• 杭电_ACM_A + B Problem II


    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
    View Code
     1 #include <stdio.h>
     2 #include <string>
     3 #define MAX 1005
     4 int main()
     5 {
     6     int T, a[MAX], b[MAX], c[MAX], i, j, k, carry;
     7     char sa[MAX], sb[MAX];
     8     scanf("%d", &T);
     9     for (j = 1; j <= T; j++)
    10     {
    11         scanf("%s %s", sa, sb);
    12         memset(a, 0, sizeof(a));
    13         memset(b, 0, sizeof(b));
    14         carry = 0;
    15         //a[0] is the length of a, it's more convenient
    16         a[0] = strlen(sa);
    17         b[0] = strlen(sb);
    18         //translate the char to number and inverse
    19         for (i = 1; i <= a[0]; i++)
    20         {
    21             a[i] = sa[a[0] - i] - '0';
    22         }
    23         for (i = 1; i <= b[0]; i++)
    24         {
    25             b[i] = sb[b[0] - i] - '0';
    26         }
    27         //get the longer one for result
    28         c[0] = a[0] > b[0] ? a[0] : b[0];
    29         //add one by one and note the carry
    30         for (i = 1; i <= c[0]; i++)
    31         {
    32             c[i] = (a[i] + b[i] + carry) % 10;
    33             carry = (a[i] + b[i] + carry) / 10;
    34         }
    35         //if the last carry is not zero, you must add the one.
    36         if (carry != 0)
    37         {
    38             c[0]++;
    39             c[c[0]] = carry; 
    40         }
    41         //formatted printing
    42         printf("Case %d:\n", j);
    43         for (k = a[0]; k >= 1; k--)
    44             printf("%d", a[k]);
    45         printf(" + ");
    46         for (k = b[0]; k >= 1; k--)
    47             printf("%d", b[k]);
    48         printf(" = ");
    49         for (k = c[0]; k >= 1; k--)
    50             printf("%d", c[k]);
    51         printf("\n");
    52         if (j != T)
    53             printf("\n");
    54     }
    55     return 0;
    56 }

    firstly, making your mind clearly.

    secondly, in formatted printting, you should care about two lines or one line. 

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  • 原文地址:https://www.cnblogs.com/chuanlong/p/2752975.html
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