提交记录 3280


用户 题目 状态 得分 用时 内存 语言 代码长度
hannibal 1004. 【模板题】高精度乘法 Accepted 100 435.016 ms 107840 KB C++ 3.14 KB
提交时间 评测时间
2018-07-11 11:59:51 2020-07-31 21:14:57
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <math.h>
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]);
        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;
}
const int MAXN = 3000050;
complex x1[MAXN],x2[MAXN];
char str1[MAXN/2],str2[MAXN/2];
char st1[MAXN/2], st2[MAXN/2];
int sum[MAXN];
int main(){
    int flag = 0;
    scanf("%s%s",st1,st2);
        if((st1[0]== '-' && st2[0] != '-')||(st1[0] != '-' && st2[0] == '-')) {
            flag = 1;
            if(st1[0] == '-') {
                strcpy(str1, st1+1);
                strcpy(str2, st2);
            }
            if(st2[0] == '-') {
                strcpy(str1, st1);
                strcpy(str2, st2+1);
            }

        }
        else if(st1[0] == '-' && st2[0] == '-') {
            strcpy(str1, st1+1);
            strcpy(str2, st2+1);
        }
        else {
            strcpy(str1, st1);
            strcpy(str2, st2);
        }
        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);
        fft(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--;
        if(flag){
            if(len == 0) {
                if(sum[len] != 0) printf("-");
            }
            else printf("-");
        }
        for(int i = len;i >= 0;i--)
            printf("%c",sum[i]+'0');
        printf("\n");
    return 0;
}

CompilationN/AN/ACompile OKScore: N/A

Testcase #1435.016 ms105 MB + 320 KBAcceptedScore: 100


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