【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,ntemplate<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 ;}