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详细说明： 1002 A + B Problem II Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 69615 Accepted Submission(s): 12678 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 Author Ignatius.L #include #include int main(){ char str1[1001], str2[1001]; int t, i, len_str1, len_str2, len_max, num = 1, k; scanf("%d", &t); getchar(); while(t--){ int a[1001] = {0}, b[1001] = {0}, c[1001] = {0}; scanf("%s", str1); len_str1 = strlen(str1); for(i = 0; i <= len_str1 - 1; ++i) a[i] = str1[len_str1 - 1 - i] - '0'; scanf("%s",str2); len_str2 = strlen(str2); for(i = 0; i <= len_str2 - 1; ++i) b[i] = str2[len_str2 - 1 - i] - '0'; if(len_str1 > len_str2) len_max = len_str1; else len_max = len_str2; k = 0; for(i = 0; i <= len_max - 1; ++i){ c[i] = (a[i] + b[i] + k) % 10; k = (a[i] + b[i] + k) / 10; } if(k != 0) c[len_max] = 1; printf("Case %d:\n", num); num++; printf("%s + %s = ", str1, str2); if(c[len_max] == 1) printf("1"); for(i = len_max - 1; i >= 0; --i){ printf("%d", c[i]); } printf("\n"); if(t >= 1) printf("\n"); } return 0; } ...展开收缩
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