质数分解

发表于 更新于 分类于 本文字数： 206 阅读时长 ≈ 1 分钟

对数据大于一万的就筛到该数的平方根

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#include <stdbool.h>
#include <stdio.h>
#include<math.h>

#define LL long long
#define IN freopen("in.txt", "r", stdin)
#define OUT freopen("out.txt", "w", stdout)
#define scan(x) scanf("%d", &x)
#define sqr(x) (x) * (x)
#define f(i, a, b) for (int i = a; i <= b; i++)
#define pn(x) printf("%d\n", x)
#define pr1(x) printf("Case %d: ", x)
#define pn1(x) printf("Case %d:\n", x)
#define pr2(x) printf("Case #%d: ", x)
#define pn2(x) printf("Case #%d:\n", x)
#define lowbit(x) (x & (-x))
#define MAXN 200000001

int prime[10000000];
bool vis[MAXN];
int cnt = 0;
void Euler_prime(int n)
{
    vis[1] = 1;
    for (int i = 2; i <= n; ++i)
    {
        if (!vis[i])
            prime[cnt++] = i;
        for (int j = 0; j < cnt; ++j)
        {
            if (i * prime[j] > n) //判断是否越界
                break;
            vis[i * prime[j]] = 1; //筛数
            if (i % prime[j] == 0)
                break;
        }
    }
}

int main()
{
    int n,x;
    scan(n);
    if(n>10000){
        x = (int)sqrt(n);
    }
    else
        x = n;
    Euler_prime(x);
    f(i, 0, cnt)
    {
        if (n % prime[i] == 0)
        {
            printf("%d", n/prime[i]);
            return 0;
        }
    }
    return 0;
}