A * B Problem Plus(fft)

题目连接：http://acm.hdu.edu.cn/showproblem.php?pid=1402

hdu_1402:A * B Problem Plus

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 15419    Accepted Submission(s):
3047

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.

 

Output

For each case, output A * B in one line.

 

Sample Input

1
2
1000
2

 

Sample Output

2
2000

 

Author

DOOM III

 

题解： 练习用fft实现大数的乘法达到O(nlog(n))的算法,两个数相乘看成是两个多项式的乘法，这样多项式中的x=10,fft套用模板即可

给出代码：

 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #include<cmath>
 using namespace std;
 const int MAX=;
 const double PI=acos(-1.0),eps=1e-;;
 double cof1[MAX], cof2[MAX];
 int n, k, permutation[MAX];
 char s1[MAX],s2[MAX];
 int ans[MAX];
 struct complex {//复数
     double r, v;
     complex operator + (complex& obj) {
         complex temp;
         temp.r = r + obj.r;
         temp.v = v + obj.v;
         return temp;
     }
     complex operator - (complex& obj) {
         complex temp;
         temp.r = r - obj.r;
         temp.v= v - obj.v;
         return temp;
     }
     complex operator * ( complex& obj) {
         complex temp;
         temp.r = r*obj.r - v*obj.v;
         temp.v = r*obj.v + v*obj.r;
         return temp;
     }
 } p1[MAX], p2[MAX], omiga[MAX], result1[MAX], result2[MAX];
 void caculate_permutation(int s, int interval, int w, int next) {
     if(interval==n) {
         permutation[w] = s;
         return ;
     }
     caculate_permutation(s,interval*, w, next/);
     caculate_permutation(s+interval, interval*, w+next, next/);
 }
 void fft(complex transform[], complex p[]) {
     int i, j, l, num, m;
     complex temp1, temp2;
     for(i=; i<n; i++)transform[i] = p[ permutation[i] ] ;
     num = , m = n;
     for(i=; i<=k; i++) {
         for(j=; j<n; j+=num*)
             for(l=; l<num; l++)
                 temp2 = omiga[m*l]*transform[j+l+num],
                         temp1 = transform[j+l],
                                 transform[j+l] = temp1 + temp2,
                                                  transform[j+l+num] = temp1 - temp2;
         num*=,m/=;
     }
 }
 void polynomial_by(int n1,int n2) {//多项式乘法，cof1、cof2保存的是a[0],a[1]..a[n-1]的值（a[i]*x^i）
     int i;
     double angle;
     k = , n = ;
     while(n<n1+n2-)k++,n*=;
     for(i=; i<n1; i++)p1[i].r = cof1[i], p1[i].v = ;
     while(i<n)p1[i].r = p1[i].v = , i++;
     for(i=; i<n2; i++)p2[i].r = cof2[i], p2[i].v = ;
     while(i<n)p2[i].r = p2[i].v = , i++;
     caculate_permutation(,,,n/);
     angle = PI/n;
     for(i=; i<n; i++)omiga[i].r = cos(angle*i), omiga[i].v = sin(angle*i);
     fft(result1,p1);
     fft(result2,p2);
     for(i=; i<n; i++)result1[i]= result1[i]*result2[i];
     for(i=; i<n; i++)omiga[i].v = -omiga[i].v;
     fft(result2, result1);
     for(i=; i<n; i++)result2[i].r/=n;
     i = n -;
     while(i&&fabs(result2[i].r)<eps)i--;
     n = i+;
     while(i>=) ans[i]=(int)(result2[i].r+0.5), i--;
 }
 int main() {
     while(scanf("%s",s1)!=EOF){
         scanf("%s",s2);
         int n1=strlen(s1),n2=strlen(s2);
         for(int i=;i<n1;i++){
             cof1[i]=s1[n1--i]-'';
         }
         for(int i=;i<n2;i++){
             cof2[i]=s2[n2--i]-'';
         }
         memset(ans,,sizeof(ans));
         polynomial_by(n1,n2);
         for(int i=;i<n;i++){
             if(ans[i]>=){
                 ans[i+]+=ans[i]/;
                 ans[i]%=;
             }
         }
         while(ans[n]>){
             if(ans[n]>=){
                 ans[n+]+=ans[n]/;
                 ans[n]%=;
             }
             n++;
         }
         for(int i=n-;i>=;i--){
             putchar(''+ans[i]);
         }
         puts("");
     }
     return ;
 }