POJ 2756 Autumn is a Genius 大数加减法

Description

Jiajia and Wind have a very cute daughter called Autumn. She is so clever that she can do integer additions when she was just 2 years old! Since a lot of people suspect that Autumn may make mistakes, please write a program to prove that Autumn is a real genius.
Input

The first line contains a single integer T, the number of test cases. The following lines contain 2 integers A, B(A, B < 32768) each. The size of input will not exceed 50K.
Output

The output should contain T lines, each with a single integer, representing the corresponding sum.
Sample Input

1
1 2

Sample Output

3

Hint

There may be '+' before the non-negative number!

本题题目没明白说明有多大的数，主要是A, B < 32768迷惑人，好像不是大数，只是后面 The size of input will not exceed 50K 的这句话就说明是大数了能够为接近无穷大的负数。

事实上50K就应该开多大的数组呢？50 * 1024 / 8 == 6400，所以会有6400个数位。

这里直接使用C++的vector或者string，然后输入使用buffer，那么就能够无论数位有多大了。

大数加法比較easy。假设是减法那么题目就比較麻烦了。眼下还想不到比較简洁的解法。要特殊处理符号，并且本题是两个数都可能是负数，那么就要分开情况讨论了，符号不同。绝对值大小不同都须要不同处理。情况分的好。那么程序就会相对简洁点。

#include <stdio.h>
#include <string>
#include <algorithm>
#include <vector>
using namespace std;

const int MAX_BUF = 512;
int id = 0, len = 0;
char buf[MAX_BUF];

char getFromBuf()
{
if (id >= len)
{
len = fread(buf, 1, MAX_BUF, stdin);
id = 0;
}
return buf[id++];
}

void getNum(vector<short> &num)
{
char c = getFromBuf();
while (('\n' == c || ' ' == c) && len) c = getFromBuf();
while ('\n' != c && ' ' != c && len)
{
num.push_back(c-'0');
c = getFromBuf();
}
}

void addBigNum(vector<short> &rs, vector<short> &A, vector<short> &B)
{
rs.clear();
if (A.empty())
{
rs = B;
return;
}
if (B.empty())
{
rs = A;
return;
}
int an = A[0] == '+'-'0' || A[0] == '-'-'0'? 1:0;
int bn = B[0] == '+'-'0' || B[0] == '-'-'0'? 1:0;
short carry = 0;
int i = (int)A.size()-1;
int j = (int)B.size()-1;
for (; i >= an || j >= bn || carry; i--, j--)
{
short a = i >= an? A[i] : 0;
short b = j >= bn? B[j] : 0;
carry += a + b;
rs.push_back(carry % 10);
carry /= 10;
}
reverse(rs.begin(), rs.end());
}

void minusBigNum(vector<short> &rs, vector<short> &A, vector<short> &B)
{
rs.clear();
if (B.empty())
{
if (A.empty()) return;
int i = A[0] == '-'-'0' || A[0] == '+'-'0'?

1 : 0;
for (; i < (int)A.size(); i++) rs.push_back(A[i]);
return ;
}
int an = A[0] == '-'-'0' || A[0] == '+'-'0' ? 1 : 0;
int bn = B[0] == '-'-'0' || B[0] == '+'-'0' ? 1 : 0;

short carry = 0;
int i = (int)A.size() - 1;
int j = (int)B.size() - 1;
for (; i >= an || j >= bn || carry != 0; i--, j--)
{
short a = i >= an ?

A[i] : 0;
short b = j >= bn ? B[j] : 0;
if (a > b)
{
short sum = a - b - carry;
carry = 0;
rs.push_back(sum);
}
else if (a < b)
{
short sum = 10 - (b - a) - carry;
carry = 1;
rs.push_back(sum);
}
else
{
short sum = 0;
if (carry) sum = 9;
rs.push_back(sum);
}
}
reverse(rs.begin(), rs.end());
}

int cmp(vector<short> &A, vector<short> &B)
{
int an, bn;
if (A.empty()) an = 0;
else an = A[0] == '-'-'0' || A[0] == '+'-'0'?

A.size()-1 : A.size();
if (B.empty()) bn = 0;
else bn = B[0] == '-'-'0' || B[0] == '+'-'0'?

B.size()-1 : B.size();

if (an > bn) return 1;
if (an < bn) return -1;
if (!an) return 0;

int i = A[0] == '-'-'0' || A[0] == '+'-'0'? 1 : 0;
int j = B[0] == '-'-'0' || B[0] == '+'-'0'? 1 : 0;

int res = 0;
for (; i < (int)A.size(); i++, j++)
{
if (A[i] < B[j]) res = -1;
else if (A[i] > B[j]) res = 1;
if (res != 0) break;
}
return res;
}

int main()
{
int T;
scanf("%d", &T);
while (T--)
{
vector<short> A, B, rs;
getNum(A);
getNum(B);

bool Asign = true, Bsign = true;
if (!A.empty() && A[0] == '-'-'0') Asign = false;
if (!B.empty() && B[0] == '-'-'0') Bsign = false;
if (Asign == Bsign)
{
addBigNum(rs, A, B);
if (!Asign) putchar('-');
}
else
{
int c = cmp(A, B);
if (0 == c) rs.push_back(0);
else if (c < 0)
{
minusBigNum(rs, B, A);
if (!Bsign) putchar('-');
}
else
{
minusBigNum(rs, A, B);
if (!Asign) putchar('-');
}
}
for (int i = 0; i < (int)rs.size(); i++)
{
printf("%d", rs[i]);
}
putchar('\n');
}
return 0;
}