hdu 1002 大数相加

原题链接：http://acm.hdu.edu.cn/showproblem.php?pid=1002

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

用字符串实现：

#include<iostream>#include<string>#include<cstring>using namespace std;string add(string s1,string s2);int main(){    int n,t=1;    string s1,s2,sum;    cin>>n;    while(n--)    {        cin>>s1>>s2;        sum = add(s1,s2);        cout << "Case " << t << ":" << endl ;        cout << s1 << " + " << s2 <<  " = " << sum << endl;        t++;        if(n>0)cout<<endl;    }    return 0;}string add(string s1,string s2){    if (s1 == "0" && s2 == "0") return "0"; //0的处理    if (s1 == "0") return s2;//0的处理    if (s2 == "0") return s1;//0的处理    string max = s1,min = s2;    if (s1.length()<s2.length())    {        max = s2;        min = s1;    }    int a = max.length()-1,b = min.length()-1;    for(int i=b;i>=0;i--)        {max[a--] += min[i] -'0';}    for(int i = max.length()-1;i>0;i--)    {        if(max[i]>'9')            {                max[i] -= 10;                max[i-1]++;            }    }    if(max[0]>'9')//位数相同，最高位大于9        {max[0] -=10;        max = '1'+max;}    return max;}