提交记录 3251


用户 题目 状态 得分 用时 内存 语言 代码长度
marcel 1004. 【模板题】高精度乘法 Accepted 100 441.63 ms 110376 KB C++ 2.57 KB
提交时间 评测时间
2018-07-10 19:03:19 2020-07-31 21:13:09
#include<stdio.h>
#include<string.h>
#include<math.h>
#include<algorithm>
#define eps 1e-6
using namespace std;

const int maxn = 3000050;
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]);
        k = len/2;
        while( j >= k)
        {
            j -= k;
            k /= 2;
        }
        if(j < k) j += k;
    }
}
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],num3[maxn];
complex x1[maxn], x2[maxn];
int ans[maxn];
int main() {
    int flag=0;
    scanf("%s",num3);
    if(num3[0]=='-') strcpy(num1,num3+1),flag++;
    else strcpy(num1,num3);
    scanf("%s",num3);
    if(num3[0]=='-') strcpy(num2,num3+1),flag++;
    else strcpy(num2,num3);
        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--;
        if(flag==1) printf("-");
        for(int i = len; i >= 0; i--) printf("%c", ans[i]+'0');
        //puts("");
    return 0;
}

CompilationN/AN/ACompile OKScore: N/A

Testcase #1441.63 ms107 MB + 808 KBAcceptedScore: 100


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