HDU----A + B Problem II

电玩女神 2022-06-15 12:51 150阅读 0赞

# A + B Problem II #

**Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)  
Total Submission(s): 359992    Accepted Submission(s): 69867**

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 = 1111111111111111110

#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 = ",str1,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--)
            {
                  printf("%d",a[t]);
            }
            printf("\n");
            if(i!=n)
                printf("\n");
        }
        return 0;
    }