HDU-1002（简单大数加法）

A + B Problem II

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

 

 

分析：高精度大数的运算，直接套用刘汝佳老师的模板。注意坑点：每个case之间都要空一行，且最后一个case后不用空行。

 

 #include <cstdio>
 #include <cmath>
 #include <cstring>
 #include <ctime>
 #include <iostream>
 #include <istream>
 #include <ostream>
 #include <algorithm>
 #include <set>
 #include <vector>
 #include <sstream>
 #include <queue>
 #include <typeinfo>
 #include <fstream>
 #include <map>
 #include <stack>
 using namespace std;
 #define INF 100000
 typedef long long ll;
 const int maxn=;
 struct BigInteger {
   static const int BASE = ;
   static const int WIDTH = ;
   vector<int> s;

   BigInteger(long long num = ) { *this = num; }
   BigInteger operator = (long long num) {
     s.clear();
     do {
       s.push_back(num % BASE);
       num /= BASE;
     } while(num > );
     return *this;
   }
   BigInteger operator = (const string& str) {
     s.clear();
     int x, len = (str.length() - ) / WIDTH + ;
     for(int i = ; i < len; i++) {
       int end = str.length() - i*WIDTH;
       int start = max(, end - WIDTH);
       sscanf(str.substr(start, end-start).c_str(), "%d", &x);
       s.push_back(x);
     }
     return *this;
   }
   BigInteger operator + (const BigInteger& b) const {
     BigInteger c;
     c.s.clear();
     for(int i = , g = ; ; i++) {
       if(g ==  && i >= s.size() && i >= b.s.size()) break;
       int x = g;
       if(i < s.size()) x += s[i];
       if(i < b.s.size()) x += b.s[i];
       c.s.push_back(x % BASE);
       g = x / BASE;
     }
     return c;
   }
 };

 ostream& operator << (ostream &out, const BigInteger& x) {
   out << x.s.back();
   for(int i = x.s.size()-; i >= ; i--) {
     char buf[];
     sprintf(buf, "%08d", x.s[i]);
     for(int j = ; j < strlen(buf); j++) out << buf[j];
   }
   return out;
 }

 istream& operator >> (istream &in, BigInteger& x) {
   string s;
   if(!(in >> s)) return in;
   x = s;
   return in;
 }
 set<BigInteger> s;

 int main()
 {
     int t,times=,tmp;
     BigInteger a,b;
     scanf("%d",&t);
     tmp=t;
     while(t--){
         cin>>a>>b;
         BigInteger sum=a+b;
         printf("Case %d:\n",++times);
         cout<<a<<" + "<<b<<" = "<<sum<<endl;
         if(times!=tmp) printf("\n");
     }
     return ;
 }