[原创]LightOJ 1035 Intelligent Factorial Factorization [预处理+一半的 质因子分解]【数论】

2016-06-29 16:32:35 Tabris_ 阅读数：305

博客爬取于2020-06-14 22:44:18
以下为正文

版权声明：本文为Tabris原创文章，未经博主允许不得私自转载。
https://blog.csdn.net/qq_33184171/article/details/51783994

题目链接：LightOJ 1035 Intelligent Factorial Factorization

Intelligent Factorial Factorization
Time Limit:500MS Memory Limit:32768KB 64bit IO Format:%lld & %llu
Submit

Status
Description
Given an integer N, you have to prime factorize N! (factorial N).

Input
Input starts with an integer T (≤ 125), denoting the number of test cases.

Each case contains an integer N (2 ≤ N ≤ 100).

Output
For each case, print the case number and the factorization of the factorial in the following format as given in samples.

Case x: N = p1 (power of p1) * p2 (power of p2) * …

Here x is the case number, p1, p2 … are primes in ascending order.

Sample Input
3
2
3
6
Sample Output
Case 1: 2 = 2 (1)
Case 2: 3 = 2 (1) * 3 (1)
Case 3: 6 = 2 (4) * 3 (2) * 5 (1)

题目大意 就是把N！ 质因子分解一下

题解： 因为数据量很小 所以预处理一下所有N！ 的质因子个数即可

这种办法 应该可以做到n<=1000
n<10000的时候用容器存一下 而且时间在大一丢丢 应该还是能做的

因为数据量太小 所有算法都是最最最最最暴力的

附本题代码

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# include <stdio.h>
# include <stdlib.h>
# include <string.h>
# include <limits.h>
# include <malloc.h>
# include <ctype.h>
# include <math.h>
# include <string>
# include <iostream>
# include <algorithm>
# include <map>
# include <vector>
# include <set>

using namespace std;

# define LL unsigned long long int

int Is_or[500];
void Prime()
{
    int n=200;
    int k=0;
    memset(Is_or,1,sizeof(Is_or));
    Is_or[0]=Is_or[1]=0;
    for(int i=2; i<n; i++)
    {
        if(Is_or[i])
        {
            for(int j=i+i; j<n; j+=i)
            {
                Is_or[j]=0;
            }
        }
    }
    return ;
}

int has[105][105];

void init()
{
    int num;
    memset(has,0,sizeof(has));
    for(int i=2;i<=100;i++)
    {
        for(int j=2;j<=i;j++)
        {
            num=i;
            if(num%j==0&&Is_or[j])
                while(num%j==0)
                    has[i][j]++,num/=j;
        }
    }

    for(int i=1;i<=100;i++)
    {
        for(int j=1;j<=100;j++)
        {
            has[i][j]+=has[i-1][j];
        }
    }

    return ;
}

int main()
{
    Prime();
    init();
    int t,tt=0,num;
    LL p,l;
    scanf("%d",&t);
    while(t--)
    {
        int n,m;
        scanf("%d",&n);
        printf("Case %d: %d = ",++tt,n);

        printf("%d (%d)",2,has[n][2]);
        for(int i=3;i<=n;i++)
        {
            if(has[n][i])
            printf(" * %d (%d)",i,has[n][i]);
        }
        printf("\n");
    }
    return 0;
}