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\n",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\n%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\n\n",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 1005int 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 1010int 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:\n",u);
                      printf("%s + %s = %d\n",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:\n",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("\n\n");
  else printf("\n");                   //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\n",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\n",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\n",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\n",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\n", 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\n", 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\n",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\n",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\n",a+b);
  printf("\n");
  }
  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",&n))  //行数
  {
  while(n--)
  {
  scanf("%d",&m);   //加数个数
  sum = 0;
  while(m--)
  {
  scanf("%d",&i);
  sum += i;
  }
  printf("%d\n",sum);
  if(m!=0)
  printf("\n",sum);
  }
  }
  }