HDU 1999 不可摸数【类似筛法求真因子和】

不可摸数

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 15591    Accepted Submission(s): 4077

Problem Description

s(n)是正整数n的真因子之和，即小于n且整除n的因子和.例如s(12)=1+2+3+4+6=16.如果任何
数m，s(m)都不等于n,则称n为不可摸数.

 

 

Input

包含多组数据，首先输入T,表示有T组数据.每组数据1行给出n(2<=n<=1000)是整数。

 

 

Output

如果n是不可摸数，输出yes，否则输出no

 

 

Sample Input

3 2 5 8

 

 

Sample Output

yes yes no

 

 

Author

Zhousc@ECJTU

 

 

Source

ECJTU 2008 Spring Contest

 

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<set>
#include<map>
#include<sstream>
#include<queue>
#include<cmath>
#include<list>
#include<vector>
#include<string>
using namespace std;
#define long long ll
const double PI = acos(-1.0);
const double eps = 1e-6;
const int inf = 0x3f3f3f3f;
const int N = 100005;
int n, m, tot;
int a[10550]={0};
int r[550], c[550];
int x, y, pr, pc;
int ok(int n)
{
    int sum = 0;
    for(int i=1 ;i<n; i++)
    {
        if(n % i == 0)
        {
            sum += i;
        }
    }
    return sum;
}

int main()
{

    int t, f = 0;
    cin >> t;
    while(t--)
    {
        //f = 0;
        cin >> n;
        int i = 2;
        while(i++ <= 3000)
        {
            if(ok(i) == n)
            {
                f = 1; break;
            }
        }
        printf("%s\n",f?"no":"yes");
    }
    return 0;
}

 

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<set>
#include<map>
#include<sstream>
#include<queue>
#include<cmath>
#include<list>
#include<vector>
#include<string>
using namespace std;
#define long long ll
const double PI = acos(-1.0);
const double eps = 1e-6;
const int inf = 0x3f3f3f3f;
const int N = 500005;
int n, m, tot;
int a[N];
int mp[N];
int x, y, pr, pc;

void init()
{
    for(int i=1; i<=N; i++)
    {
        for(int j=i+i; j<=N; j+=i)
        {
            a[j] += i;
        }
    }
    for(int j=1; j<=N; j++)
    {
        if(a[j] <= 1000) mp[a[j]] = 1; //hash标记
    }
}

int main()
{

    int t;
    cin >> t;
    init();
    while(t--)
    {
        cin >> n;
        printf("%s\n",mp[n]?"no":"yes");
    }
    return 0;
}