一个ACM思路

/*
Problem DescriptionI 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 OutputCase 1:
1 + 2 = 3

Case 2:
112233445566778899 + 998877665544332211 = 1111111111111111110

*/

 

//-----------------------------------------------------------------------------//Copyright (c) 2009 eryar All rights reserved.////File:main.cpp//Author:[e-mail]//Date:2009-8-5 23:29//Version:1.0v////Description:Given two integers A and B, //your job is to calculate the Sum of A + B.////=============================================================================#include <string>#include <iostream>using namespace std;// functions declarationvoidcalculate(string& sOne, string& sTwo);int main(int argc, char *argv[]){stringsNumberOne;stringsNumberTwo;cin>>sNumberOne>>sNumberTwo;//---------------------------------------calculate(sNumberOne, sNumberTwo);return 0;}/*Parameter:string& , string &Return:noneDescription:*/voidcalculate(string& sOne, string& sTwo){charcOne[1000];charcTwo[1000];intiOne[1000];intiTwo[1000];intiAnswer[1000];// 把字符串转换成字符数组strcpy(cOne, sOne.c_str());strcpy(cTwo, sTwo.c_str());// 把字符数组转换到整型变量for (int i = 0; i < sOne.length(); ++i) {iOne[i]=cOne[sOne.length() -1 - i]- '0';}for (int j = 0; j < sTwo.length(); ++j) {iTwo[j]=cTwo[sTwo.length() - 1 - j]- '0';}// 计算intk = (sOne.length() > sTwo.length()) ? sOne.length() : sTwo.length();for ( int l = 0; l < k ; ++l) {iAnswer[l]=iOne[l]+iTwo[l];if (iAnswer[l] >= 10) {iAnswer[l]=iAnswer[l] - 10;iOne[l+1]+=1;iAnswer[l+1]=1;}}// 输出for (int m=k; m>=0; --m) {cout<<iAnswer[m];}} // calculate