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 ;
}