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 = 2000050;
    //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