杭电OJ 输入输出练习汇总

• 主题

  Calculate a + b

• 杭电OJ-1000

  • Input

    Each line will contain two integers A and B. Process to end of file.

  • Output

    For each case, output A + B in one line.

  • Mine

    #include <stdio.h>
    
    int main()
    {
        int a,b;
        while(~scanf("%d %d",&a,&b))   //多次输入a和b。
        {
            printf("%d
    ",a+b);
        }
    }
    
    /*** 
    while(~scanf("%d %d",&a,&b)) 多次输入a和b。
    这句话中的“~”符号可以理解为“重复”，代码含义是反复执行scanf(“%d %d”,&a,&b) 语句，直到语句接收不到有效结果。换一种说法就是while语句会在括号中的判断为真的情况执行语句，那么对于scanf函数而言，判断为真也就是接收到了有效数据。而~符号代表无限重复，直到scanf语句不能取到有效的值为止（while的括号中判断为假），循环跳出。
    ***/
    

  • Review

    题目: 接收两个整数并返回两个数的和。
    需要注意的是题目中说明了每行两个数据，但并没有说明多少行。

    换一种常用说法叫：“多组数据”，是常见的要求。但没有C语言算法书会写明接收多组数据的方式。

• 杭电OJ-1001

  • Description

    calculate SUM(n) = 1 + 2 + 3 + ... + n.

  • Input

    The input will consist of a series of integers n, one integer per line.

  • Outpu

    For each case, output SUM(n) in one line, followed by a blank line. You may assume the result will be in the range of 32-bit signed integer.

  • Mine

    #include<stdio.h>
    
    int main()
    {
        int x,n,sum;
        scanf("%d",&x);
        scanf("%d",&n);
        for(x=1;x<=n;x++)
        {
            sum=0;
            sum=sum+x;
        }
       printf("1
    %d",sum);
        
    }
    

  • Writeup

    #include <stdio.h>
    int main()
    {
        int a;
        int sum=0;
        while((scanf("%d",&a))!=EOF){
            for(int i=0;i<=a;i++)
                sum = sum+i;
            printf("%d
    
    ",sum);
            sum = 0;
        }
        return 0;
    }
    

  • Review

    • 我的问题在于没有考虑连续读取的可能性，好像对输入输出有误解T-T
    • 有一个疑问：输入输出需要和sample一样的格式吗？
    • 本身是一个前N项和的累加问题，如果用公式法也是no accept,原因是S=(1+n)*n/2中乘法容易造成溢出，而循环累加的好处在于溢出的可能比较小。

• 杭电OJ-1002

  • Description

    Given two integers A and B, your job is to calculate the Sum of A + B.

  • Input

    The first line of the input contains an integer T(1<=T<=20) which means the number of test cases. Then T lines follow, each line consists of two positive integers, A and B. Notice that the integers are very large, that means you should not process them by using 32-bit integer. You may assume the length of each integer will not exceed 1000.

  • Output

    For each test case, you should output two lines. The first line is "Case #:", # means the number of the test case. The second line is the an equation "A + B = Sum", Sum means the result of A + B. Note there are some spaces int the equation. Output a blank line between two test cases.

  • 手写代码@V1-20210805

    #include<stdio.h>
    #include<string.h>
    #define max 1005
    
    int main()
    {
        int a[max],b[max],c[max];
        while(~scanf("%s1 %s2",s1,s2))
        {
            int i,j,k;
            i=0；j=0;
            k=strlen(s1)>strlen(s2)?strlen(s1):strlen(s2)
            
            for(i;i<=k;i++)  //k有无定义的必要？
            {
                a[]=s1;   //好像不太合适？字符数组能直接赋值给数组吗？查一下书
                b[]=s2;   //似乎需要循环读入？
                
                c[i]=a[i]+c[i];
                if(c[i]>=10)
                {
                    c[i]=c[i]%10;
                    c[i+1]++;         //s2的长度是必要的吗？
                }
            }
            for(j=0;j<=k;j++)
            {
                printf("%d",a[j]);   //哪里有点奇怪
            }
        }
        
        
    }
    

  • v1现在存在的问题

    • 读入的顺序和相加的顺序不太对？要处理一下？[n-i]好像可行？
    • 输出的时候应该是逆序？
    • 需要加一个读入的限制条件：读取正整数？
    • 题目要求的输入输出 用一个循环？

  • 手写代码@V2-20210806

    #include<stdio.h>
    #include<string.h>
    #define max 1010
    
    int main()
    {
    	int T,u,n,i,j;
        char s1[max],s2[max],c[max];
    	scanf("%d",&T);
    	while(T>=1 && T<=20)
    	{
    		for(u=0;u<=T;u++)
    		{
    			scanf("%s %s",s1,s2);
    		}	
    		
    		 n =strlen(s1)>strlen(s2)?strlen(s1):strlen(s2);
    		 
    		 for(i=2 ; i<=n ; i++)
                     {
                        c[n-i]=s1[n-i]+s2[n-i];
                            if(c[i]>=10)
                                {
                                    c[n-i]=c[n-i]%10;
                                    c[n-i-1]++;         //没有考虑不是同位数的情况
                                }
                      } 
                     
    		
                     for(j=0;j<=n;j++)
                     {
                        printf("case %d:
    ",u);
                        printf("%s + %s = %d
    ",s1,s2,c[j]);   
                     }     
                
    	}
    	
    }
    

  • writeup-0806

    #include<stdio.h>
    #include<string.h>
    #define max 1000+10
    
    /**
     * 1. define the variable
    **/
    
    int a[max],b[max];    
    char str1[max],str2[max];  
    
    int main(){
    
    	int m;    //test number T?
    	int k=1;
    	scanf("%d",&m);  // read T?
    
    /**
     *2.read number and make sure input.
      this part is to read the big number and covert to array
    **/
    
    	while(m--){  //this circle ie funny! it's better than mine which use more variable u.              
                         
            scanf("%s %s",str1,str2);  //read big number
            memset(a,0,sizeof(a));    
            memset(b,0,sizeof(b));  //memset() 函数可以说是初始化内存的“万能函数”，通常为新申请的内存进行初始化工作。
    
            int i,j;
            for(i=0,j=strlen(str1)-1;i<strlen(str1);i++){     //it's similar to my code ,i use the i=1 to replace len 
            a[j--]=str1[i]-'0';   //string1 covert to array1?
    	}
     
    	for(i=0,j=strlen(str2)-1;i<strlen(str2);i++){      //two strlen conditions,less code ,nice~
    	b[j--]=str2[i]-'0';    //string2 to array2?  why not combine with above 'for cicle' together ?
    	}
    /**
     * 3. add two big number 
    **/   
    	for(i=0;i<max;i++){
    		a[i]+=b[i];
    		if(a[i]>=10){
    			a[i]-=10;     //mine : a[i]=a[i]%10
    			a[i+1]+=1;    //similar~
    		}
    	}
    /**
     * 4.output
     * 
    **/
    	printf("Case %d:
    ",k++);   // k has been defined and value=1?
    	printf("%s + %s = ",str1,str2);   
        for(i=max-1;(i>=0)&&(a[i]==0);i--); //reverse output？(i>=0)&&(a[i]==0) what's mean? only deal the 10?
    	if(i>=0){
    		for(;i>=0;i--){
    			printf("%d",a[i]);
    		}
    	}
    	else printf("0");
    	if(m!=0) printf("
    
    ");
    	else printf("
    ");                   //not clear..
    	}
    	return 0;
    }
    

  • Review

    • 大数加法问题一般考虑数组进行存储，然后按位相加满十进一。
    • 再看Writeup时发现大家再提java,import java.util.Scanner，以及大数需要import java.math.BigInteger,且BigInterger相加不是"a+b"，而是"a.add(b)",就可以很好的解决大数问题。

  import java.util.Scanner;
  import java.math.BigInteger;
  public class Main{
      public static void main(String args[]){
          BigInteger a,b;
          int T;
          int n=1;
          Scanner in = new Scanner(System.in);
          T=in.nextInt();
          while(T>0){
              a=in.nextBigInteger();
              b=in.nextBigInteger();
              System.out.println("Case "+n+":");
              System.out.println(a+" + "+b+" = "+a.add(b));
              if(T!=1) System.out.println();
              T--;
              n++;
          }
      }
  }
  

• 杭电OJ-1089

  • Input

    The input will consist of a series of pairs of integers a and b, separated by a space, one pair of integers per line.

  • Output

    For each pair of input integers a and b you should output the sum of a and b in one line, and with one line of output for each line in input.

  • **Mine **- accepted

    #include<stdio.h>
    
    int main()
    {
    
        int a,b;
        while(~scanf("%d %d",&a,&b))
        {
            printf("%d
    ",a+b);
        
        }
        return 0;
    
    }
    

• 杭电OJ-1090

  • Input

    Input contains an integer N in the first line, and then N lines follow. Each line consists of a pair of integers a and b, separated by a space, one pair of integers per line.

  • Output

    For each pair of input integers a and b you should output the sum of a and b in one line, and with one line of output for each line in input.

  • Mine

    #include<stdio.h>
    
    int i,a,b,c;
    int main()
    {
    	while(~scanf("%d",&i))
    	{
    		for(c=1;c<=i;c++)
            {
                scanf("%d %d",&a,&b);
                printf("%d
    ",a+b);
    		}
    	}
    }
    

• 杭电OJ-1091

  • Input

    Input contains multiple test cases. Each test case contains a pair of integers a and b, one pair of integers per line. A test case containing 0 0 terminates the input and this test case is not to be processed.

  • Output

    For each pair of input integers a and b you should output the sum of a and b in one line, and with one line of output for each line in input.

  • Mine

    #include<stdio.h>
    
    int a,b;
    int main()
    {
    	while(~scanf("%d %d",&a,&b))
    	{
    		if(a==0 && b==0)
    		{return 0;}
    		else
    		{
    			printf("%d
    ",a+b);
    		}
    	}
    }
    

• 杭电OJ-1092

  • Input

    Input contains multiple test cases. Each test case contains a integer N, and then N integers follow in the same line. A test case starting with 0 terminates the input and this test case is not to be processed.

  • Output

    For each group of input integers you should output their sum in one line, and with one line of output for each line in input.

  • Mine

    #include<stdio.h>   //始终是wrong...
    
    int main()
    {
    	int a,n,sum;
        while(scanf("%d",&n)!=EOF && n!=0)
        {
            while(n--)
            {
            	scanf("%d",&a);
            	sum=0;
            	sum+=a;
            }   
           printf("%d
    ",sum);    
        }
        return 0;
    }
    

  • Writeup

    #include<stdio.h>
    int main()
    {
    	int a[10000];
    	int n, i, s;
    	while (scanf("%d", &n) && n)  
    //输入正确scanf返回1，n!=0继续输入
    	{
    		for (i = s = 0; i < n; i++)
    			scanf("%d", &a[i]);
    		for (i = 0; i < n; i++)
    		{
    			s = s + a[i];
    		}
    		printf("%d
    ", s);
    	}
    	return 0;
    }
    
    

  • Review

    没有考虑到大数的可能，存在溢出问题，用数组存放更为合理

• 杭电OJ-1093

  • Input

    Input contains an integer N in the first line, and then N lines follow. Each line starts with a integer M, and then M integers follow in the same line.

  • Output

    For each group of input integers you should output their sum in one line, and with one line of output for each line in input.

  • Mine

    #include<stdio.h>   //wrong answer...
    
    int main()
    {
    	int i,j,k,s;
    	int a[1000];
    	while(~scanf("%d",&i))
    	{
    		for(i;i>=1;i--)
    		{
    			while(scanf("%d",&j)!=EOF && j)
    			{
    				for(j;j>=0;j--)
                    {
                       scanf("%d", &a[i]); 
                    }
                    for (k= 0; k< j; k++)
                    {
                        s += a[i];
                    }
    		printf("%d
    ", s);
    			}
    		}
    	}
    }
    
    

  • writeup

    #include<stdio.h>
    int main()
    {
    	int n,i,m,sum;    //n是行数，m是加数个数
    	scanf("%d",&n);
    	while(n--)  //简洁！
    	{
    		sum=0;	
    	    scanf("%d",&m);
    		while(m--)    //两个-- 简洁欸！
    		{
    			scanf("%d",&i);
    			sum=sum+i;
    		}	
    		printf("%d
    ",sum);	
    	}
    	return 0;
    }	
    

  • Review

    所以上一道题不是大数的原因？？ (((φ(◎ロ◎;)φ)))

• 杭电OJ-1094

  • Input

    Input contains multiple test cases, and one case one line. Each case starts with an integer N, and then N integers follow in the same line.

  • Output

    For each test case you should output the sum of N integers in one line, and with one line of output for each line in input.

  • Mine

    #include<stdio.h>
    
    int main()
    {
    	int n,m,sum;
    	while(~scanf("%d",&n))
        {
            sum = 0;
            while(n--)
            {
                scanf("%d",&m);
                sum += m;
            }
            printf("%d
    ",sum);
        }
    }
    

• 杭电OJ-1095

  • Input

    The input will consist of a series of pairs of integers a and b, separated by a space, one pair of integers per line.

  • Output

    For each pair of input integers a and b you should output the sum of a and b, and followed by a blank line.

  • Mine

    #include<stdio.h>
    int main()
    {
    	int a,b;
    	while(~scanf("%d %d",&a,&b))
    	{
    		printf("%d
    ",a+b);
    		printf("
    ");
    	}
    	return 0;
    }	
    	
    
    

• 杭电OJ-1096

  • Input

    Input contains an integer N in the first line, and then N lines follow. Each line starts with a integer M, and then M integers follow in the same line.

  • Output

    For each group of input integers you should output their sum in one line, and you must note that there is a blank line between outputs.

  • Mine

    #include<stdio.h>
    int main()
    {
    	int n,m,i;
    	int sum;
    	while(~scanf("%d
    ",&n))  //行数
    	{
    		while(n--)
    		{
    			scanf("%d",&m);   //加数个数
    			sum = 0;
    			while(m--)
    			{
    				scanf("%d",&i);
    				sum += i;
    			}
    			printf("%d
    ",sum);
    			if(m!=0)
    			printf("
    ",sum);
    		}
    	}
    }