LightOJ 1341 Aladdin and the Flying Carpet 算数基本定理

题目大意:给出面积n,和最短边m,求能形成的矩形的个数(不能为正方形)。

题目思路:根据算数基本定理有:

1.每个数n都能被分解为:n=p1^a1*p2^a2*^p3^a3……pn^an(p为素数);

2.n的正因数的个数sum为:sum=(1+a1)*(1+a2)*(1+a3)……(1+an);

最短边为m,若m>=sqrt(n),则无解。所以m最多我10^6,可遍历找出1-m中n的因子,并用sum去减去这类因子的个数。

ps:最近一直想去证明算数基本定理,可是感觉能力不够,唉,慢慢来吧。

#include<cstdio>
#include<stdio.h>
#include<cstdlib>
#include<cmath>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<vector>
#include<queue>
#define INF 0x3f3f3f3f
#define MAX 1000005
#define mod 1000000007

using namespace std;

long long p[MAX],v[MAX],num_prime;

void GetPriem()//获得素数
{
    memset(p,0,sizeof(p));
    memset(v,0,sizeof(v));
    for(int i=2;i<MAX;i++)
    {
        if(!v[i])
        {
            p[++num_prime]=i;
            for(int j=i;j<MAX;j+=i)
                v[j]=1;
        }
    }
}

long long Ans(long long n)
{
    long long cnt,sum=1;
    for(int i=1;i<=num_prime && p[i]<=n;i++)
    {
        if(n%p[i]==0)
        {
            cnt=0;
            while(n%p[i]==0)
            {
                cnt++;
                n/=p[i];
            }
            sum*=(cnt+1);
        }
    }
    if(n > 1)
        sum*=2;
    return sum;
}

int main()
{
    GetPriem();
    int T,cnt=1,i,j;
    long long n,m;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%lld%lld",&n,&m);
        if(n/m < m)
        {
           printf("Case %d: %lld\n",cnt++,0);
            continue;
        }
        long long ans=Ans(n)/2;
        for(i=1;i<m;i++)
        {
            if(n%i==0)
                ans--;
        }
        printf("Case %d: %lld\n",cnt++,ans);
    }
    return 0;
}

时间: 2024-10-11 03:00:49

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