HDOJ1002(大数相加)

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A + B Problem II

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 356833 Accepted Submission(s): 69210

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

思路：

问题描述：大数相加超出Int范围，考虑从个位数相加开始，并将每一位存放在int类型的数组中。

注意：

1.两个数的位数可能不同；

2.结果位数可能增加；

#include <iostream>#include <string>using namespace std;int main(){int sum[1001],len1,len2,d;int n = 0,index = 1,k =0;string str1,str2;cin>>n;while(n--){d = 0;cin>>str1>>str2;for(len1= str1.size()-1,len2 = str2.size()-1,k = 0;len1 >=0 && len2 >= 0;len1--,len2--,k++){d = str1[len1] - '0' + str2[len2] - '0' + d;sum[k]=d % 10;d = d /10; }if(k ==str1.size()){while(len2 >= 0){d = str2[len2--]-'0'+d;sum[k++] = d % 10;d = d /10;}}else{while(len1 >=0){d = str1[len1--] - '0' + d;sum[k++] = d %10;d = d / 10;}}if(d != 0){sum[k++] = d;}cout<<"Case "<<index<<":"<<endl;cout<<str1<<" + "<<str2<<" = ";for (int j = k-1;j >= 0;j--)cout<<sum[j];cout<<endl;if(0 != n)cout<<endl;++index;}return 0;}

注意：输出的结果每一行之间都需要空一行，最后一个输出不需要

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