PAT Grade B 1021 digit statistics 15 (points)

topic

Given a kk $k$ -bit integer N = dk? 1 1 0 k? 1 +? + D 1 1 0 1 + d 0 N = d_{k-1}10^{k-1} + \cdots + d_1 10^ . 1 + D_0 $N$$=$$D$? K ? . 1 ? ?$.\; 1$$0$? K ? . 1 ? ?$+$$?$$+$$D$? . 1 ? ?$.\; 1$$0$? . 1 ? ?$+$$D$? 0 ? ? ( 0 ≤ DI ≤. 9 0 \ Le D_i \le 9 $0$$\le$ D ? I ? ? ≤ . 9 ,I = 0,?, K?. 1 I = 0, \ cdots, K-. 1$I$$=$$0$$,$$?$$,$$K$$?$$.\; 1$ ,DK?. 1> 0 D_ {K-. 1}> 0$d$? k ? 1 ? ?$>$$0$ ), write a program to count the number of each different digit arise. For example: GivenN = 1 0 0 3 1 1 N = 100311$N$$=$$1$$0$$0$$3$$1$$1$ , there are 2$0s,$$3$ 1s, and 1 3.

Input format:

Each input contains 1 test case, that is, a positive integer NN $N\; with$ no more than 1000 digits .

Output format:

Of NN $N$ each different numbers of bits to the digital format of the output bit in a row and in NN $N$ number appears . Required by the ascending output. D:M D M D

Input sample:

100311

Sample output:

0:2
1:3
3:1

Code

#include<iostream>
using namespace std;
int main()
{
    
    
	string NUM = "0123456789", n;
	int cal[10] = {
    
     0 }, i, j;
	cin >> n;
	for (i = 0; n[i] != '\0'; i++)
		for (j = 0; j < 10; j++)
			if (n[i] == NUM[j])
				cal[j]++;
	for (i = 0; i < 10; i++)
		if (cal[i] != 0)
			cout << i << ":" << cal[i] << endl;
	return 0;
}

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