[JZOJ5773]【NOIP2008模拟】简单数学题

Description

  话说，小X是个数学大佬，他喜欢做数学题。有一天，小X想考一考小Y。他问了小Y一道数学题。题目如下：
      对于一个正整数N，存在一个正整数T（0<T<N），使得
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的值是正整数。
      小X给出N，让小Y给出所有可能的T。如果小Y不回答这个神奇的大佬的简单数学题,他学神的形象就会支离破碎。所以小Y求你帮他回答小X的问题。

Input

一个整数N。

Output

第一个数M，表示对于正整数N，存在M个不同的正整数T，使得
aaarticlea/png;base64,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" alt=" " />
是整数。
后面是M个数，每一个数代表可能的正整数T（按从小到大的顺序排列）。

Sample Input

Sample Input1:

1

Sample Input2:

3

Sample Input3

180

Sample Output

Sample Output

0

Sample Output

1 2

Sample Output

5 120 144 160 168 176

Data Constraint

  对于5%的数据，N=1.
      对于20%的数据，N<=5.
      对于40%的数据，N<=1000000
      对于另外20%的数据，答案只有1个，且N为质数，保证对于前60%的数据，当N为质数的时候，答案都一定只有一个,对于这20%的数据，满足2<N。
      对于80%的数据，N<=10^9.
      对于100%的数据，N<=10^14.

233考场打的大暴力拿了65分。

既没有判断素数又没有打表，竟然暴力拿这么多分真是不敢相信。

放上考场yy代码：

#include <iostream>

#include <cstdio>

#include <cstring>

#include <queue>

#include <cmath>

#include <algorithm>

using namespace std;

#define  int long long

inline int read() {

    int res=;char c=getchar();bool f=;

    while(!isdigit(c)) {if(c=='-')f=;c=getchar();}

    while(isdigit(c))res=(res<<)+(res<<)+(c^),c=getchar();

    return f?-res:res;

}

int tmp[], cnt;

signed main(){

    freopen("math.in", "r", stdin);

    freopen("math.out", "w", stdout);

    int n = read();

    if (n == ) return puts(""), ;

    int ans = ;

    if (n <= 6e7)

    {

        for (register int i =  ; i < n ; i ++)

            if ((n - i / ) % (n - i) ==  and (i % ) == ) tmp[++ans] = i;

        printf("%lld ", ans);

        for (register int j =  ; j <= ans ; j ++) printf("%lld ", tmp[j]);

        return ;

    }

    int tt = n -  * sqrt(n);

    tt = max(tt, (int));

    if (tt %  == ) tt--;

    for (register int i = tt ; i < n ; i += )

        if ((n - i / ) % (n - i) == ) tmp[++ans] = i;

    printf("%lld ", ans);

    for (register int i =  ; i <= ans ; i ++) printf("%lld ", tmp[i]);

    return ;

}

暴力

其实它给的暴力分是40分(逃。。

然后聊聊正解233.

设...算了不写了233.

放个网址就跑题解

然后有不少细节要注意，比如枚举i的时候一定要判断是不是奇数，和最后答案是不是奇数。

#include <iostream>

#include <cstdio>

#include <cstring>

#include <cmath>

#include <algorithm>

using namespace std;

#define int long long

inline int read() {

    int res=;char c=getchar();bool f=;

    while(!isdigit(c)) {if(c=='-')f=;c=getchar();}

    while(isdigit(c))res=(res<<)+(res<<)+(c^),c=getchar();

    return f?-res:res;

}

int tmp[*], cnt;

signed main()

{

    int n = read();

    n <<= ;

    int ans = ;

    for (register int i =  ; i <= sqrt(n) ; i ++)

    {

        if (n % i == )

        {

            if (i %  == ) tmp[++ans] = n / i * max(((i - ) / ), (int));

            int t = n / i;

            if (i != t and t %  == )

                tmp[++ans] = n / t * max(((t - ) / ), (int));

        }

    }

    int tt = ans;

    sort(tmp + , tmp +  + ans);

    for (register int i =  ; i <= tt ; i ++) if (tmp[i] %  == ) tmp[i] = , ans--;

    printf("%lld ", ans);

    for (register int i =  ; i <= tt ; i ++) if(tmp[i]) printf("%lld ", tmp[i]);

    return ;

}