代码

#include<cstdio>
#include<cstring>
#include<cmath>
#include<cstdlib>
#include<algorithm>
#include<vector>
#include<map>
#include<queue>
#include<stack>
#include<string>
#include<map>
#include<set>
#include<ctime>
#define eps 1e-6
#define LL long long
#define pii (pair<int, int>)
//#pragma comment(linker, "/STACK:1024000000,1024000000")
using namespace std;

const int maxn = 1000050;
//const int INF = 0x3f3f3f3f;
const double PI = acos(-1.0);
//复数结构体
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);
    }
};
/*
 * 进行FFT和IFFT前的反转变换。
 * 位置i和 （i二进制反转后位置）互换
 * len必须取2的幂
 */
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;
}
char num1[maxn], num2[maxn];
complex x1[maxn], x2[maxn];
int ans[maxn];
int main() {
    //freopen("input.txt", "r", stdin);
	scanf("%s%s",num1,num2);
		memset(ans, 0, sizeof(ans));
		int len = 1, len1 = strlen(num1), len2 = strlen(num2);
		while(len<len1+len2+1) len <<= 1;
		for(int i = 0; i < len1; i++) x1[len1-1-i] = complex((double)(num1[i]-'0'), 0);
		for(int i = len1; i < len; i++) x1[i] = complex(0, 0);
		fft(x1, len, 1);
		for(int i = 0; i < len2; i++) x2[len2-1-i] = complex((double)(num2[i]-'0'), 0);
		for(int i = len2; i < len; i++) x2[i] = complex(0, 0);
		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++) ans[i] = (int)(x1[i].r+0.5);
		for(int i = 1; i < len; i++) {
			ans[i] += ans[i-1]/10;
			ans[i-1] %= 10;
		}
		while(len>0 && !ans[len]) len--;
		for(int i = len; i >= 0; i--) printf("%c", ans[i]+'0');
		puts("");
    return 0;
}

评测结果