NSYSTEM(Numeral System) spoj problem solution

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HEY this is easy problem simply follow the rule......... keep patient..........this may take some time....... i give 45 minute in it...... then i got AC in 1st attempt..... MY AC C++ SOLUTION IS::::::::: #include<bits/stdc++.h> using namespace std; int main(){ int t; scanf("%d\n",&t); while(t--) { char name1[1000],name2[1000]; scanf("%s %s",name1,name2); // printf("%s %s\n",name1,name2); int ans1=0,ans2=0,len1,len2,i; len1=strlen(name1); len2=strlen(name2); for(i=0;i<len1;i++) { if(name1[i]>=48 && name1[i]<=57) { if(name1[i+1]=='m') ans1+=1000*(name1[i]-'0'); if(name1[i+1]=='c') ans1+=100*(name1[i]-'0'); if(name1[i+1]=='x') ans1+=10*(name1[i]-'0'); if(name1[i+1]=='i') ans1+=1*(name1[i]-'0'); i++; } else { if(name1[i]=='m') ans1+=1000; if(name1[i]=='c') ans1+=100; if(name1[i]=='x') ans1+=10; if(name1[i]=='i') ans1+=1; } } for(i=0;i<len2;i++) { if(name2[i]>=48 && name2[i]<=57) { if(name2[i+1]=='m') ans2+=1000*(name2[i]-'0'); if(name2[i+1]=='c') ans2+=100*(name2[i]-'0'); if(name2[i+1]=='x') ans2+=10*(name2[i]-'0'); if(name2[i+1]=='i') ans2+=1*(name2[i]-'0'); i++; } else { if(name2[i]=='m') ans2+=1000; if(name2[i]=='c') ans2+=100; if(name2[i]=='x') ans2+=10; if(name2[i]=='i') ans2+=1; } } int sum=ans1+ans2; // printf("%d %d =%d\n",ans1,ans2,sum); char arr[10]; int count=0; while(sum) { arr[count++]=(sum%10)+'0'; sum/=10; } /* for(i=count-1;i>=0;i--) { printf("%c ",arr[i]); }*/ int p=0; char str[1000]; for(i=count-1;i>=0;i--) { if(arr[i]!='0' && arr[i]!='1') { str[p++]=arr[i]; } if(i==3 && arr[i]!='0') str[p++]='m'; if(i==2 && arr[i]!='0') str[p++]='c'; if(i==1 && arr[i]!='0') str[p++]='x'; if(i==0 && arr[i]!='0') str[p++]='i'; } str[p]='\0'; printf("%s\n",str); } return 0; }