bzoj2179 FFT快速傅立叶

题目链接：bzoj2179
题目大意：
给出两个n位10进制整数x和y，你需要计算x*y。
n<=60000

题解：
RT
把一位看成系数直接上FFT。
注意输出前处理进位和前导0去掉什么的。

还不会非递归的。。

#include<cstdio>#include<cstdlib>#include<cstring>#include<cmath>#include<iostream>#include<algorithm>using namespace std;#define maxn 60010*2const double pi=acos(-1);char str[maxn];int ans[maxn];struct Complex{    double x,y;    Complex() {x=0;y=0;}    Complex(double x,double y):x(x),y(y){}      friend Complex operator + (Complex x, Complex y) {return Complex(x.x+y.x,x.y+y.y);}    friend Complex operator - (Complex x, Complex y) {return Complex(x.x-y.x,x.y-y.y);}    friend Complex operator * (Complex x, Complex y) {return Complex(x.x*y.x-x.y*y.y,x.x*y.y+x.y*y.x);}}a[maxn],b[maxn];void DFT(Complex *c,int ln,int t){    if (ln==1) return;    Complex A0[ln>>1],A1[ln>>1];    for (int i=0;i<=ln;i+=2) A1[i>>1]=c[i+1],A0[i>>1]=c[i];    Complex wn(cos(2*pi/ln),t*sin(2*pi/ln)),w(1,0);    DFT(A0,ln>>1,t);DFT(A1,ln>>1,t);    for (int i=0;i<(ln>>1);i++,w=w*wn)    {        c[i]=A0[i]+w*A1[i];        c[i+(ln>>1)]=A0[i]-w*A1[i];    }}void FFT(int n){    int i,fn=1;while (fn<=2*n) fn<<=1;    DFT(a,fn,1);DFT(b,fn,1);    for (i=0;i<=fn;i++) a[i]=a[i]*b[i];    DFT(a,fn,-1);    for (i=0;i<=2*n;i++) ans[i]=(int)(a[i].x/fn+0.5);}int main(){    //freopen("a.in","r",stdin);    //freopen("a.out","w",stdout);    int n,i;    scanf("%d",&n);n--;    scanf("%s",str);    for (i=0;i<=n;i++) a[n-i].x=(double)str[i]-'0';    scanf("%s",str);    for (i=0;i<=n;i++) b[n-i].x=(double)str[i]-'0';    FFT(n);    for (i=0;i<2*n;i++)    {        ans[i+1]+=ans[i]/10;        ans[i]%=10;     }    int fn=2*n;while (!ans[fn] && fn) fn--;    for (i=fn;i>=0;i--) printf("%d",ans[i]);    printf("\n");    return 0;}