HDU 1402 A*B(FFT 模板题)

大整数乘法
将每一位数字看做系数，作FFT,变为点值表达，然后再逆变换回去，这时每个系数就是对应结果的十进制位的数，把它变为十进制整数即可

#include 
#include 
#include 
#include
#include
#include
#include
#define maxn 140000
using namespace std;
typedef long long LL;
typedef complex<double> CLDD;

const double PI = acos(-1);

 CLDD a[maxn],b[maxn],A[maxn],B[maxn];
 int ans[maxn];

void strToInt(char * str,int len,CLDD * x)
{
    int i = 0;
    len--;
    while(len>=0){
        x[i] = CLDD(str[len]-'0',0);
        i++;len--;
    }
}
char str[50010];

int rev(int id,int len)
{
    int res = 0;
    for(int i=0; (1<1;
        if(id&(1<1;
    }
    return res;
}

void FFT(CLDD * c,CLDD* ans,int len,int DFT)
{
    for(int i=0 ; ifor(int s = 1 ; (1<int m = (1<cos(2*PI*DFT/m),sin(2*PI*DFT/m));
        for(int k=0 ; k1,0);
            for(int j = 0 ; j<(m>>1) ; ++j)
            {
                CLDD u = ans[k+j];
                CLDD t = w*ans[k+j+(m>>1)];
                ans[k+j] = u+t;
                ans[k+j+(m>>1)] = u-t;
                w*=wm;
            }
        }
    }
     if(DFT==-1)for(int i=0 ; iint main()
{
    //freopen("H:\\c++\\file\\stdin.txt","r",stdin);
   // int T;scanf("%d",&T);
    int sa = 0,sb =0;
    while(scanf("%s",str)!=EOF)
    {
        sa = 0,sb = 0;
        memset(a,0,sizeof(a));
        memset(b,0,sizeof(b));
        memset(A,0,sizeof(A));
        memset(B,0,sizeof(B));
        int len = strlen(str);
        strToInt(str,len,a);
        while((1<scanf("%s",str);
        len = strlen(str);
        strToInt(str,len,b);
        while((1<1<<(max(sb,sa)+1);
        FFT(a,A,len,1);

        FFT(b,B,len,1);
        for(int i = 0 ; i1);
        for(int i=0 ; i// cout<<"ans["<
            ans[i] = (int)(B[i].real()+0.2);
        }
        for(int i=0 ; i1 ; ++i)
        {
            ans[i+1]+=ans[i]/10;
            ans[i]%=10;
        }
        int start = len-1;
        while(start >0 && ans[start]==0)start--;
        for(int i = start; i>=0 ; --i)
            printf("%d",ans[i]);

        putchar('\n');
    }
    return 0 ;
}