【HDU】1402 A * B Problem Plus 【FFT】

传送门：【HDU】1402 A * B Problem Plus

题目分析：

这就是大数乘法题，问两个大数相乘的结果，由于O(n2)的算法复杂度太大，所以我们用FFT来优化他。关于FFT网上资料很多，我就不多说啦。

这是我做的第一道FFT，FFT是看算法导论学来的，前面几篇文章是从july大神那边转载来的，感觉都讲的很不错，简单易懂~

// whn6325689
// Mr.Phoebe
// http://blog.csdn.net/u013007900
#include <algorithm>
#include <iostream>
#include <iomanip>
#include <cstring>
#include <climits>
#include <complex>
#include <fstream>
#include <cassert>
#include <cstdio>
#include <bitset>
#include <vector>
#include <deque>
#include <queue>
#include <stack>
#include <ctime>
#include <set>
#include <map>
#include <cmath>
#include <functional>
#include <numeric>
#pragma comment(linker, "/STACK:1024000000,1024000000")

using namespace std;

#define eps 1e-9
#define PI acos(-1.0)
#define INF 0x3f3f3f3f
#define LLINF 1LL<<62
#define speed std::ios::sync_with_stdio(false);

typedef long long ll;
typedef long double ld;
typedef pair<ll, ll> pll;
typedef complex<ld> point;
typedef pair<int, int> pii;
typedef pair<pii, int> piii;
typedef vector<int> vi;

#define CLR(x,y) memset(x,y,sizeof(x))
#define CPY(x,y) memcpy(x,y,sizeof(x))
#define clr(a,x,size) memset(a,x,sizeof(a[0])*(size))
#define cpy(a,x,size) memcpy(a,x,sizeof(a[0])*(size))

#define mp(x,y) make_pair(x,y)
#define pb(x) push_back(x)
#define lowbit(x) (x&(-x))

#define MID(x,y) (x+((y-x)>>1))
#define ls (idx<<1)
#define rs (idx<<1|1)
#define lson ls,l,mid
#define rson rs,mid+1,r
#define root 1,1,n

template<class T>
inline bool read(T &n)
{
    T x = 0, tmp = 1;
    char c = getchar();
    while((c < '0' || c > '9') && c != '-' && c != EOF) c = getchar();
    if(c == EOF) return false;
    if(c == '-') c = getchar(), tmp = -1;
    while(c >= '0' && c <= '9') x *= 10, x += (c - '0'),c = getchar();
    n = x*tmp;
    return true;
}
template <class T>
inline void write(T n)
{
    if(n < 0)
    {
        putchar('-');
        n = -n;
    }
    int len = 0,data[20];
    while(n)
    {
        data[len++] = n%10;
        n /= 10;
    }
    if(!len) data[len++] = 0;
    while(len--) putchar(data[len]+48);
}
//-----------------------------------

const int MAXN=200010;

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)//rev=-1表示逆变换
{
    for(int i=1,j,k,t; i<n; i++)  //进行蝶型变换
    {
        for(j=0,k=n>>1,t=i; 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=Complex(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;
}

char s1[MAXN],s2[MAXN];
Complex x1[MAXN],x2[MAXN];
int num[MAXN];


int main()
{
    while(~scanf("%s%s",s1,s2))
    {
        int n1=strlen(s1);
        int n2=strlen(s2);
        int n=1;
        while(n<n1+n2) n<<=1;       //进行FFT的级数大小
        for(int i=0; i<n1; i++) x1[i]=Complex(s1[n1-i-1]-'0',0); //初始化数组
        for(int i=n1; i<n; i++) x1[i]=Complex(0,0);
        for(int i=0 ; i<n2 ; i++) x2[i]=Complex(s2[n2-i-1]-'0',0) ;
        for(int i=n2; i<n; i++) x2[i]=Complex(0,0);
        FFT(x1,n,1);
        FFT(x2,n,1);
        for(int i=0; i<n; i++) x1[i]=x1[i]*x2[i];
        FFT(x1,n,-1);
        int t=0;
        for(int i=0; i<n; i++,t/=10)
        {
            t+=(int)(x1[i].r+0.1);
            num[i]=t%10;
        }
        for(; t; t/=10) num[n++]=t%10;
        while(n>1 && !num[n-1]) --n;
        for(int i=n-1; i>=0; i--) printf("%d",num[i]);
        printf("\n");
    }
    return 0 ;
}