hdu1402A * B Problem Plus

链接：http://acm.hdu.edu.cn/showproblem.php?pid=1402

题意：计算A*B。

分析：由于A，B很大，长度可能达50000，所以一般的高精度乘法是不行的。所以要用FFT，这个叫快速傅里叶变换。算法导论第30章有讲，讲得挺清晰的，还有就是picks的博客也不错：http://picks.logdown.com/posts/177631-fast-fourier-transform。O(nlogn)

代码：

#include<map>
#include<set>
#include<cmath>
#include<queue>
#include<bitset>
#include<math.h>
#include<cstdio>
#include<vector>
#include<string>
#include<cstring>
#include<iostream>
#include<algorithm>
#pragma comment(linker, "/STACK:102400000,102400000")
using namespace std;
const int N=50010;
const int MOD1=1000007;
const int MOD2=100000009;
const int MAX=1000000000;
const double EPS=0.00000001;
typedef long long ll;
const ll MOD=1000000007;
const ll INF=10000000010;
const double pi=acos(-1.0);
typedef double db;
typedef unsigned long long ull;
char s[N],t[N];
struct Complex{
    db r,i;
    Complex() {}
    Complex(db r,db 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);
    }
}a[3*N],b[3*N];
void FFT(Complex x[],int n,int rev) {
    int i,j,k,t,ds;
    Complex w,u,wn;
    for (i=1;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(x[i],x[j]);
    }
    for (i=2,ds=1;i<=n;ds=i,i<<=1) {
        w=Complex(1,0);wn=Complex(cos(rev*2*pi/i),sin(rev*2*pi/i));
        for (j=0;j<ds;j++,w=w*wn)
            for (k=j;k<n;k+=i) {
                u=w*x[k+ds];x[k+ds]=x[k]-u;x[k]=x[k]+u;
            }
    }
    if (rev==-1) for (i=0;i<n;i++) x[i].r/=n;
}
int num[3*N];
void solve() {
    int i,k=0,n=1,les,let;
    les=strlen(s);
    let=strlen(t);
    while (n<les+let) n<<=1;
    for (i=0;i<les;i++) a[i]=Complex(s[les-i-1]-'0',0);
    for (i=les;i<n;i++) a[i]=Complex(0,0);
    for (i=0;i<let;i++) b[i]=Complex(t[let-i-1]-'0',0);
    for (i=let;i<n;i++) b[i]=Complex(0,0);
    FFT(a,n,1);FFT(b,n,1);
    for (i=0;i<n;i++) a[i]=a[i]*b[i];
    FFT(a,n,-1);
    for (i=0;i<n;i++) {
        k+=(int)(a[i].r+0.1);
        num[i]=k%10;k/=10;
    }
    while (k) { num[n++]=k%10;k/=10; }
    while (n>1&&!num[n-1]) n--;
    for (i=n-1;i>=0;i--) printf("%d", num[i]);
    printf("\n");
}
int main()
{
    while (scanf("%s%s", s, t)!=EOF) solve();
    return 0;
}