<hdu - 1002> A + B Problem II-白红宇

<hdu - 1002> A + B Problem II

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发布时间：2019-03-10

本文共 3662 字，大约阅读时间需要 12 分钟。

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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：

1 2

112233445566778899 998877665544332211

Sample Output：

Case 1:

1 + 2 = 3

Case 2:

112233445566778899 + 998877665544332211 = 1111111111111111110

解题思路：刚开始做这道题的时候，以为是A+B的题也难不到哪去，谁知道改了好几个小时。开始试了int，不行，又改long，不行，然后改long long，还是不行。只能另外想一种方法了：用大位数加法，用数组装数字，然后从个位开始逐位相加，过十进一。开始的时候想肯定是用数组了，肯定不会是字符数组。但是做着做着发现不行，数组的长度无法确定，想起来字符数组有个strlen的用法，就改用字符数组了。

以下是具体代码：

1 #include 
       
          2 #include 
        
           3 #include 
         
            4 #define N 1005  5 using namespace std;  6   7 int count = 0;  8 int a[N];  9 int b[N]; 10 int he[N]; 11 char sza[N]; 12 char szb[N]; 13  14 void calcuation() { 15     int flag = 1; 16     int len1,len2; 17     len1 = strlen(sza); 18     len2 = strlen(szb); 19     int i,j,k,l; 20     i = 0; 21     while(sza[i] != '\0') { 22         a[i] = sza[i] - '0'; 23         i++; 24     } 25     i = 0; 26     while(szb[i] != '\0') { 27         b[i] = szb[i] - '0'; 28         i++; 29     } 30     if(len1 < len2)  {l = len1-1;flag = 0;} 31     if(len1 > len2)  {l = len2-1;flag = -1;} 32     if(len1 == len2) l = len1-1; 33     k = 0; 34     int num = 0; 35     int m,n; 36     m=len1-1; 37     n=len2-1; 38     while(l >= 0) { 39         he[k] += a[m--] + b[n--]; 40          41         if(he[k] >= 10) { 42             num = he[k] / 10; 43             he[k] = he[k] % 10; 44             he[k+1] = num; 45             num = 0; 46         } 47         k++; 48         l--; 49     } 50     i = k; 51     j = k; 52     if(flag == 0) {//len1 < len2 53         for(k = len2-1-i;k >= 0; k--) { 54             he[j] += b[k]; 55             j++; 56         } 57     } 58     if(flag == -1) {//len1 > len2 59         for(k = len1-1-i;k >= 0; --k) { 60             he[j] += a[k]; 61             j++; 62         } 63     } 64 } 65  66 void find_print() { 67     int loc; 68     calcuation(); 69     printf("Case %d:\n",count); 70 //    cout << "Case";//此处考虑到用cout输出需要的行数多，故使用printf输出 71 //    cout << count; 72 //    cout << ":\n"; 73     int i; 74     i = 0; 75     while(sza[i]!='\0') { 76         printf("%c",sza[i]); 77         i++; 78     } 79     cout << " + "; 80     i = 0; 81     while(szb[i]!='\0') { 82         printf("%c",szb[i]); 83         i++; 84     } 85     cout << " = "; 86     for(i = N; i >= 0; i--) { 87         if(he[i] != 0) { 88             loc = i; 89             break; 90         } 91     } 92     for(i = loc;i >= 0; i--) { 93         printf("%d",he[i]); 94     } 95     cout << endl; 96 } 97  98 int main() { 99     int n;100     cin >> n;101     while(n--) {102         count++;103         memset(he,0,sizeof(he));104         memset(a,0,sizeof(a));105         memset(b,0,sizeof(b));106         cin >> sza >> szb;107         find_print();108         if(n != 0)///根据题意最后用换行109             cout << endl;110     }111     return 0;112 }

View Code

方法仅供参考，我知道肯定不是最简单的方法，有时间会再改善的，谢谢码友支持，欢迎评论。

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