//麻烦的水题//A + B Problem II ------一Z

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本文介绍了一种处理大整数相加问题的算法实现，通过逐位计算的方法解决了传统32位整数溢出的问题，并提供了一个具体的C语言代码示例。

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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 = 1111111111111111110

解题思路：
相当于自己写一个加法计算器，一位一位算，%10是当前位上的数，/10是进位。注意即使一个数的所有位数都加完，另一个数还要和剩下的进位一位一位继续计算；当两个数的位数都加完时，还要处理最后的进位。

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

char num1[1010],num2[1010];
int ans[1010]={0};

int main()
{
    int t,len1,len2,lenb,lens,flag;
    scanf("%d",&t);
    for(int j=1;j<=t;j++)
    {
        char temp=getchar();
        scanf("%s",num1);
        scanf("%s",num2);
        len1=strlen(num1);
        len2=strlen(num2);
        if(len1>len2)
        {
            lenb=len1;
            lens=len2;
            flag=1;
        }
        else
        {
            lenb=len2;
            lens=len1;
            flag=2;
        }
        int add=0;
        for(int i=1;i<=lenb;i++)
        {
            if(i<=lens)
            {
                ans[i]=(num1[len1-i]-'0'+num2[len2-i]-'0'+add)%10;
                add=(num1[len1-i]-'0'+num2[len2-i]-'0'+add)/10;
            }
            else
            {
                if(flag==1)
                {
                    ans[i]=(num1[len1-i]-'0'+add)%10;
                    add=(num1[len1-i]-'0'+add)/10;
                }
                else
                {
                    ans[i]=(num2[len2-i]-'0'+add)%10;
                    add=(num2[len2-i]-'0'+add)/10;
                }
            }
        }
        if(add!=0)
        {
            ans[lenb+1]=add;
            lenb=lenb+1;
        }
        printf("Case %d:\n",j);
        printf("%s + %s = ",num1,num2);
        for(int i=lenb;i>1;i--)
        {
            printf("%d",ans[i]);
        }
        printf("%d\n",ans[1]);
        if(j<t)
        {
            printf("\n");
        }
    }
}