c++ - Count how many times elements in an array are repeated -

the program i'm trying write allows me enter 10 numbers , should tell me number x repeated x times , on.

i've been trying problem result follows:

for example...{1,1,1,1,4,6,4,7,4}

the number 1 repeated 4 times

the number 1 repeated 3 times

the number 1 repeated 2 times

the number 1 repeated 1 times

the number 4 repeated 3 times

the number 6 repeated 1 times

the number 4 repeated 2 times

the number 7 repeated 1 times

the number 4 repeated 1 times

the problem checks next number following numbers without skipping it, or without knowing has written before

#include <iostream> #include <string> using namespace std; int main() {     int x[10];     (int i=0;i<10;i++) {         cin>>x[i];     }      (int i=0;i<9;i++) {         int count=1;         (int j=i+1;j<10;j++) {              if (x[i]==x[j]) count++;         }         cout<<"the number "<<x[i]<<" repeated "<<count<<" times"<<"\n";     } } 

the problem code re-process numbers you've processed. if there occurrence of 1 @ position 0 , occurrence of 1 @ position 5, process 1 @ position 5 again when there in loop.

so need way decide if number has been processed or not. easy way add second array (initially values set 0) , whenever process number mark positions element occurs. before processing element check if it's been processed , nothing if that's case.

also, try indent code :)

c++ code:

int main( void ) {     const int n = 10;      int a[n];     for(int = 0; < n; i++)         cin >> a[i];      int seen[n];     for(int = 0; < n; i++)         seen[i] = 0;      for(int = 0; < n; i++) {         if(seen[i] == 0) {             int count = 0;             for(int j = i; j < n; j++)                 if(a[j] == a[i]) {                     count += 1;                     seen[j] = 1;                 }             cout << a[i] << " occurs " << count << " times" << endl;         }     }      return 0; }