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GDUFE ACM-1002

2024-08-16 23:00:55   222人阅读

题目：http://acm.gdufe.edu.cn/Problem/read/id/1002

 

A+B(Big Number Version)

Time Limit: 2000/1000ms (Java/Others)

Problem 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 400.

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:

3
1 2
112233445566778899 998877665544332211
33333333333333333333333333 100000000000000000000

Sample Output:

Case 1:
1 + 2 = 3

Case 2:
112233445566778899 + 998877665544332211 = 1111111111111111110

Case 3:
33333333333333333333333333 + 100000000000000000000 = 33333433333333333333333333


思路：400位数啊，明显是unusigned long long int是不够大的，所以就要自己写一个。所以我就想，把最后一位数相加，然后把十位数加到前一位数上（如果没有十位数，那就是0），最后把整个数输出来，因为不知道具体有多少位数，所以两个数的和我是倒着储存的（能看懂我的意思吗==）

难度：感觉有一定的难度，想了很长时间，主要是写的时候觉得有难度，想出来不算很难吧。要把字符串转变成整数数组。

代码：
 1 #include<stdio.h>
 2 #include<string.h>
 3 int main()
 4 {
 5     int n;
 6     while(scanf("%d",&n)!=EOF)
 7     {
 8         char ai[400],bi[400];
 9         int i,j,d,e,k,q=0,a[400],b[400],c[400];
10         while(n--)
11         {
12             q++;
13            getchar();
14            scanf("%s",ai);
15            scanf("%s",bi);
16            d=strlen(ai);
17            e=strlen(bi);
18            for(i=0;i<d;i++)
19             a[i]=ai[i]-‘0‘;
20            for(j=0;j<e;j++)
21             b[j]=bi[j]-‘0‘;
22             k=0;
23             c[0]=a[d-1]+b[e-1];
24             if(d>1||e>1)
25            for(k=1,i=d-2,j=e-2;;i--,j--,k++)
26            {
27                if(i>=0&&j>=0)
28                 c[k]=c[k-1]/10+a[i]+b[j];
29               else if(i>=0&&j<0)
30                c[k]=c[k-1]/10+a[i];
31               else if(i<0&&j>=0)
32                 c[k]=c[k-1]/10+b[j];
33                else if(i<0&&j<0)
34                {if(c[k-1]>=10)
35                 c[k]=1;
36                 else k--;break;}
37            }
38            printf("Case %d:\n",q);
39            printf("%s + %s = ",ai,bi);
40            for(;k>=0;k--)
41            {c[k]=c[k]%10;
42             printf("%d",c[k]);
43             }
44             printf("\n");
45             if(n>0)
46                 printf("\n");
47         }
48     }
49     return 0;
50 }

 

GDUFE ACM-1002


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