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=1010;
struct BigInteger {
  static const int BASE = 100000000;
  static const int WIDTH = 8;
  vector<int> s;

  BigInteger(long long num = 0) { *this = num; } 
  BigInteger operator = (long long num) {
    s.clear();
    do {
      s.push_back(num % BASE);
      num /= BASE;
    } while(num > 0);
    return *this;
  }
  BigInteger operator = (const string& str) {
    s.clear();
    int x, len = (str.length() - 1) / WIDTH + 1;
    for(int i = 0; i < len; i++) {
      int end = str.length() - i*WIDTH;
      int start = max(0, 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 = 0, g = 0; ; i++) {
      if(g == 0 && 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()-2; i >= 0; i--) {
    char buf[20];
    sprintf(buf, "%08d", x.s[i]);
    for(int j = 0; 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=0,tmp;
    BigInteger a,b;
    scanf("%d",&t);
    tmp=t;
    while(t--){
        cin>>a>>b;
        BigInteger sum=a+b;
        printf("Case %d:
",++times);
        cout<<a<<" + "<<b<<" = "<<sum<<endl;
        if(times!=tmp) printf("
");
    }
    return 0;
}