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.
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 }
