# PAT自测_打印沙漏、素数对猜想、数组元素循环右移、数字加倍重排、机器洗牌

```00-自测1. 打印沙漏(20)

*****
***
*
***
*****

19 *

*****
***
*
***
*****
2```
```#include<stdio.h>
#include<math.h>
int main(void)
{
int n, half, i, j;
char c;

scanf("%d %c", &n, &c);
half = sqrt((n+1)/2.0);
for(i = half; i >= 1; i--)
{
for(j = 1; j <= half - i; j++)
putchar(‘ ‘);
for(j = 1; j <= 2*i - 1; j++)
putchar(c);
putchar(‘\n‘);
}
for(i = 2; i <= half; i++)
{
for(j = 1; j <= half - i; j++)
putchar(‘ ‘);
for(j = 1; j <= 2*i - 1; j++)
putchar(c);
putchar(‘\n‘);
}
printf("%d\n", n - (2*half*half - 1));
return 0;
}

/*

1 + 2·3 + 2·5 + 2·7 + 2·(2n-1) = 2(1+3+5+...+2n-1) - 1 = 2n^2 - 1.

*/```
```00-自测2. 素数对猜想 (20)

20

4```
```// 最容易想到的方法当然是暴力求解了:~ 11MS
#include<stdio.h>
#include<math.h>
int IsPrime(int num);
int main(void)
{
int n, count, i;

scanf("%d", &n);

count = 0;
for(i = 2; i < n-1; i++)
if(IsPrime(i) && IsPrime(i+2))
count++;

printf("%d\n", count);

return 0;
}

int IsPrime(int num)
{
int i, bound = sqrt(num);
for(i = 2; i <= bound; i++)
if(num % i == 0)
return 0;

return 1;
}```
```// 升级版：素数筛，1MS
#include<stdio.h>
#include<string.h>
#define MAXN 100000
int arr[MAXN+10];
int prime[MAXN+10];
int main(void)
{
int n, i, j, count;

memset(arr, 0, sizeof(arr));
scanf("%d", &n);
for(i = 2; i*i <= n; i++)
if(arr[i] == 0)
for(j = i*i; j <= n; j += i)
arr[j] = 1;

j = 0;
count = 0;
for(i = 2; i <= n; i++)
if(arr[i] == 0)
prime[j++] = i;
for(i = 1; i < j; i++)
if(prime[i] - prime[i-1] == 2)
count++;
printf("%d\n", count);
return 0;
}```
```00-自测3. 数组元素循环右移问题 (20)

6 2
1 2 3 4 5 6

5 6 1 2 3 4```
```// 注意到循环一圈的话，相当于数组元素没有移动，因此可以利用模运算来尽可能的减少数据移动的次数
#include<stdio.h>
int main(void)
{
int arr[100];
int n, m, i, j, temp, cir;

scanf("%d%d", &n, &m);
for(i = 0; i < n; i++)
scanf("%d", &arr[i]);

cir = m % n;
for(i = 0; i < cir; i++)
{
temp = arr[n-1];
for(j = n-2; j >= 0; j--)
arr[j+1] = arr[j];
arr[0] = temp;
}

for(i = 0; i < n; i++)
{
if(i == 0)
printf("%d", arr[i]);
else
printf(" %d", arr[i]);
}

return 0;
}```
```00-自测4. Have Fun with Numbers (20)

Notice that the number 123456789 is a 9-digit number consisting exactly the numbers from 1 to 9, with no duplication. Double it we will obtain 246913578, which happens to be another 9-digit number consisting exactly the numbers from 1 to 9, only in a different permutation. Check to see the result if we double it again!

Now you are suppose to check if there are more numbers with this property. That is, double a given number with k digits, you are to tell if the resulting number consists of only a permutation of the digits in the original number.

Input Specification:

Each input file contains one test case. Each case contains one positive integer with no more than 20 digits.

Output Specification:

For each test case, first print in a line "Yes" if doubling the input number gives a number that consists of only a permutation of the digits in the original number, or "No" if not. Then in the next line, print the doubled number.

Sample Input:
1234567899
Sample Output:
Yes
2469135798
```
```#include<stdio.h>
#include<string.h>
void bbsort(char[], int);
int main(void)
{
char str[25], dest[25], temp[25];
int length, i, c, change;

scanf("%s", str);
length = strlen(str);
for(i = 0; i < length / 2; i++)
{
change = str[i];
str[i] = str[length-1-i];
str[length-1-i] = change;
}

c = 0;
for(i = 0; i < length; i++)
{
dest[i] = (str[i]-‘0‘)*2 + c;
c = dest[i] / 10;
dest[i] = dest[i] % 10 + ‘0‘;
}
if(c != 0)
{
dest[i] = c + ‘0‘;
dest[i+1] = ‘\0‘;
printf("No\n");
}
else
{
dest[i] = ‘\0‘;
i = 0;
while((temp[i] = dest[i]) != ‘\0‘)
i++;
bbsort(str, length);
bbsort(temp, length);
for(i = 0; i < length; i++)
if(temp[i] != str[i])
break;
if(i == length)
printf("Yes\n");
else
printf("No\n");
}

length = strlen(dest);
for(i = length - 1; i >= 0; i--)
putchar(dest[i]);
return 0;
}

void bbsort(char s[], int length)
{
int temp, outer, inner;

for(outer = length - 1; outer > 0; outer--)
for(inner = 0; inner < outer; inner++)
if(s[inner] < s[inner+1])
{
temp = s[inner];
s[inner] = s[inner+1];
s[inner+1] = temp;
}

}
/*

1.读入字符数组，因为存储形式是从高位到低位，所以要做一下反转
2.加倍,注意这里我采用了字符的形式存储数字，因为单纯用数字存储的话，0是一个问题，可能引起麻烦，因为它和‘\0‘相等
3.如果有进位，直接输出"No"，否则，分别对加倍之前和加倍之后的数组排序，接下来逐一比较即可
*/```
```00-自测5. Shuffling Machine (20)

Shuffling is a procedure used to randomize a deck of playing cards. Because standard shuffling techniques are seen as weak, and in order to avoid "inside jobs" where employees collaborate with gamblers by performing inadequate shuffles, many casinos employ automatic shuffling machines. Your task is to simulate a shuffling machine.

The machine shuffles a deck of 54 cards according to a given random order and repeats for a given number of times. It is assumed that the initial status of a card deck is in the following order:

S1, S2, ..., S13, H1, H2, ..., H13, C1, C2, ..., C13, D1, D2, ..., D13, J1, J2

where "S" stands for "Spade", "H" for "Heart", "C" for "Club", "D" for "Diamond", and "J" for "Joker". A given order is a permutation of distinct integers in [1, 54]. If the number at the i-th position is j, it means to move the card from position i to position j. For example, suppose we only have 5 cards: S3, H5, C1, D13 and J2. Given a shuffling order {4, 2, 5, 3, 1}, the result will be: J2, H5, D13, S3, C1. If we are to repeat the shuffling again, the result will be: C1, H5, S3, J2, D13.

Input Specification:

Each input file contains one test case. For each case, the first line contains a positive integer K (<= 20) which is the number of repeat times. Then the next line contains the given order. All the numbers in a line are separated by a space.

Output Specification:

For each test case, print the shuffling results in one line. All the cards are separated by a space, and there must be no extra space at the end of the line.

Sample Input:
2
36 52 37 38 3 39 40 53 54 41 11 12 13 42 43 44 2 4 23 24 25 26 27 6 7 8 48 49 50 51 9 10 14 15 16 5 17 18 19 1 20 21 22 28 29 30 31 32 33 34 35 45 46 47
Sample Output:
S7 C11 C10 C12 S1 H7 H8 H9 D8 D9 S11 S12 S13 D10 D11 D12 S3 S4 S6 S10 H1 H2 C13 D2 D3 D4 H6 H3 D13 J1 J2 C1 C2 C3 C4 D1 S5 H5 H11 H12 C6 C7 C8 C9 S2 S8 S9 H10 D5 D6 D7 H4 H13 C5
```
```#include<stdio.h>
int main(void)
{
int arr[55], temp[55], dest[55];
int n, i;

scanf("%d", &n);
for(i = 1; i <= 54; i++)
scanf("%d", &arr[i]);
for(i = 1; i <= 54; i++)
dest[i] = i;
while(n--)
{
for(i = 1; i <= 54; i++)
temp[arr[i]] = dest[i];

for(i = 1; i <= 54; i++)
dest[i] = temp[i];
}
for(i = 1; i <= 54; i++)
{
if(dest[i] == 53)
printf("J1");
else if(dest[i] == 54)
printf("J2");
else if(dest[i] <= 13)
printf("S%d", dest[i]);
else if(13 < dest[i] && dest[i] <= 26)
printf("H%d", dest[i]-13);
else if(26 < dest[i] && dest[i] <= 39)
printf("C%d", dest[i] - 26);
else
printf("D%d", dest[i] - 39);

if(i != 54)
printf(" ");

}
return 0;
}
/*

1.开两个数组，一个顺序存储1-54，每个数字代表相应的牌，另一个用于存储给出的洗牌序列
2.根据给定的洗牌次数进行洗牌，再开一个数组，存储每一论洗牌的中间结果，最后再复制到源数组中
3.根据洗完牌后的数组数字代表的相应的牌进行输出
*/

(END_XPJIANG)```

## PAT：1008. 数组元素循环右移问题 (20) AC

#include<stdio.h> #include<stdlib.h> int main() { int n,move,cnt=0; scanf("%d%d",&n,&move); int* arr; arr=(int*)calloc(n,sizeof(int)); for(int i=0 ; i<n ; ++i) scanf("%d",&arr[i]); move%=n; //取余可以保证最多移动一轮 for

## pat00-自测2. 素数对猜想 (20)

00-自测2. 素数对猜想 (20) 时间限制 400 ms 内存限制 65536 kB 代码长度限制 8000 B 判题程序 Standard 作者 CHEN, Yue 让我们定义 dn 为:dn = pn+1 - pn,其中 pi 是第i个素数.显然有 d1=1 且对于n>1有 dn 是偶数.“素数对猜想”认为“存在无穷多对相邻且差为2的素数”. 现给定任意正整数N (< 105),请计算不超过N的满足猜想的素数对的个数. 输入格式:每个测试输入包含1个测试用例,给出正整数N. 输出格式:

## PAT (Basic Level) Practise （中文）1027. 打印沙漏(20)

1027. 打印沙漏(20) 时间限制 200 ms 内存限制 32000 kB 代码长度限制 8000 B 判题程序 Standard 作者 CHEN, Yue 本题要求你写个程序把给定的符号打印成沙漏的形状.例如给定17个“*”,要求按下列格式打印 ***** *** * *** ***** 所谓“沙漏形状”,是指每行输出奇数个符号:各行符号中心对齐:相邻两行符号数差2:符号数先从大到小顺序递减到1,再从小到大顺序递增:首尾符号数相等. 给定任意N个符号,不一定能正好组成一个沙漏.要求打印出

## PAT：1027. 打印沙漏(20) AC

#include int main() { int n; char xing; scanf("%d %c",&n,&xing); int num=1,i=1; while(1) { num = num + 2 * (i + 2); //一开始就超出,不改变i if(num > n) break; i += 2; } int MAX=i; for(int k=0 ; kPAT:1027. 打印沙漏(20) AC