HDOJ -- 1002 大数A+B

发布时间：2021-01-18 07:28:23 所属栏目：大数据来源：网络整理

导读：A + B Problem II Time Limit: 2000/1000 MS (Java/Others)????Memory Limit: 65536/32768 K (Java/Others) Problem Description I have a very simple problem for you. Given two integers A and B,your job is to calculate the Sum of A + B. ? Input

A + B Problem II

Time Limit: 2000/1000 MS (Java/Others)????Memory Limit: 65536/32768 K (Java/Others) Problem Description I have a very simple problem for you. 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.
?
Sample Input

  
  
   
   2
1 2
112233445566778899 998877665544332211

?
Sample Output

  
  
   
   Case 1:
1 + 2 = 3

Case 2:
112233445566778899 + 998877665544332211 = 2222222222222221110

?题解：对于此题做法有两种：其一，使2字符串的中的字符数字减去'0'，逐个相加大于等于10的可以使本位减10，下一位自增1，后面的处理就非常简单了；其二，便是读入字符串后先让各个字符减'0'，一一对应存入整形数组中;之后再相加。对于2种方法大致是相同的，都要从后面向前加，逢十进位，以及数组初始化均要初始为0，方便运算。下面用法一，代码如下:

#include<stdio.h>
#include<string.h>
int main()
{
    int n,i,len1,len2,j,k,pi,t,flag;
    char str1[1010],str2[1010];
    scanf("%d",&n);
    for(i=1;i<=n;i++)
    {
        int a[1200]={0};
        flag=0;
        printf("Case %d:n",i);
        scanf("%s%s",str1,str2);//以字符串形式读入
        len1=strlen(str1);
        len2=strlen(str2);
        printf("%s + %s = ",str2);
        j=len1-1;
        k=len2-1;
        pi=0;
        while(j>=0&&k>=0)//开始相加
        {
            if(a[pi]+(str1[j]-'0')+(str2[k]-'0')>=10)//相加后大于10的
            {
                a[pi]=a[pi]+(str1[j]-'0')+(str2[k]-'0')-10;
                a[pi+1]++;
            }
            else
                a[pi]=a[pi]+(str1[j]-'0')+(str2[k]-'0');
            pi++;
            k--;
            j--;
        }
        if(j>=0)
        {
            for(t=j;t>=0;t--)
            {
                a[pi]=a[pi]+(str1[t]-'0');
                pi++;
            }
        }
        else if(k>=0)
        {
            for(t=k;t>=0;t--)
            {
                a[pi]=a[pi]+str2[t]-'0';
                pi++;
            }
        }
        else if(a[pi]!=0)//对于位数相同2个数加后最高位大于10的
            pi++;
        for(t=pi-1;t>=0;t--)
        {
            if(a[t]==0&&flag==0)//处理一次啊前导0，估计属于无用的一步
                continue;
            else
            {
                flag=1;
                printf("%d",a[t]);
            }
           
        }
        printf("n");
        if(i!=n)//对于2组之间加空行的情况
            printf("n");
    }
    return 0;   
}

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