A + B Problem II
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 426568 Accepted Submission(s): 82846
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
深夜做水题(更新:完全不是水题!!本以为是个青铜没想到是王者!!)。一开始想只用stack,后来发现太麻烦,于是加上了deque。提交的时候又接连出现两次错误,一次Compilation Error,
error C2679: 二进制“>>”: 没有找到接受“std::string”类型的右操作数的运算符(或没有可接受的转换)
查了一下发现是头文件要写成#include<string>而不能是#include<cstring>和#include<string.h>。看来以后写成#include<string>保险点。
还有一次错误是Presentation Error(罪魁祸首就是它!,为了这个错我懵逼了半个多小时!),除了最后一组测试是一个换行,其他都是两个换行,ACM的题还是太讲究了。
①利用栈先进后出(LIFO)的特点做运算
②为了防止sum的第一位进位可以求出sum后加个前导0再处理进位。最后要去掉前导0
测试数据补充
0001 1000
0 0
000 0000
9999 1
1 9999
99900 00999
00999 99900
AC代码:
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 #include<iostream> 2 #include<string> 3 #include<stack> 4 #include<deque> 5 using namespace std; 6 string Sum(string a,string b) 7 { 8 stack<int>num1; 9 stack<int>num2; 10 deque<int>sum; 11 string ans; 12 int i,la=a.length(),lb=b.length(); 13 for(i=0;i<la;i++) num1.push(a[i]-'0'); 14 for(i=0;i<lb;i++) num2.push(b[i]-'0'); 15 while(!num1.empty()||!num2.empty()) 16 { 17 if(num1.empty()) 18 { 19 sum.push_front(num2.top()); 20 num2.pop(); 21 } 22 else if(num2.empty()) 23 { 24 sum.push_front(num1.top()); 25 num1.pop(); 26 } 27 else 28 { 29 sum.push_front(num1.top()+num2.top()); 30 num1.pop();num2.pop(); 31 } 32 } 33 sum.push_front(0);//嵌入一个0 34 for(i=sum.size()-1;i>=0;i--) //处理进位 35 if(sum[i]>=10) 36 { 37 sum[i]-=10; 38 sum[i-1]+=1; 39 } 40 if(sum.front()>=10) 41 { 42 sum.front()-=10; 43 sum.push_front(1); 44 } 45 while(!sum.empty()) 46 { 47 ans+=char(sum.front()+'0'); 48 sum.pop_front(); 49 } 50 while(ans.find("0",0)==0&&ans.length()>1) ans=ans.substr(1);//去掉前导0 51 return ans; 52 } 53 int main() 54 { 55 int T,i=1; 56 cin>>T; 57 string A,B; 58 while(T--) 59 { 60 cin>>A>>B; 61 cout<<"Case "<<i++<<':'<<endl; 62 cout<<A<<" + "<<B<<" = ";//注意空格 63 cout<<Sum(A,B)<<endl; 64 if(T>0) cout<<endl;//除了最后一组测试是一个换行,其他都是两个换行 65 } 66 }