题目大意：
求两个大整数的乘积, 两个大整数长度都不超过50000, 多组数据, 时限1s
大致思路:
作为FFT算法的一个开头的题, 恩还是仔细写了一下
这个题就是把整数视作是两个多项式, 每一位就是一项, 那么就相当于是两个最高次数不超过50000的多项式乘积之后在x = 10出的值, 那么这样就很简单了, 直接处理出其多项式然后用FFT计算即可
来自kuangbin的模板：
#include<iostream>
#include<algorithm>
#include<cstring>
#include<cstdio>
using namespace std;
const double PI = acos(-1.0);
struct Complex
{double r, i;Complex(double _r = 0.0, double _i = 0.0){r = _r; i = _i;}Complex operator +(const Complex &b){return Complex(r+b.r, i+b.i);}Complex operator -(const Complex &b){return Complex(r-b.r, i-b.i);}Complex operator * (const Complex &b){return Complex(r*b.r - i*b.i, r*b.i+i*b.r);}
};
void change(Complex y[], int len)
{int i, j, k;for(i = 1, j = len/2;i < len-1; i++){if(i < j)swap(y[i],y[j]);//交换互为小标反转的元素，i<j保证交换一次//i做正常的+1，j左反转类型的+1,始终保持i和j是反转的k = len/2;while( j >= k){j -= k;k /= 2;}if(j < k) j += k;}
}
/** 做FFT* len必须为2^k形式，* on==1时是DFT，on==-1时是IDFT*/
void fft(Complex y[],int len,int on)
{change(y,len);for(int h = 2; h <= len; h <<= 1){Complex wn(cos(-on*2*PI/h),sin(-on*2*PI/h));for(int j = 0;j < len;j+=h){Complex w(1,0);for(int k = j;k < j+h/2;k++){Complex u = y[k];Complex t = w*y[k+h/2];y[k] = u+t;y[k+h/2] = u-t;w = w*wn;}}}if(on == -1)for(int i = 0;i < len;i++)y[i].r /= len;
}
const int maxn = 200010;
Complex x1[maxn], x2[maxn];
char str1[maxn], str2[maxn];
int sum[maxn];
int main()
{while(scanf("%s%s", str1, str2) == 2){int len1 = strlen(str1);int len2 = strlen(str2);int len = 1;while(len < len1*2 || len < len2*2)len <<= 1;for(int i = 0; i < len1; i++)x1[i] = Complex(str1[len1-1-i]-'0', 0);for(int i = len1; i < len; i++)x1[i] = Complex(0, 0);for(int i = 0; i < len2; i++)x2[i] = Complex(str2[len2-1-i]-'0', 0);for(int i = len2; i < len; i++)x2[i] = Complex(0, 0);//求DFTfft(x1,len,1);fft(x2,len,1);for(int i = 0; i < len; i++)x1[i] = x1[i] * x2[i];fft(x1, len, -1);for(int i = 0; i < len; i++)sum[i] = (int)(x1[i].r + 0.5);for(int i = 0; i < len; i++){sum[i+1] += sum[i]/10;sum[i] %= 10;}len = len1+len2-1;while(sum[len] <= 0 && len > 0) len--;for(int i = len; i >= 0; i--)printf("%c", sum[i] + '0');printf("\n");}return 0;
}