题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1402
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
#define maxn 65536+5
const double pi=acos(-1.0);
struct cp{double a,b;cp operator +(const cp &o)const {return (cp){a+o.a,b+o.b};}cp operator -(const cp &o)const {return (cp){a-o.a,b-o.b};}cp operator *(const cp &o)const {return (cp){a*o.a-b*o.b,b*o.a+a*o.b};}cp operator *(const double &o)const {return (cp){a*o,b*o};}cp operator !() const{return (cp){a,-b};}
}w[maxn];///复数的表示手段
int pos[maxn];
void fft_init(int len){int j=0;while((1<<j)<len)j++;j--;///得出最终的jfor(int i=0;i<len;i++)pos[i]=pos[i>>1]>>1|((i&1)<<j);///动态规划搞定翻转函数
}
void fft(cp *x,int len,int sta){for(int i=0;i<len;i++) if(i<pos[i]) swap(x[i],x[pos[i]]);w[0]=(cp){1,0};for(unsigned i=2;i<=len;i<<=1){cp g=(cp){cos(2*pi/i),sin(2*pi/i)*sta};for(int j=i>>1;j>=0;j-=2) w[j]=w[j>>1];for(int j=1;j<i>>1;j+=2) w[j]=w[j-1]*g;for(int j=0;j<len;j+=i){cp *a=x+j,*b=a+(i>>1);for(int l=0;l<i>>1;l++){cp o=b[l]*w[l];b[l]=a[l]-o;a[l]=a[l]+o;}}}if(sta==-1) for(int i=0;i<len;i++) x[i].a/=len,x[i].b/=len;///求逆矩阵最后还要处理一下
}
cp x[maxn],y[maxn],z[maxn];
void FFT(int *a,int *b,int n,int m,int *c){int len=1;while(len<=(n+m)>>1) len<<=1;fft_init(len);for(int i=n/2;i<len;i++)x[i].a=x[i].b=0;for(int i=m/2;i<len;i++)y[i].a=y[i].b=0;for(int i=0;i<n;i++)(i&1?x[i>>1].b:x[i>>1].a)=a[i];for(int i=0;i<m;i++)(i&1?y[i>>1].b:y[i>>1].a)=b[i];fft(x,len,1),fft(y,len,1);for(int i=0;i<len/2;i++){int j=len-1&len-i;z[i]=x[i]*y[i]-(x[i]-!x[j])*(y[i]-!y[j])*(w[i]+(cp){1,0})*0.25;}for(int i=len/2;i<len;i++){int j=len-1&len-i;z[i]=x[i]*y[i]-(x[i]-!x[j])*(y[i]-!y[j])*((cp){1,0}-w[i^len>>1])*0.25;}fft(z,len,-1);for(int i=0;i<n+m;i++)if(i&1) c[i]=(int)(z[i>>1].b+0.5);else c[i]=(int)(z[i>>1].a+0.5);
}
char s1[maxn],s2[maxn];
int a[maxn],b[maxn],c[maxn<<1];
int main(){while(~scanf("%s%s",s1,s2)){memset(c,0,sizeof(c));int n=strlen(s1),m=strlen(s2);for(int i=0;i<n;i++)a[i]=s1[n-1-i]-'0';for(int i=0;i<m;i++)b[i]=s2[m-1-i]-'0';FFT(a,b,n,m,c);for(int i=0;i<n+m-1;i++){c[i+1]+=c[i]/10;c[i]%=10;}int len=n+m-1;while(c[len]==0&&len>0)len--;for(int i=len;i>=0;i--)printf("%c",c[i]+'0');printf("\n");}return 0;
}