PAT/string processing question set (1), pat question set

PAT/string processing question set (1), pat question set

B1006. output integer in another format (15)

Description:

Let's use the letter B to represent "Hundred", the letter S to represent "Ten", and "12... n (<10) is a single digit. In another format, a positive integer of no more than three digits is output. For example, 234 should be output as BBSSS1234 because it has two "hundreds", three "Ten", and four digits.

Input:

Each test input contains one test case and a positive integer n (<1000 ).

Output:

The output of each test case occupies one line and outputs n in the specified format.

Sample Input1:

234

Sample Output1:

BBSSS1234

Sample Input2:

23

Sample Output2:

SS123

 1 #include <cstdio> 2  3 int main() 4 { 5     int n; 6     scanf("%d", &n); 7  8     int num = 0, ans[5]; 9     while(n != 0) {10         ans[num++] = n%10;11         n /= 10;12     }13 14     for(int i=num-1; i>=0; --i) {15         if(i == 2) {16             for(int j=0; j<ans[i]; ++j)17                 printf("B");18         } else if(i == 1) {19             for(int j=0; j<ans[i]; ++j)20                 printf("S");21         } else {22             for(int j=1; j<=ans[i]; ++j)23                 printf("%d", j);24         }25     }26 27     return 0;28 }

 

B1021. count (15)

Description:

Given a k-bit integer N = dk-1 * 10k-1 +... + d1 * 101 + d0 (0 <= di <= 9, I = 0 ,..., k-1, dk-1> 0), compile a program to count the number of occurrences of each individual digit. For example, given N = 100311, there are 2 0, 3 1, and 1 3.

Input:

Each input contains one test case, that is, a positive integer N with no more than 1000 bits.

Output:

For each digit in N, output the digit D and the number of times M appears in N in a row in the format of D: M. The output must be in ascending order of D.

Sample Input:

100311

Sample Output:

0: 2
1: 3
3:1

 1 #include <cstdio> 2 #include <cstring> 3  4 #define MaxSize 1010 5 char List[MaxSize]; 6  7 int main() 8 { 9     //freopen("E:\\Temp\\input.txt", "r", stdin);10 11     gets(List);12 13     int len = strlen(List), ans[10] = {0};14     for(int i=0; i<len; ++i)15         ++ans[List[i]-'0'];16 17     for(int i=0; i<10; ++i) {18         if(ans[i] != 0)19             printf("%d:%d\n", i, ans[i]);20     }21 22     return 0;23 }