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
1 #include <stdio.h> 2 #include <string.h> 3 int main() 4 { 5 6 int x, sa[1001], sb[1001], k = 1, i, j, m, n, t, d; 7 char a[1001], b[1001]; 8 scanf("%d", &x); 9 while( x-- ){ 10 11 //int sa[1001] = {0}; 12 //int sb[1001] = {0};#1 13 memset(sa, 0, sizeof(sa)); 14 memset(sb, 0, sizeof(sb)); 15 scanf("%s", a); 16 scanf("%s", b); 17 m = strlen(a); 18 n = strlen(b); 19 for( i = 0, j = m - 1; j >= 0; j-- ) 20 { 21 22 //sa[i++] = a[j]; 23 sa[i++] = a[j] - '0'; 24 25 } 26 for( i = 0, j = n - 1; j >= 0; j-- ) 27 { 28 29 //sb[i++] = b[j]; 30 sb[i++] = b[j] - '0'; 31 } 32 t = m >= n ? m : n;/*找出最长的长度*/ 33 for( i = 0; i < t; i++ ) 34 { 35 36 sa[i] += sb[i]; 37 if( sa[i] >= 10 ) 38 { 39 40 sa[i] %= 10; 41 sa[i + 1]++; 42 43 } 44 45 } 46 //d = (sa[i] == 1) ? t : t - abs(m - n) ; 47 d = (sa[i] == 1) ? t : t - 1;/*如果最高位还进1位，那就从它的上位开始输出*/ 48 printf("Case %d: %s + %s = ", k++, a, b); 49 while( d >= 0 ){ 50 51 //printf("%d", d--); 52 printf("%d", sa[d--]); 53 54 } 55 putchar(' '); 56 //putchar(' ');细节问题：最后一次测试无空行 57 //return 0; 58 if( x == 0 ) 59 continue; 60 putchar(' '); 61 62 } 63 return 0; 64 65 }
note:
#1只有定义变量才可以全部赋值，但只能是0。一般不喜欢这样做，所以调用头文件<string.h>里的memset函数，它不仅可以全盘赋值，而且赋值自由。
#2 细节问题：因为Output a blank line between two test cases，所以最后一组数据后面没有空行