OJ input and output exercise

1. practice 1

1.1 subject 【 No number of input groups , No exit requirements 】

1.2 Code C++

#include<iostream>
int main (){
    
    int i,d,n;
    while(scanf("%d %d",&i,&d)!=EOF){
    
    n=i+d;
    printf("%d\n",n);}
    return 0;
}

1.3 matters needing attention

1. adopt scanf(“%d %d”,&i,&d)!=EOF To control input stop ,(EOF Is the error identifier )
2. The input variable scanf(“%d %d”,&i,&d) Yes & Address value symbol , When something is passed in , It is sent to the address to save ; And output printf(“%d\n”,n) There is no need to &, The value of the output variable .
3. Because it will be assigned to i、d, Addition is assigned to n, So there's no need to int i=0;int d=0;int n=0; This initializes the assignment .
4. Note that the output should wrap , Input without line breaks

2. practice 2

2.1 subject 【 Tell the number of input groups 】

2.2 Code C++

#include<iostream>
using namespace std;
int main(){
    
    int a,b,c, n;
    scanf("%d",&a);
    for (int t1=0;t1<a;t1++){
    
        scanf("%d %d",&b,&c);
        n=b+c;
        printf("%d\n",n);
    }
}

2.3 matters needing attention

1. Comma English

3. practice 3

3.1 subject 【0 0 sign out 】

3.2 Code C++

#include<iostream>
using namespace std;
int main (){
    
    int a,b,c;
     scanf("%d %d",&a,&b);
    while (a!=0||b!=0){
    
        c=a+b;
        printf("%d\n",c);
        scanf("%d %d",&a,&b);
    }  
}

3.3 matters needing attention

1. while Conditions inside a!=0||b!=0, perhaps !(a==0 &&b ==0)

4. practice 4

4.1 subject 【 One line for each group of data , The first integer in each line is the number of integers n;0 sign out 】

4.2 Code C++

#include<iostream>
using namespace std;
int main(){
    
    int a=1; 
    int b[1000]={
    0};
    int sum=0;
    scanf("%d ",&a);
    while(a!=0){
    
        for(int t1=0;t1<a;t1++){
    
        	if(t1!=a-1){
    
        		scanf("%d ",&b[t1]);
                sum=sum+b[t1];
			}else{
      
			scanf("%d",&b[t1]);
            sum=sum+b[t1];
			} 
          
        }
        printf("%d\n",sum);
        scanf("%d ",&a);
        sum=0;
        int b[1000]={
    0};
    }
}

4.3 matters needing attention

1. There is no space after the last input
2. Array input plus &

5. practice 5

5.1 subject 【 One line for each group of data , The first integer in each line is the number of integers n; Number of notification groups 】

5.2 Code C++

#include<iostream>
using namespace std;
int main(){
    
    int a=1; 
    int b[1000]={
    0};
    int sum=0;
    int c=0;
    scanf("%d",&c);
    scanf("%d ",&a);
    while(c!=0){
    
        for(int t1=0;t1<a;t1++){
    
        	if(t1!=a-1){
    
        		scanf("%d ",&b[t1]);
                sum=sum+b[t1];
			}else{
      
			scanf("%d",&b[t1]);
            sum=sum+b[t1];
			} 
          
        }
        printf("%d\n",sum);
        scanf("%d ",&a);
        sum=0;
        int b[1000]={
    0};
        c--;
    }
}

5.3 matters needing attention

6. practice 6

6.1 subject 【 One line for each group of data , The first integer in each line is the number of integers n; No number of input groups 】

6.2 Code C++

#include<iostream>
using namespace std;
int main(){
    
    int a=1; 
    int b[1000]={
    0};
    int sum=0;
    while( scanf("%d ",&a)!=EOF){
    
        for(int t1=0;t1<a;t1++){
    
        	if(t1!=a-1){
    
        		scanf("%d ",&b[t1]);
                sum=sum+b[t1];
			}else{
      
			scanf("%d",&b[t1]);
            sum=sum+b[t1];
			} 
        }
        printf("%d\n",sum);
        sum=0;
        int b[1000]={
    0};
    }
}

6.3 matters needing attention

7. practice 7

7.1 subject 【 One line for each group of data , The number of each line is variable ; No number of input groups 】

7.2 Code C++

#include<iostream>
int main(){
    
    int a;
    int sum=0;
    while(scanf("%d",&a)!=EOF){
    
        sum=sum+a;
        if(getchar() == '\n'){
    
            printf("%d\n",sum);
            sum=0;// It can't be int sum =0;
        }
    }
}

7.3 matters needing attention

1. Newline capture through getchar() == ‘\n’

8. practice 8

8.1 subject 【 String sort , Tell me the number 】

8.2 Code C

#include <stdio.h>
#include <string.h>
int main(){
    
    int n;
    int ch[100][100];
    scanf("%d",&n);
    for(int i=0;i<n;i++){
    
        scanf("%s",&ch[i]);
    }
    qsort(ch,n,sizeof(ch[0]),strcmp);
    for(int j=0;j<n;j++){
    
        printf("%s ",ch[j]);
    }
    return 0;
}

8.3 matters needing attention

1. qsort English analysis ：Quick sort, Translation is quick sorting , Its internal implementation is realized by fast sorting algorithm .
   function ： Sort any incoming data , Make it an ordered sequence .【 String sort !!!】
2. The string is stored in a two-dimensional array /string
3. sort Cannot sort string

9. practice 9

9.1 subject 【 String sort , No number , No group number 】

9.2 Code C

Fang 1:

#include<stdio.h>
#include<string.h>
int main(){
    
    char ch[1000][1000]={
    0};
    int t1;
    int t=0;
    while(scanf("%s",&ch[t++])!=EOF){
    
        if(getchar()=='\n'){
    
            qsort(ch,t,sizeof(ch[0]),strcmp);
            for(t1=0;t1<t;t1++){
    
                if(t1!=t-1){
     printf("%s ",ch[t1]);}
                else{
    printf("%s\n",ch[t1]);}
            }
          t=0;
        }  
    }
    return 0;
}

Fang 2:

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

int main()
{
    
    char str[100][100] = {
    0};
    int number = 0;
    
    while (scanf("%s", str[number++]) != EOF) {
    
        if (getchar() == '\n') {
    
            qsort(str, number, sizeof(str[0]), strcmp);
            for (int i = 0; i < number; i++) {
    
                printf("%s ", str[i]);
            }
            printf("\n");
            number = 0;
        }
    }
    
    return 0;
}

9.3 matters needing attention

1. Initialization of string char str[100][100] = {0};

10. practice 10

10.1 subject 【 String sort , No number , No group number , There are commas between strings 】

10.2 Code C

#include<stdio.h>
#include<string.h>
int main(){
    
    char ch[1000][1000]={
    0};
    int t1;
    int t=0;
    while(scanf("%[^,\n]",&ch[t++])!=EOF){
    
        if(getchar()=='\n'){
    
            qsort(ch,t,sizeof(ch[0]),strcmp);
            for(t1=0;t1<t;t1++){
    
                if(t1!=t-1){
     printf("%s,",ch[t1]);}
                else{
    printf("%s\n",ch[t1]);}
            }
          t=0;
        }  
    }
    return 0;
}

10.3 matters needing attention

1. Regular expressions ,scanf(“%[^,\n]”,&ch[t++])!=EOF

11. practice 11

11.1 subject 【 Pay attention to the data range 】

11.2 Code

#include<iostream>
using namespace std;
int main(){
    
    long int a=0;
    long int sum =0;
    while(scanf("%ld",&a)!=EOF){
    
   		 sum=sum+a;
       if(getchar()!='\n'){
    

       }else{
    
         printf("%ld\n",sum);  
         sum=0;
       }
    }
    return 0;
}

11.3 matters needing attention

1. sum=sum+a; To be in getchar()!='\n’ front
2. The number size exceeds $10^7$ when , Use long int as well as %ld

Brush problem