题目1471: A+B without carry

 

题目描述

Xiao Ming always tends to ignore the carry when he does decimal addition with paper and pencil.For example,15+17,Xiao Ming will answer 22,because he ignores the carry from the single digits.5+7=12,and the digit 1 is the carry. 

 

 

 

输入

The input will consist of a series of pairs of integers a and b(both less than 1000000000),separated by a space, one pair of integers per line.

 

 

输出

For each pair of input integers a and b you should output the correct answer of the sum of a and b,a space character and Xiao Ming's answer of the sum of a and b in one line,and with one line of output for each line in input.If Xiao Ming's answer begins with zero,don't output unnecessary zero.

 

 

样例输入

15 16

1 999

31 71

 

 

样例输出

31 21

1000 990

102 2

 

 

提示 [+]

*** 提示已隐藏，点击上方 [+] 可显示 ***

 

 

 

 

/*********************************
*   总结：
**********************************/
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
int main()
{
  int a,b,len1,len2,index,i,j;
  char str1[11],str2[11],c[11];
  //freopen("C:\\Users\\SJF\\Desktop\\acm.txt","r",stdin);
  while(scanf("%s %s",str1,str2)!=EOF){
    len1 = strlen(str1);
    len2 = strlen(str2);
    index = 0;
    //小明的计算过程
    for(i = len1-1,j = len2-1;i >= 0 && j >= 0;i--,j--){
      int sum = str1[i] - '0' + str2[j] - '0';
      //忽略进位
      if(sum > 9){
        sum -= 10;
      }
      c[index++] = sum + '0';
    }
    while(i >= 0){
      c[index++] = str1[i];
      i--;
    }
    while(j >= 0){
      c[index++] = str2[j];
      j--;
    }
    //去掉前导0
    index = index -1;
    while(c[index] == '0' && index > 0){
      index--;
    }
    //正确答案
    printf("%d ",atoi(str1) + atoi(str2));
    //小明的答案
    for(i = index;i >= 0;i--){
      printf("%c",c[i]);
    }
    printf("\n");


  }
  return 0;
}

 

 

*   总结：
**********************************/
#include<stdio.h>
#include<string.h>
#include<stdlib.h>

int main()
{
  int a,b,index;
  int c[12];
  //freopen("C:\\Users\\SJF\\Desktop\\acm.txt","r",stdin);
  while(scanf("%d %d",&a,&b)!=EOF){
    //正确答案
    printf("%d ",a + b);
    //求小名的答案
    if(a == 0 && b == 0){
      printf("%d",a + b);
    }
    index = 0;
    while(a || b){
      int sum = a % 10 + b % 10;
      if(sum > 9){
        sum -= 10;
      }
      c[index++] = sum;
      a /= 10;
      b /= 10;
    }
    //去掉前导0
    index = index - 1;
    while(c[index] == 0 && index > 0){
      index--;
    }
    //输出答案
    for(int i = index;i >= 0;i--){
      printf("%d",c[i]);
    }
    printf("\n");
  }
  return 0;
}

第二种方法时忘记了a = 0 b = 0的情况wrong了好几次.........

 

这种方法很有意思，自己没有想到.........

 

#include <stdio.h>
int a,b;
void run()
{
        int c,k;
        c=a+b;
        printf("%d ",c);//这个程序是由正常的和来反求不进位的和（或者直接按位相加也是可以的 ）。 
        k=1;
        while(k<=a||k<=b)
        {
                if(a/k%10+b/k%10>9)
                        c-=k*10;//在某一位上的进位被忽略，相当于总和减小了这一位权值的十倍，个位进位被忽略，总和就少了10。 
                k*=10;
        }
        printf("%d\n",c);
}
int main()
{
        while(scanf("%d%d",&a,&b)!=EOF)
                run();
        return 0;
}