A + B Problem II

Problem descriptioni have a very simple problem for you. given two integers A and B, your job is to calculate the sum of A + B. inputthe 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 shoshould not process them by using 32-bit integer. you may assume the length of each integer will not exceed 1000. outputfor 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 input21 2112233445566778899 998877665544332211 sample outputcase + 2 = 3 Case 2: 112233445566778899 + 998877665544332211 = 1111111111111111110

1 # include <stdio. h> 2 # include <string. H & gt; 3 # define n 1001 4 5 Int main () {6 int t; 7 char a [n]; 8 char B [N]; 9 int sum [N]; 10 char temp [N]; 11 int a_length; 12 INT B _length; 13 int I; 14 int length; // length difference 15 int first; 16 int second; 17 int flag; 18 int flag2; 19 int time; 20 21 scanf ("% d", & T); 22 time = 1; // time indicates the number of input data groups 23 24 while (t --) {25 flag = 0; 26 flag2 = 0; 27 scanf ("% S % s",, b); 28 29 printf ("Case % d: \ n", time); 30 time ++; 31 printf ("% S + % s =", a, B ); 32 33 a_length = strlen (a); 34 B _length = strlen (B); 35 36 IF (a_length <B _length) {// ensure that the length of a is greater than or equal to the length of B 37 strcpy (temp, a); 38 strcpy (a, B); 39 strcpy (B, temp ); 40} 41 42 a_length = strlen (a); 43 B _length = strlen (B); 44 length = A_length-B_length; 45 for (I = B _length-1; I> = 0; I --) {// move array B back 46 B [I + Length] = B [I]; 47} 48 49 for (I = 0; I <length; I ++) // Add 'before array B' 0 '50 B [I] = '0'; 51 52 B [a_length] = '\ 0'; // array B add end symbol 53 54 for (I = A_length-1; i> = 0; I --) {55 First = A [I]-'0'; 56 second = B [I]-'0 '; 57 sum [I] = (first + second) % 10; 58 59 if (I> 0) 60 a [I-1] = A [I-1]-'0' + (first + second)/10 + '0 '; 61} 62 63 first = A [0]-'0'; 64 second = B [0]-'0'; 65 66 flag = (first + second)/10; 67 68 if (FLAG) 69 printf ("% d", flag); 70 71 for (I = 0; I <a_length; I ++) {72 If (FLAG) {73 flag2 = 1; 74 printf ("% d", Sum [I]); 75} 76 77 else {78 If (sum [I]! = 0) {79 flag2 = 1; 80 printf ("% d", sum [I]); 81} 82 83 else {84 If (flag2 = 1) 85 printf ("% d", sum [I]); 86} 87} 88 89} 90 91 If (flag2 = 0) 92 printf ("0 "); 93 94 printf ("\ n"); 95 96 If (T! = 0) 97 printf ("\ n"); 98} 99 100 return 0; 101}

 

 

A + B Problem II