A + B Problem II
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 203220 Accepted Submission(s): 39047
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
解析:其實就是實現加法運算,數組模擬大數加法,輸入的時候採用字符串輸入,一定要考慮到如果兩數的長度不同的情況,還有加法進位的處理,最後還有輸出的+=左右都有空格,最後那組數據不再輸出空行,需要考慮的情況還真多,wa了幾次哈!
貼一下自己的代碼
#include<iostream>
#include <string>
#include<algorithm>
using std::endl;
using std::cin;
using std::cout;
using std::string;
int main()
{
#ifdef LOCAL
freopen("input.txt" , "r" , stdin);
freopen("output.txt" , "w" , stdout);
#endif
int T;
cin >> T;
string a , b , sum;
for(int cases = 1; cases<=T; ++cases)
{
sum.clear();
cin >> a >> b;
cout << "Case " << cases << ":" << endl;
cout << a <<" + "<<b << " = ";
int c = 0 ;
reverse(a.begin() , a.end());
reverse(b.begin() , b.end());
int alength = a.length();
int blength = b.length();
int length = alength;
//使兩個字符串的長度相等,補齊零
if(alength < blength)
{
for(int i=alength; i<blength; ++i)
{
a += '0';
}
length = blength;
}
if(blength < alength)
{
for(int i=blength; i<alength; ++i)
{
b +='0';
}
length = alength;
}
//進行加法計算,c爲進位
for(int i=0; i<length; ++i)
{
int temp = (a[i] - '0') + (b[i] - '0') +c;
c = temp/10;
sum += (temp%10 + '0');
}
//最高位的進位如果不爲0的話則進位
if(c!=0)
{
sum += (c +'0');
}
//結果轉置
reverse(sum.begin() , sum.end());
cout << sum << endl;
//控制最後的那組數據不再輸出空行
if(cases != T)
{
cout << endl;
}
}
}