L1-003 single digit statistics

The original title is

Given a  k  An integer  N=dk−1​10k−1+⋯+d1​101+d0​ (0≤di​≤9, i=0,⋯,k−1, dk−1​>0), Please write a program to count the number of times each different digit number appears . for example ： Given  N=100311, Then there are 2 individual 0,3 individual 1, and 1 individual 3.

Input format ：

Each input contains 1 Test cases , I.e. no more than one 1000 Bit positive integer  N.

Output format ：

Yes  N  Each different digit in , With  D:M  Output the digit in one line  D  And in  N  Is the number of times  M. Required press  D  Ascending output of .

sample input ：

100311

sample output ：

0:2
1:3
3:1

Code

#include <bits/stdc++.h>
using namespace std;
int main()
{
	char s[1010];// Single character array s 
	int a[10]={0};// initialization a The array is full of 0, Used to count the number of occurrences of each digit  
	int t = 0;
	scanf("%s",s);
	for(int i = 0; i < strlen(s); i++)//strlen(s) It's an array s The length of  
	{
		t = s[i] - '0';//S[i] - '0' Change to an integer  
		a[t]++;
	}	
	for(int i = 0; i < 10; i++)
	{
		if(a[i] != 0)
		printf("%d:%d\n",i,a[i]);//a The subscript i Is the number that appears ,a[i]  Is the number of occurrences  
	}
	return 0;
}