HDOJ A + B Problem II 大数的加法

A + B Problem II

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 3870 Accepted Submission(s): 1365 

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

21 2112233445566778899 998877665544332211

 

Sample Output

Case 1:1 + 2 = 3Case 2:112233445566778899 + 998877665544332211 = 1111111111111111110

 

Author

Ignatius.L

 

#include  #include  #include  #include  using namespace std;int main(){int t;cin >> t;int l = 1;while (t--){if (l > 1)cout << endl;string s1, s2;int sum[1010] = {0};int a[1010] = {0};int b[1010] = {0};cin >> s1;cin >> s2;for (int i = 0; i < s1.size(); i++){a[i] = s1[s1.size()-i-1] - '0';}for (int i = 0; i < s2.size(); i++){b[i] = s2[s2.size()-i-1] - '0';}int i = max(s1.size(), s2.size());for (int j = 0; j < i; j++){a[j] += b[j];a[j + 1]+= a[j] / 10;a[j] = a[j] % 10;}cout << "Case " << l++ << ":" << endl;cout << s1 << " + " << s2 << " = ";int q;for (int j = 1001; j >= 0; j--)//去前导0{if (a[j] != 0) { q = j; break; }}while (q>=0){cout << a[q--];}cout << endl;}return 0;}

A + B Problem II