【城会玩系列】hdu 5175 Misaki's Kiss again【思维】【公式转换】

Misaki's Kiss again

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1558    Accepted Submission(s): 389

Problem Description

After the Ferries Wheel, many friends hope to receive the Misaki's kiss again,so Misaki numbers them 1,2...N−1,N,if someone's number is M and satisfied the GCD(N,M) equals to N XOR M,he will be kissed again.

Please help Misaki to find all M(1<=M<=N).

Note that:
GCD(a,b) means the greatest common divisor of a and b.
A XOR B means A exclusive or B

 

Input

There are multiple test cases.

For each testcase, contains a integets N(0<N<=1010)

 

Output

For each test case,
first line output Case #X:,
second line output k means the number of friends will get a kiss.
third line contains k number mean the friends' number, sort them in ascending and separated by a space between two numbers

 

Sample Input

3
5
15

 

Sample Output

Case #1:
1
2
Case #2:
1
4
Case #3:
3
10 12 14

Hint
In the third sample, gcd(15,10)=5 and (15 xor 10)=5, gcd(15,12)=3 and (15 xor 12)=3,gcd(15,14)=1 and (15 xor 14)=1

一道很有意识的题，暴力走了一发10^10无力吐槽，到网上找题解，才发现出题人真会玩~，会玩是真会玩。

怎么会玩呢？这里我们一步一步详解。

首先题目要求是这样的：

摩天轮后，一些朋友希望再次得到Misaki的吻，所以Misaki把他们分别编号从1到$N$，如果他们中有人的编号是$M$，而且$gcd(N,M)=N$ xor $M$,那么他以可以得到一个吻。

如果满足了gcd(N,M)==N^M，相对的output就要++，这里我们知道N^N==1，所以我们把M换成N^X，然后我们公式就转换为：

gcd（N，N^X）==X，我们知道，gcd是用来求最大公约数的，所以从中得出结论：x是N,N^X的最大公约数，那么X，一定是N的约数。

所以我们这里要找到N的约数，然后再判断是否有满足公式的N^X，然后就能得出结果啦~；

注意数据较大，要用long long：

#include<cstdio>
#include<cmath>
using namespace std;
#define ll long long int
ll f[100000];
ll ans[100000];
ll gcd(ll x,ll y)
{
    return y?gcd(y,x%y):x;
}
int main()
{
    int kase=0;
    ll n;
    while(~scanf("%I64d",&n))
    {
        int cont=0;
        for(ll i=1;i<=(ll)sqrt(n);i++)
        {
            if(n%i==0)
            {
                f[cont++]=i;
            }
        }
        for(ll i=(ll)sqrt(n);i>1;i--)
        {
            if(n%i==0)
            {
                f[cont++]=n/i;
            }
        }
        int cont2=0;
        for(int i=cont-1;i>=0;i--)
        {
            if(f[i]==gcd(n,n^f[i])&&(n^f[i]!=0&&(n^f[i])<=n))
            {
                ans[cont2++]=(n^f[i]);
            }
        }
        printf("Case #%d:\n%d\n",++kase, cont2);
        if(cont2 == 0)
            printf("\n");
        for(int i = 0; i < cont2; i++)
        {
            if(i != cont2 - 1)
                printf("%I64d ", ans[i]);
            else
                printf("%I64d\n", ans[i]);
        }

    }
}