HDU 1402 A * B problem Plus (FFT + large number multiplication)

A * B Problem PlusTime limit:2000/1000 MS (java/others) Memory limit:65536/32768 K (java/others)
Total submission (s): 13456 Accepted Submission (s): 2401


Problem descriptioncalculate A * B.

Inputeach line would contain the integers A and B. Process to end of file.

Note:the length of each integer would not exceed 50000.

outputfor, Output A * B in one line.

Sample Input

1210002

Sample Output

22000

After doing this, I have a little sense of the FFT.

#include <iostream> #include <string> #include <cstdio> #include <cstring> #include <cmath    >using namespace Std;const double Pi=acos ( -1.0); const int Maxn=50010;struct complex{Double A, B; Complex () {} Complex (double _a,double _b): A (_a), B (_b) {} Complex operator + (const Complex &c) {return    Complex (A+C.A,B+C.B);    } Complex Operator-(const Complex &c) {return Complex (A-C.A,B-C.B);    } Complex operator * (const Complex &c) {return Complex (A*C.A-B*C.B,A*C.B+B*C.A);    }}num1[maxn*4],num2[maxn*4];string str1,str2;int cnt,ans[maxn*4];void Ready () {cnt=1;    int len1=str1.size (), len2=str2.size ();    while (Cnt<2*len1 | | cnt<2*len2) cnt<<=1;    for (int i=len1-1,j=0;i>=0;i--, j + +) Num1[j]=complex (str1[i]-' 0 ', 0);    for (int i=len1;i<=cnt;i++) Num1[i]=complex (0,0);    for (int i=len2-1,j=0;i>=0;i--, j + +) Num2[j]=complex (str2[i]-' 0 ', 0); for (int i=len2;i<=cnt;i++) Num2[i]=coMplex (0,0);}        void Change (Complex s[],int len) {for (int i=1,j=len/2;i<len-1;i++) {if (i<j) swap (s[i],s[j]);        int K=LEN/2;            while (j>=k) {j-=k;        k/=2;    } if (j<k) j+=k;    }}void FFT (Complex s[],int len,int on) {change (S,len);        for (int i=2;i<=len;i<<=1) {Complex wn=complex (cos (-on*2*pi/i), sin (-on*2*pi/i));            for (int j=0;j<len;j+=i) {Complex W (1,0);                 for (int k=j;k<j+i/2;k++) {Complex u=s[k];                 Complex T=W*S[K+I/2];                 S[k]=u+t;                 S[k+i/2]=u-t;            W=w*wn;    }}} if (On==-1) {for (int i=0;i<len;i++) S[i].a/=len;    }}void deal () {FFT (num1,cnt,1);    FFT (num2,cnt,1);    for (int i=0;i<cnt;i++) num1[i]=num1[i]*num2[i]; FFT (num1,cnt,-1);}    void Solve () {for (int i=0;i<cnt;i++) ans[i]= (int) (num1[i].a+0.5); for (int i=0;i<cnt;i++) {ANS[I+1]=ANS[I+1]+ANS[I]/10;    ans[i]=ans[i]%10;    } cnt=str1.size () +str2.size ()-1;    while (ans[cnt]==0 && cnt>0) cnt--;    for (int i=cnt;i>=0;i--) Putchar (ans[i]+ ' 0 '); Putchar (' \ n ');}          int main () {while (CIN&GT;&GT;STR1&GT;&GT;STR2) {ready ();          Deal ();    Solve (); } return 0;}

HDU 1402 A * B problem Plus (FFT + large number multiplication)