HDU 1402 A * B Problem Plus FFT

Posted 2020-09-26 Meek

A * B Problem Plus

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

 

 

题解：

　　FFT入门

　　注意几组数据

　　0 0

　　0 5

       0005 000006

#include<bits/stdc++.h>
using namespace std;
#pragma comment(linker, "/STACK:102400000,102400000")
#define ls i<<1
#define rs ls | 1
#define mid ((ll+rr)>>1)
#define pii pair<int,int>
#define MP make_pair
typedef long long LL;
const long long INF = 1e18+1LL;
const double pi = acos(-1.0);
const int N = 1e5+10, M = 1e3+20,inf = 2e9,mod = 1e9+7;


struct Complex {
    double r , i ;
    Complex () {}
    Complex ( double r , double i ) : r ( r ) , i ( i ) {}
    Complex operator + ( const Complex& t ) const {
        return Complex ( r + t.r , i + t.i ) ;
    }
    Complex operator - ( const Complex& t ) const {
        return Complex ( r - t.r , i - t.i ) ;
    }
    Complex operator * ( const Complex& t ) const {
        return Complex ( r * t.r - i * t.i , r * t.i + i * t.r ) ;
    }
} ;

void FFT ( Complex y[] , int n , int rev ) {
    for ( int i = 1 , j , t , k ; i < n ; ++ i ) {
        for ( j = 0 , t = i , k = n >> 1 ; k ; k >>= 1 , t >>= 1 ) j = j << 1 | t & 1 ;
        if ( i < j ) swap ( y[i] , y[j] ) ;
    }
    for ( int s = 2 , ds = 1 ; s <= n ; ds = s , s <<= 1 ) {
        Complex wn = Complex ( cos ( rev * 2 * pi / s ) , sin ( rev * 2 * pi / s ) ) , w ( 1 , 0 ) , t ;
        for ( int k = 0 ; k < ds ; ++ k , w = w * wn ) {
            for ( int i = k ; i < n ; i += s ) {
                y[i + ds] = y[i] - ( t = w * y[i + ds] ) ;
                y[i] = y[i] + t ;
            }
        }
    }
    if ( rev == -1 ) for ( int i = 0 ; i < n ; ++ i ) y[i].r /= n ;
}

Complex s[N*4],t[N*4];
char a[N],b[N];
int ans[N];
int main() {
    while(scanf("%s%s",a,b)!=EOF) {
        int n = strlen(a);
        int m = strlen(b);
        int flag2 = 0, flag1 = 0;
        for(int i = 0; i < n; ++i) {
            if(a[i] == ‘0‘ && !flag1) {
                continue;
            }
            else a[flag1++] = a[i];
        }
        for(int i = 0; i < m; ++i) {
            if(b[i] == ‘0‘ && !flag2) {
                continue;
            }
            else b[flag2++] = b[i];
        }
        n = flag1;
        m = flag2;
       // cout<<n<<" "<<m<<endl;
       if(n == 0 || m == 0) {
        puts("0");
        continue;
       }
        int n1 = 1;
        while(n1 <= m+n-2) n1<<=1;
        for(int i = 0; i < n; ++i) {
            s[i] = Complex(a[i] - ‘0‘,0);
        }
        for(int i = n; i < n1; ++i) {
            s[i] = Complex(0,0);
        }
        for(int i = 0; i < m; ++i) {
            t[i] = Complex(b[i] - ‘0‘,0);
        }
        for(int i = m; i < n1; ++i) {
            t[i] = Complex(0,0);
        }
        FFT(s,n1,1);
        FFT(t,n1,1);
        for(int i = 0; i < n1; ++i) s[i] = s[i]*t[i];
        FFT(s,n1,-1);
        int f = 1;
        int last = 0,cnt = 0;
        for(int i = n+m-2; i >= 0; --i) {
            int x = (int) (s[i].r+0.1) + last;
           //printf("%d\n",x);
            last = 0;
            if(x > 9) {
                last+=x/10;
                ans[++cnt] = x % 10;
            }
            else ans[++cnt] = x;
        }
        if(last) {
            ans[++cnt] = last;
        }
        for(int i = cnt; i >= 1; --i) printf("%d",ans[i]);
        printf("\n");
    }
    return 0;
}