UOJ #34. 多项式乘法

#34. 多项式乘法

这是一道模板题。

给你两个多项式，请输出乘起来后的多项式。

输入格式

第一行两个整数 nn 和 mm，分别表示两个多项式的次数。

第二行 n+1n+1 个整数，表示第一个多项式的 00 到 nn 次项系数。

第三行 m+1m+1 个整数，表示第二个多项式的 00 到 mm 次项系数。

输出格式

一行 n+m+1n+m+1 个整数，表示乘起来后的多项式的 00 到 n+mn+m 次项系数。

样例一

input

1 2
1 2
1 2 1

output

1 4 5 2

explanation

(1+2x)⋅(1+2x+x2)=1+4x+5x2+2x3(1+2x)⋅(1+2x+x2)=1+4x+5x2+2x3。

限制与约定

0≤n,m≤1050≤n,m≤105，保证输入中的系数大于等于 00 且小于等于 99。

时间限制：1s1s

空间限制：256MB

 

分析

FFT/NTT模板题。

 

code

递归：

 1 #include<cstdio>
 2 #include<algorithm>
 3 #include<cstring>
 4 #include<cmath>
 5 #include<iostream>
 6 
 7 using namespace std;
 8 const int N = 300100;
 9 const double eps = 1e-8;
10 const double pi = acos(-1.0);
11 typedef long long LL;
12 
13 struct Complex {
14     double x,y;
15     Complex() {x=0,y=0;}
16     Complex(double xx,double yy) {x=xx,y=yy;}
17 
18 }A[N],B[N];
19 
20 Complex operator + (Complex a,Complex b) {
21     return Complex(a.x+b.x,a.y+b.y);
22 }
23 Complex operator - (Complex a,Complex b) { 
24     return Complex(a.x-b.x,a.y-b.y);
25 }
26 Complex operator * (Complex a,Complex b) {
27     return Complex(a.x*b.x-a.y*b.y,a.x*b.y+b.x*a.y);
28 }
29 
30 void FFT(Complex *a,int n,int ty) {
31     if (n==1) return ;
32     Complex a1[n>>1],a2[n>>1];
33     for (int i=0; i<=n; i+=2) {
34         a1[i>>1] = a[i],a2[i>>1] = a[i+1];
35     }
36     FFT(a1,n>>1,ty);
37     FFT(a2,n>>1,ty);
38     Complex w1 = Complex(cos(2.0*pi/n),ty*sin(2.0*pi/n));
39     Complex w = Complex(1.0,0.0);
40     for (int i=0; i<(n>>1); i++) {
41         Complex t = w * a2[i];
42         a[i+(n>>1)] = a1[i] - t;
43         a[i] = a1[i] + t;
44         w = w * w1;
45     }
46 }
47 int main() {
48     int n,m;
49     scanf("%d%d",&n,&m);
50     for (int i=0; i<=n; ++i) scanf("%lf",&A[i].x);
51     for (int i=0; i<=m; ++i) scanf("%lf",&B[i].x);
52     int fn = 1;
53     while (fn <= n+m) fn <<= 1;
54     FFT(A,fn,1);
55     FFT(B,fn,1);
56     for (int i=0; i<=fn; ++i) 
57         A[i] = A[i] * B[i];
58     FFT(A,fn,-1);
59     for (int i=0; i<=n+m; ++i) 
60         printf("%d ",(int)(A[i].x/fn+0.5));
61     return 0;
62 }
View Code

 

非递归

 1 #include<cstdio>
 2 #include<algorithm>
 3 #include<cstring>
 4 #include<cmath>
 5 #include<iostream>
 6 
 7 using namespace std;
 8 const int N = 300100;
 9 const double eps = 1e-8;
10 const double Pi = acos(-1.0);
11 typedef long long LL;
12 
13 struct Complex {
14     double x,y;
15     Complex() {x=0,y=0;}
16     Complex(double xx,double yy) {x=xx,y=yy;}
17 
18 }A[N],B[N];
19 
20 Complex operator + (Complex a,Complex b) {
21     return Complex(a.x+b.x,a.y+b.y);
22 }
23 Complex operator - (Complex a,Complex b) { 
24     return Complex(a.x-b.x,a.y-b.y);
25 }
26 Complex operator * (Complex a,Complex b) {
27     return Complex(a.x*b.x-a.y*b.y,a.x*b.y+b.x*a.y);
28 }
29 void FFT(Complex *a,int n,int ty) {
30     // 按递归时最底层的顺序翻转 
31     for (int i=0,j=0; i<n; ++i) {
32         if (i < j) swap(a[i],a[j]);
33         for (int k=n>>1; (j^=k)<k; k>>=1);
34     }
35     // 当前正在求次数界为m的多项式。m=1的已经在上面求出来了，所以现在在求次数界为2的多项式。
36     for (int m=2; m<=n; m<<=1) {  
37         Complex w1 = Complex(cos(2*Pi/m),ty*sin(2*Pi/m));
38         for (int i=0; i<n; i+=m) { // 当前求的多项式下标为[i,i+m-1] 
39             Complex w = Complex(1,0); 
40             for (int k=0; k<(m>>1); ++k) { // 由[i,i+(m/2)-1]和[i+(m/2),i+m-1] 求出[i,i+m-1]的多项式 
41                 Complex t = w * a[i+k+(m>>1)];
42                 Complex u = a[i+k];
43                 a[i+k] = u + t;
44                 a[i+k+(m>>1)] = u - t;
45                 w = w * w1;
46             }
47         }
48     }
49 }
50 int main() {
51     int n,m;
52     scanf("%d%d",&n,&m);
53     for (int i=0; i<=n; ++i) scanf("%lf",&A[i].x);
54     for (int i=0; i<=m; ++i) scanf("%lf",&B[i].x);
55     int fn = 1;
56     while (fn <= n+m) fn <<= 1;
57     FFT(A,fn,1);
58     FFT(B,fn,1);
59     for (int i=0; i<=fn; ++i) 
60         A[i] = A[i] * B[i];
61     FFT(A,fn,-1);
62     for (int i=0; i<=n+m; ++i) 
63         printf("%d ",(int)(A[i].x/fn+0.5));
64     return 0;
65 }
View Code

 

NTT

注意代码中的longlong，取模。

 1 #include<cstdio>
 2 #include<algorithm>
 3 #include<cstring>
 4 #include<cmath>
 5 #include<iostream>
 6 
 7 using namespace std;
 8 
 9 typedef long long LL;
10 const int N = 300100;
11 const int P = 998244353;
12 int A[N],B[N];
13 
14 int ksm(int a,int b) {
15     int ans = 1;
16     while (b) {
17         if (b & 1) ans = (1ll * ans * a) % P;
18         a = (1ll * a * a) % P;
19         b >>= 1;
20     }
21     return ans % P;
22 }
23 void NTT(int *a,int n,int ty) {
24     // 按递归时最底层的顺序翻转 
25     for (int i=0,j=0; i<n; ++i) {
26         if (i < j) swap(a[i],a[j]);
27         for (int k=n>>1; (j^=k)<k; k>>=1);
28     }
29     // 当前正在求次数界为m的多项式。m=1的已经在上面求出来了，所以现在在求次数界为2的多项式。
30     for (int m=2; m<=n; m<<=1) {  
31         int w1 = ksm(3,(P-1)/m);
32         if (ty == -1) w1 = ksm(w1,P-2);
33         for (int i=0; i<n; i+=m) { // 当前求的多项式下标为[i,i+m-1] 
34             int w = 1; 
35             for (int k=0; k<(m>>1); ++k) { // 由[i,i+(m/2)-1]和[i+(m/2),i+m-1] 求出[i,i+m-1]的多项式 
36                 int t = 1ll * w * a[i+k+(m>>1)] % P;
37                 int u = a[i+k];
38                 a[i+k] = (u + t) % P;
39                 a[i+k+(m>>1)] = (u - t + P) % P;
40                 w = 1ll * w * w1 % P;
41             }
42         }
43     }
44 }
45 int main() {
46     int n,m;
47     scanf("%d%d",&n,&m);
48     for (int i=0; i<=n; ++i) scanf("%d",&A[i]),A[i] = (A[i] + P) % P;
49     for (int i=0; i<=m; ++i) scanf("%d",&B[i]),B[i] = (B[i] + P) % P;
50     int len = 1;
51     while (len <= n+m) len <<= 1;
52     NTT(A,len,1);
53     NTT(B,len,1);
54     for (int i=0; i<len; ++i) A[i] = 1ll * A[i] * B[i] % P;
55     NTT(A,len,-1);
56     int inv = ksm(len,P-2);
57     for (int i=0; i<len; ++i) A[i] = 1ll * A[i] * inv % P;
58     for (int i=0; i<=n+m; ++i) printf("%d\n",A[i]);
59     return 0;
60 }
View Code

 

 

posted @ 2018-02-19 18:57  MJT12044  阅读(243)  评论(0)  编辑  收藏  举报