Code Study

00_1. Code Review.cpp

/* To study Other good Codes

0밀기.cpp

#include <iostream>
#include <vector>
#include <cstdio>

using namespace std;

char change (int x){
  switch(x){
		case 1: return '1'; break;
		case 2: return '2'; break;
		case 3: return '3'; break;
		case 4: return '4'; break;
		case 5: return '5'; break;
		case 6: return '6'; break;
		case 7: return '7'; break;
		case 8: return '8'; break;
		case 9: return '9'; break;
	}
}

int main(){
	int input[100];
	int n= 0; int inputing; int temp;

	scanf("%d", &inputing);
  /* To store the input by digits, two way. The first is doing in this way, 
  and the second is use char array, as the char array only one character by each room 
  */

	for (int i=0; inputing > 0; i++){
		input[n++] = inputing % 10;
		inputing /= 10;
	}

	for (int i=0; i<n/2;i++){
		temp = input[n-1-i];
		input[n-1-i] = input[i];
		input[i] = temp;
	}
	
	vector<char>vt;
	for (int i=0; i<n; i++){
		if (input[i] != 0){
			vt.push_back(change(input[i]));
		}
	}
	cout << n << endl;
	cout << vt.size() << endl;

	for (int i =0; i<=n-vt.size();i++){
		vt.push_back('0');
	}

	for (int i=0;i<vt.size();i++){
		printf("%c",vt[i]);
	}
	
	return 0;
}

1_친구찾기.cpp

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <vector>
using namespace std;

char input[1000000];

int main()
{
    vector<int> vt;
    vector<int> vt2;
    int diff = 999999999;

    fgets(input, 1000000, stdin);
    /* This operator is the key to take acceptance of variables as the user wants */
    
    /* --------------------fget -----------------------
    char * fgets ( char * str, int num, FILE * stream );
    Read chracters from stream ==> stores them into C-string until num-1 character has been read
    Either New line reached ==> Then saved what has been read into the string.
    
    str is the adress that where the string read is copied.
    
    fget is used for accepting the stream having the space or tab
    fget = cin>>char a[anynum] with space. 
    
    -------------------------------------------- */

    char *p = strtok(input, " \n\t,");
    /* char * strtok ( char * str, const char * delimiters );
    it return the pointer that the first group of splited string */

    do
    {
        vt.push_back(atoi(p));
    } while (p = strtok(NULL, " \n\t,"));
    
    /* strtok always needs do-while phrase to print the group
    When the strtok is declared, from the next time it is OKAY to use NULL instread of "input" 
    
    by do-while we can scan whole splited strings 
    */ 

    int n;
    scanf("%d",&n);

    for (int i = 0; i < vt.size(); i++)
    {
        int dif2 = abs(n - vt[i]);
        if (dif2 < diff)
        {
            vt2.clear();
            vt2.push_back(vt[i]); // this is how to input the value into the vector 
            diff = dif2;
        }
        else if (dif2 == diff)
        {
            vt2.push_back(vt[i]);
        }
    } /* Unconditionally the roop started for the first time 
    and from the second time, vt2 became the vector with single value or the vector having all the same value. 
    In both cases, the value will be the biggest difference */ 

    for (int i = 0; i < vt2.size(); i++)
    {
        printf(i ? " %d" : "%d", vt2[i]);
        /* Logical Operator. "Condition ? A:B. True, A. False, B 
        This commands print out only one variable because it has only a single value
        */
    }
    return 0;
}

2_가장긴문자열.cpp

#include <stdio.h>
#include <string.h>
#include <stdlib.h>

int main()
{
  char *arr=NULL;
	char str[100],*a[10];
	int i=0,j=0,r=0;
	char *t;

	fgets(str, sizeof(str), stdin);

	arr=strtok(str," ");

	while(arr!=NULL)
	{
		if(r<strlen(arr))
		{
			t=arr;
			r=strlen(arr);
		}
		a[i]=arr;
		arr=strtok(NULL," ");
		i++;
		
	}

	printf("%s", t);	
	return 0;
}

3_동우_문자열포함.cpp

#include <iostream>
#include <cstring>
#include <cstdlib>
 
using namespace std;
 
int main(){
    
    char arrA[100];
    char arrB[100];
    char temp[100];
    
    cin >> arrA;
    cin >> arrB;
    
    if (strlen(arrB) > strlen(arrA)){
       strcpy(temp,arrA);
       strcpy(arrA,arrB);
       strcpy(arrB,temp); 
    }
   
    int lenA = strlen(arrA),
        lenB = strlen(arrB);
    int idx = -1;
    for ( int i = 0 ; i < lenA-lenB ; i++) {
        int match = 1;
        for ( int j = 0 ; j < lenB ; j++ )
        {
            if ( arrA[i+j] != arrB[j] ) {
               match = 0;
               break;
            }  
        }
        if ( match ) {
           idx = i;
           break;
        }
    }
    //
    if ( idx != -1 )
    {
       for ( int i = 0 ; i < lenA ; i++)
       {
           if ( i == idx )
           {
              i += lenB;
           }
           cout << arrA[i];    
       }
    } else {
      cout << arrA;
    }
    return 0;
}

4_strtok-study.cpp

#include <iostream>
#include <cstdlib>
#include <cstring>
#include <cstdio>

using namespace std;

int main(){
  char a[] = "Hello world";
	cout << a << endl;

	char *p;
	p = strtok(a," "); // strtok return the pointer 

	cout << p << endl; // call whole array 
	while (p = strtok(NULL, " ")){
		cout << p << endl;
	}
}

Technique for getting string.cpp

/* This is the code review for remembering how to get the string as an input 
in the form of char array */

string input;
getline (cin, input);

trycatch_middle_subset.cpp

#include <iostream>
#include <math.h>
using namespace std;

int n;  //집합내 원소들의 총 개수
char* items;  //집합을 문자로 저장할 공간

void powerSet(char* aSet, int aSetLen, int current)
{ 
	if(current==n) 
	{ 
		cout << "{";
		for(int i=0; i<aSetLen; i++){
			cout << aSet[i] << " ";
			}
		cout << "}" << endl;
	}
	else { 
		current++;
		powerSet(aSet, aSetLen, current);
		current--;
		aSet[aSetLen++]=items[current++];
		powerSet(aSet, aSetLen, current);
	}
}	


int main(int argc, char **argv)
{
	cout << "Input the SIZE of Set : " ;
	cin >> n;
	if(n < 0) {
		cout << " size wrong !! " << endl;
		return 0;
	}

	items = new char[n];

	for(int i = 0; i < n; i++) 
	{
		cout << "Insert a element - " << i+1 << " : ";
		cin >> items[i];
	}

	char* aSet;	
	aSet = new char[n];
	powerSet(aSet, 0, 0);

	cout << "all subsets No : 2^" << n << "=" << pow(2.0,n) << endl;
	return 0;
}

두배열합치기.cpp

#include <iostream>
#include <cmath>
#include <ctime>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <vector>
#include <conio.h>

using namespace std;

int main() {
  char input1[100];  char input2[100];	char seps = '\n';
	vector<int>vt;
	char *p1; char *p2;

	fgets(input1, 100, stdin);	

	p1 = strtok(input1," ");	vt.push_back(atoi(p1));
	while (p1 != NULL){
		p1 = strtok(NULL, " " " \n ");
		vt.push_back(atoi(p1));
	}

	fgets(input1, 100, stdin);	
  /* 여기서 중요한 것은 fgets를 어떻게 사용하는가. 
  왜인지 모르겠으나, fgets 는 배열  하나에 값을 다 받더라. 
  따라서, fgets를 갱신하는 방식으로 코드를 짰다. 

  */
  
  p1 = strtok(input1," ");	vt.push_back(atoi(p1));

	while (p1 != NULL){
		p1 = strtok(NULL, " " " \n ");
		vt.push_back(atoi(p1));
	}

	sort(vt.begin(), vt.end());

	for(int i=2; i<vt.size(); i++){
		cout << vt[i] << " ";
	}

  	return 0;

  }

오름차순.cpp

#include<stdio.h>
#include<algorithm>
int n,a[99],i;
int main(){
  while(scanf("%d",&a[n])>0)n++;
  /* 
  1. Maybe it would be better to use scanf instead of cin
  2. &a[n]<0 means that if enter put in, it is terminated // 동우야 이거 맞아? 
  3. We can get the number of input as we want by &a[n]>0
  */
	std::sort(a,a+n);
	while(i<n)printf("%d ",a[i++]);
}

온도변환.cpp

#include <cstdio>

int main()
{
    int n;
    scanf("%dF",&n);
    /* This is the high tech using scanf. If we want to taled the form of input, the wisest choice 
    is to use scanf in such a way 
    */
    printf("%dC",(n-30)/2);
}

일번찾기.cpp

#include<stdio.h>
char x,y;
int main(){
  while(scanf("%c",&x)>0) /* while (stopping condition) */
		if(++y&&x&1)
			return 0 & printf("%d",y);
	} /* 동우야 이 코드가 지금, X가 들어올 때마다 단계가 하나씩 올라가고 있는거지? 

중복누락 1,2.cpp

#include <algorithm>
#include <cstdio>

using namespace std;

int data[100], n;

int main(void)
{
    scanf("%d", &n);
    for(int i=0;i<n;i++) scanf("%d", data + i);
    sort(data, data + n);
    for(int i=0;i<n;i++) if(data[i] != i + 1) { printf("%d %d\n", i + 1, data[i]); break; }
    return 0;
}


#include <cstdio>

int main()
{
    int n,k,i,cnt[20000]={0,};
    scanf("%d",&n);
    for(i=0;i<n;i++)
    {
        scanf("%d",&k);
        cnt[k]++;
    }
    for(i=1;i<=n;i++) if(!cnt[i]) printf("%d ",i);
    for(i=1;i<=n;i++) if(cnt[i]==2) printf("%d",i);
}


#include <iostream>

using namespace std;

int main() {
  const int MAX_N = 50;
	int list[MAX_N] = {0};
	int N, n, val;
	int drop = 0, dupl = 0;

	cin >> N;
	for(n = 1; n <= N; ++n) {
		cin >> val;
		++list[val];
	}
	for(n = 1; n <= N; ++n) {
		if(list[n] == 0) drop = n;
		else if(list[n] == 2) dupl = n;

		if(drop && dupl) break;
	}

	cout << drop << ' ' << dupl << endl;

	return 0;
}