1002 A + B Problem II

A + B Problem II

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 353608    Accepted Submission(s): 68625

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

代码

#include <iostream>#include <string>using namespace std;string sum(string s1, string s2){    if (s1.length()<s2.length())    {        string temp = s1;        s1 = s2;        s2 = temp;    }    int i, j;    for (i = s1.length() - 1, j = s2.length() - 1; i >= 0; i--, j--)    {        s1[i] = char(s1[i] + (j >= 0 ? s2[j] - '0' : 0));   //注意细节          if (s1[i] - '0' >= 10)        {            s1[i] = char((s1[i] - '0') % 10 + '0');            if (i) s1[i - 1]++;            else s1 = '1' + s1;        }    }    return s1;}int main(){    int n;    int k = 0, m = 0;    cin >> n;    while (k++ < n)    {        //题意要求保留原来的大数         string a, b, s;        cin >> a >> b;        cout << "Case " << k << ":" << endl;        cout << a << " + " << b << " = ";        if (a.length() < b.length())        {            string temp = a;            a = b;            b = temp;        }        s = a;        //length后的-1不能忘记         for (int i = a.length() - 1, j = b.length() - 1; i >= 0; --i, --j)        {            if (j >= 0)                s[i] = s[i] + b[j] - '0';            if (s[i] > '9')            {                s[i] = s[i] - 10;                if (i)                    ++s[i - 1];                else                    //最高位需要进位的时候，在最前面+1位                     s = '1' + s;            }        }        //s = sum(a, b);                 cout << s << endl;        if (++m < n)            cout << endl;    }    return 0;}

大数运算，写个模板就可以了，HDU上输出格式尤其要注意，经常性的格式出错