Codefroces 784 愚人节题目（部分）

A. Numbers Joke

time limit per test

2 seconds

memory limit per test

64 megabytes

input

standard input

output

standard output

Input

The input contains a single integer a (1 ≤ a ≤ 30).

Output

Output a single integer.

Example

Input

3

Output

27

史密夫数列

史密夫数（或作史密斯数）是指在某个进位下，它每一位数的数字的和相等於它因数分解（不用指数记数）中每一个因数的位数的和。

在十进位下，202就是一个史密夫数，因 2 + 0 + 2 = 4，202的因数分解为2 × 101，2 + 1 + 0 + 1 = 4。

所有质数也不会当作是史密夫数，因质数的因数只有它自己和1，必定不符合以上的要求。

史密夫数是由美国利哈伊大学的阿尔伯特·维兰斯基（Albert Wilansky）发现，而第一个被发现的史密夫数是4937775，因为Wilansky该位表兄弟的电话号码是4937775。

在十进位下的史密夫数如下（OEIS中的数列A006753）：

4, 22, 27, 58, 85, 94, 121, 166, 202, 265, 274, 319, 346, 355, 378, 382,
391, 438, 454, 483, 517, 526, 535, 562, 576, 588, 627, 634, 636, 645,
648, 654, 663, 666, 690, 706, 728, 729, 762, 778, 825, 852, 861, 895,
913, 915, 922, 958, 985, 1086

#include<iostream>
using namespace std;
int a[]={,,,,,,,,,,,
  ,,,,,,,,,,
,,,,,,,,,,,};
int main()
{
  int n;
  while(cin>>n)
  {
    cout<<a[n]<<endl;
  }
  return ;
}

B. Kids' Riddle

time limit per test

2 seconds

memory limit per test

64 megabytes

input

standard input

output

standard output

Programmers' kids solve this riddle in 5-10 minutes. How fast can you do it?

Input

The input contains a single integer n (0 ≤ n ≤ 2000000000).

Output

Output a single integer.

Examples

Input

11

Output

2

Input

14

Output

0

Input

61441

Output

2

Input

571576

Output

10

Input

2128506

Output

3

转换为16进制后数圈圈

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <vector>
#include <queue>
#include <cstdlib>
#include <iomanip>
#include <cmath>
#include <ctime>
#include <map>
#include <set>
using namespace std;
#define lowbit(x) (x&(-x))
#define max(x,y) (x>y?x:y)
#define min(x,y) (x<y?x:y)
#define MAX 100000000000000000
#define MOD 1000000007
#define pi acos(-1.0)
#define ei exp(1)
#define PI 3.141592653589793238462
#define ios() ios::sync_with_stdio(false)
#define INF 0x3f3f3f3f3f
#define mem(a) (memset(a,0,sizeof(a)))
typedef long long ll;
int n,ans=;
int main()
{
    scanf("%d",&n);
    while(n>=)
    {
        if(n%==) ans+=;
        if(n%==) ans+=;
        if(n%==) ans+=;
        if(n%==) ans+=;
        if(n%==) ans+=;
        if(n%==) ans+=;
        if(n%==) ans+=;
        if(n%==) ans+=;
        n/=;
    }
    if(n==) ans+=;
    if(n==) ans+=;
    if(n==) ans+=;
    if(n==) ans+=;
    if(n==) ans+=;
    if(n==) ans+=;
    if(n==) ans+=;
    if(n==) ans+=;
    printf("%d\n",ans);
    return ;
}

C. INTERCALC

time limit per test

2 seconds

memory limit per test

64 megabytes

input

standard input

output

standard output

DO YOU EXPECT ME TO FIND THIS OUT?

WHAT BASE AND/XOR LANGUAGE INCLUDES string?

DON'T BYTE OF MORE THAN YOU CAN CHEW

YOU CAN ONLY DISTORT THE LARGEST OF MATHEMATICS SO FAR

SAYING "ABRACADABRA" WITHOUT A MAGIC AND WON'T DO YOU ANY GOOD

THE LAST STACK RUPTURES. ALL DIE. OH, THE EMBARRASSMENT!

I HAVE NO ARRAY AND I MUST SCREAM

ELEMENTS MAY NOT BE STORED IN WEST HYPERSPACE

Input

The first line of input data contains a single integer n (1 ≤ n ≤ 10).

The second line of input data contains n space-separated integers a_i (1 ≤ a_i ≤ 11).

Output

Output a single integer.

Example

Input

4
2 5 3 1

Output

4

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <vector>
#include <queue>
#include <cstdlib>
#include <iomanip>
#include <cmath>
#include <ctime>
#include <map>
#include <set>
using namespace std;
#define lowbit(x) (x&(-x))
#define max(x,y) (x>y?x:y)
#define min(x,y) (x<y?x:y)
#define MAX 100000000000000000
#define MOD 1000000007
#define pi acos(-1.0)
#define ei exp(1)
#define PI 3.141592653589793238462
#define ios() ios::sync_with_stdio(false)
#define INF 0x3f3f3f3f3f
#define mem(a) (memset(a,0,sizeof(a)))
typedef long long ll;
int n,ans=;
int main()
{
    int n,x,ans;
    cin>>n;
    cin>>x;
    ans=x;
    for(int i=;i<n;i++)
    {
        cin>>x;
        ans=max(ans,x);
    }
    printf("%d\n",ans^x);
    return ;
}

D. Touchy-Feely Palindromes

time limit per test

2 seconds

memory limit per test

64 megabytes

input

standard input

output

standard output

Input

The only line of the input contains a string of digits. The length of the string is between 1 and 10, inclusive.

Output

Output "Yes" or "No".

Examples

Input

373

Output

Yes

Input

121

Output

No

Input

436

Output

Yes

数字回文串

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <vector>
#include <queue>
#include <cstdlib>
#include <iomanip>
#include <cmath>
#include <ctime>
#include <map>
#include <set>
using namespace std;
#define lowbit(x) (x&(-x))
#define max(x,y) (x>y?x:y)
#define min(x,y) (x<y?x:y)
#define MAX 100000000000000000
#define MOD 1000000007
#define pi acos(-1.0)
#define ei exp(1)
#define PI 3.141592653589793238462
#define ios() ios::sync_with_stdio(false)
#define INF 0x3f3f3f3f3f
#define mem(a) (memset(a,0,sizeof(a)))
typedef long long ll;
//              0  1  2  3  4  5  6  7  8  9
int mirror[] = {,-,-, , , , , , , };
int main()
{
    char a[];
    scanf("%s",&a);
    int len=strlen(a);
    int i=;
    while(a[i]!='\0' && mirror[a[i]-'']==a[len-i-]-'') i++;
    puts(i==len?"Yes":"No");
}

E. Twisted Circuit

time limit per test

2 seconds

memory limit per test

64 megabytes

input

standard input

output

standard output

Input

The input consists of four lines, each line containing a single digit 0 or 1.

Output

Output a single digit, 0 or 1.

Example

Input

0
1
1
0

Output

0

数字电路，或与非

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <vector>
#include <queue>
#include <cstdlib>
#include <iomanip>
#include <cmath>
#include <ctime>
#include <map>
#include <set>
using namespace std;
#define lowbit(x) (x&(-x))
#define max(x,y) (x>y?x:y)
#define min(x,y) (x<y?x:y)
#define MAX 100000000000000000
#define MOD 1000000007
#define pi acos(-1.0)
#define ei exp(1)
#define PI 3.141592653589793238462
#define ios() ios::sync_with_stdio(false)
#define INF 0x3f3f3f3f3f
#define mem(a) (memset(a,0,sizeof(a)))
typedef long long ll;
int a,b,c,d;
int main()
{
    scanf("%d%d%d%d",&a,&b,&c,&d);
    printf("%d\n",((a^b)&(c|d))^((b&c)|(a^d)));
    return ;
}