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.
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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.
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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.