【HDU】1402 A * B Problem Plus 【FFT】

缺乏、安全感 2022-08-05 04:28 138阅读 0赞

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 ;
    }