1032.ZOJ problem

Title Description:

read into a string, the string contains zoj three characters, the number is not necessarily equal, in the order of zoj output, when a character runs out, the rest is still in the order of zoj output.

Input:

The topic contains multiple sets of use cases, one for each group of use cases, containing zoj three characters, and the end of the input when the "E" is entered.
1<=length<=100.

Output:

For each set of inputs, output a line that represents the string after processing as required.
A concrete example of the visible sample.

Sample input:

Zzooojjjzzzzooooojjjzooojje

Sample output:

Zojzojojzojzojzojzoozojojo

#include <iostream>#include<cstring>using namespacestd;intMain () {strings;  while(Getline (cin,s) && s!="E"){        intz=0, o=0, j=0;  for(intI=0; I<s.size (); i++){            if(s[i]=='Z') z++; Else if(s[i]=='O') o++; Else if(s[i]=='J') J + +; }         for(;z>0&& o>0&& j>0; z--, o--, j--) {cout<<"ZOJ"; }        if(z==0){             for(;o>0&& j>0; O--, j--) {cout<<"OJ"; }            if(o==0) {                 for(;j>0; j--) cout<<"J"; }            Else  for(;o>0; o--) cout<<"O"; }        if(o==0){             for(;z>0&& j>0; Z--, j--) {cout<<"ZJ"; }            if(z==0) {                 for(;j>0; j--) cout<<"J"; }            Else  for(;z>0; z--) cout<<"Z"; }        if(j==0){             for(;o>0&& z>0; O--, z--) {cout<<"ZO"; }            if(o==0) {                 for(;z>0; z--) cout<<"Z"; }            Else  for(;o>0; o--) cout<<"O"; } cout<<Endl; }        return 0;}

1032.ZOJ problem