PAT/进制转换习题集

B1022. D进制的A+B (20)

Description：

输入两个非负10进制整数A和B(<=2³⁰-1)，输出A+B的D (1 < D <= 10)进制数。

Input：

输入在一行中依次给出3个整数A、B和D。

Output：

输出A+B的D进制数。

Sample Input：

123 456 8

Sample Output：

1103

 1 #include <cstdio>
 2
 3 int main()
 4 {
 5     int a, b, d;
 6     scanf("%d%d%d", &a, &b, &d);
 7
 8     int sum = a+b;
 9     int ans[31], num = 0;
10     do {
11         ans[num++] = sum%d;
12         sum /= d;
13     } while(sum != 0);
14
15     for(int i=num-1; i>=0; --i)
16         printf("%d", ans[i]);
17
18     return 0;
19 }

A1019. General Palindromic Number (20)

Description：

A number that will be the same when it is written forwards or backwards is known as a Palindromic Number. For example, 1234321 is a palindromic number. All single digit numbers are palindromic numbers.

Although palindromic numbers are most often considered in the decimal system, the concept of palindromicity can be applied to the natural numbers in any numeral system. Consider a number N > 0 in base b >= 2, where it is written in standard notation with k+1 digits a_i as the sum of (a_ibⁱ) for i from 0 to k. Here, as usual, 0 <= a_i < b for all i and a_k is non-zero. Then N is palindromic if and only if a_i = a_k-i for all i. Zero is written 0 in any base and is also palindromic by definition.

Given any non-negative decimal integer N and a base b, you are supposed to tell if N is a palindromic number in base b.

Input：

Each input file contains one test case. Each case consists of two non-negative numbers N and b, where 0 <= N <= 10⁹ is the decimal number and 2 <= b <= 10⁹ is the base. The numbers are separated by a space.

Output：

For each test case, first print in one line "Yes" if N is a palindromic number in base b, or "No" if not. Then in the next line, print N as the number in base b in the form "a_k a_k-1 ... a₀". Notice that there must be no extra space at the end of output.

Sample Input1：

27 2

Sample Output1：

Yes
1 1 0 1 1

Sample Input2：

121 5

Sample Output2：

No

4 4 1

 1 #include <cstdio>
 2
 3 #define MaxSize 50
 4 int ans[MaxSize];
 5
 6 int main()
 7 {
 8     //freopen("E:\\Temp\\input.txt", "r", stdin);
 9
10     int N, b, num = 0;
11     bool flag = true;
12     scanf("%d %d", &N, &b);
13
14     do {
15         ans[num++] = N%b;
16         N /= b;
17     } while(N != 0);
18
19     for(int i=0, j=num-1; i<=(num-1)/2, j>=(num-1)/2; ++i, --j) {
20         if(ans[i] != ans[j]) {
21             flag = false;
22             break;
23         }
24     }
25
26     if(flag == true)    printf("Yes\n");
27     else printf("No\n");
28     for(int i=num-1; i>=0; --i) {
29         printf("%d", ans[i]);
30         if(i != 0)  printf(" ");
31     }
32
33     return 0;
34 }

 1 #include <cstdio>
 2
 3 bool judge(int z[], int num)
 4 {
 5     for(int i=0; i<=num/2; ++i) {
 6         if(z[i] != z[num-1-i])  return false;
 7         else return true;
 8     }
 9 }
10
11 int main()
12 {
13     int n, b, z[40], num = 0;
14     scanf("%d%d", &n, &b);
15
16     do {
17         z[num++] = n%b;
18         n /= b;
19     } while(n != 0);
20     bool flag = judge(z, num);
21
22     if(flag == true)    printf("Yes\n");
23     else printf("No\n");
24     for(int i=num-1; i>=0; --i) {
25         printf("%d", z[i]);
26         if(i != 0)  printf(" ");
27     }
28
29     return 0;
30 }