friend

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Time Limit : 1000/1000ms (Java/Other)
Memory Limit : 32768/32768K (Java/Other)
Total Submission(s) : 53
Accepted Submission(s) : 16

Problem DescriptionFriend number are defined recursively as follows.
(1) numbers 1 and 2 are friend number;
(2) if a and b are friend numbers, so is ab+a+b;
(3) only the numbers defined in (1) and (2) are friend number.
Now your task is to judge whether an integer is a friend number.

InputThere are several lines in input, each line has a nunnegative integer a, 0<=a<=2^30.

OutputFor the number a on each line of the input, if a is a friend number, output “YES!”, otherwise output “NO!”.

Sample Input3
13121
12131

Sample OutputYES!
YES!
NO!

可观察其规律，c+1=(a+1)(b+1);即一定是由2，3变换过来的，不过不能直接计算它是否是2的倍数或者3的倍数，该过程与其无关

#include<cstdio>
int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        if(n==0)
        {
            printf("NO!\n");
            continue;
        }
        if((n+1)%2==0||(n+1)%3==0)//测试该数是否是由2，3变换过来的
        {
            n++;
            while(n%2==0)
            n/=2;
            while(n%3==0)
            n/=3;
        }
        if(n==1)
            printf("YES!\n");
        else
            printf("NO!\n");
    }
    return 0;
}

最初写的内存不够。。。不过确实，十亿的内存开不起

#include<cstdio>
#include<cstring>
#include<math.h>
__int64 flag[100000000],a[100000000];
int main()
{
    int m,n;
    a[0]=1,a[1]=2;
    memset(flag,0,sizeof(flag));
    flag[a[0]]=1;
    flag[a[1]]=1;
    for(int i=2;i<10000000;)
    {
        a[i]=a[i-1]*a[i-1]+2*a[i-1];
        flag[a[i]]=1;
        i++;
        a[i]=a[i-3]*a[i-3]+2*a[i-3];
        flag[a[i]]=1;
        i++;
        a[i]=a[i-3]*a[i-4]+a[i-3]+a[i-4];
        flag[a[i]]=1;
        i++;
    }
    while(~scanf("%d",&n))
    {
        if(flag[n])
            printf("YES!\n");
        else
            printf("NO!\n");
    }   
    return 0;
}