+=p){ 33 C w(1,0); 34 for(int k=0;k<i;k++,w*=wn){ 35 C x=a[j+k],y=w*a[j+k+i]; 36 a[j+k]=x+y,a[j+k+i]=x-y; 37 } 38 } 39 } 40 } 41 char s1[50010],s2[50010]; 42 int ans[100010]; 43 int main(){ 44 //freopen("in","r",stdin); 45 /*N=read(),M=read(); 46 for(int i=0;i<=N;i++) 47 a[i]=read(); 48 for(int i=0;i<=M;i++) 49 b[i]=read();*/ 50 while(scanf("%s",s1)!=EOF){ 51 scanf("%s",s2); 52 N=strlen(s1)-1,M=strlen(s2)-1; 53 for(int i=0;i<=N;i++) 54 a[i]=s1[N-i]-'0'; 55 for(int i=0;i<=M;i++) 56 b[i]=s2[M-i]-'0'; 57 M+=N; 58 for(N=1;N<=M;N<<=1) 59 L++; 60 for(int i=0;i<N;i++) 61 R[i]=(R[i>>1]>>1)|((i&1)<<(L-1)); 62 FFT(a,1);FFT(b,1); 63 for(int i=0;i<=N;i++) 64 a[i]*=b[i]; 65 FFT(a,-1); 66 for(int i=M;i>=0;i--) 67 ans[i]=(int)(a[i].real()/N+0.5); 68 int k; 69 for(k=0;k<=M||ans[k];k++) 70 if(ans[k]>=10) 71 ans[k+1]+=ans[k]/10,ans[k]%=10; 72 while(!ans[k]&&k>0) 73 k--; 74 for(int i=k;i>=0;i--) 75 printf("%d",ans[i]); 76 puts(""); 77 memset(a,0,sizeof(a)); 78 memset(b,0,sizeof(b)); 79 memset(R,0,sizeof(R)); 80 memset(ans,0,sizeof(ans)); 81 L=0; 82 } 83 return 0; 84 }