Acdreamoj1115(数学思维称号)

意甲冠军：1,3是完美的数，假定a，b是完美的数，然后，2+a*b+2*a+2*b，结论认为，n无论是完美的数字。

解法：開始仅仅看出来2+a*b+2*a+2*b=(a+2)*(b+2)-2,没推出很多其它结论，囧。没办法，仅仅能暴力将全部的完美数求出来然后查表。正解是c+2=(a+2)*(b+2);完美数都是有质因子3或5组成的（5本身除外）；

自己暴力代码：

/******************************************************

* author:xiefubao

*******************************************************/

#pragma comment(linker, "/STACK:102400000,102400000")

#include <iostream>

#include <cstring>

#include <cstdlib>

#include <cstdio>

#include <queue>

#include <vector>

#include <algorithm>

#include <cmath>

#include <map>

#include <set>

#include <stack>

#include <string.h>

//freopen ("in.txt" , "r" , stdin);

using namespace std;

#define eps 1e-8

#define zero(_) (abs(_)<=eps)

const double pi=acos(-1.0);

typedef long long LL;

const int Max=1010;

const int INF=1000000000;

LL num[Max];

int help[Max];

LL get(int l,int r)

{

    return 2*(num[l]+1)*(num[r]+1)-num[l]*num[r];

}

struct point

{

    LL ans;

    int h;

};

bool operator<(const point& a,const point& b)

{

    return a.ans>b.ans;

}

priority_queue<point> pri;

int main()

{

    num[0]=1;

    num[1]=3;

    point p;

    p.ans=7;

    p.h=0;

    pri.push(p);

    p.ans=23;

    p.h=1;

    pri.push(p);

    for(int i=0; i<Max; i++)

        help[i]=i;

    help[0]=1;

    help[1]=2;

    int po=2;

    while(pri.top().ans<=INF)

    {

        point ptop=pri.top();

        pri.pop();

        if(num[po-1]==ptop.ans)

        {

            point p;

            p.ans=get(ptop.h,help[ptop.h]);

            p.h=ptop.h;

            pri.push(p);

            help[ptop.h]++;

            continue;

        }

        point p;

        p.ans=get(ptop.h,help[ptop.h]);

        p.h=ptop.h;

        pri.push(p);

        help[ptop.h]++;

       // cout<<ptop.ans<<" ";

        num[po++]=ptop.ans;

        p.ans=get(po-1,po-1);

        p.h=po-1;

        pri.push(p);

        help[po-1]=po;

    }

    int t;

    while(scanf("%d",&t)==1)

    {

        if(binary_search(num,num+po,t))

            puts("Yes");

        else

            puts("No");

    }

    return 0;

}