杭电ACM水题题目及代码.docx

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 = 11111111**********
 
Author

#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){
p