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POJ 2756 Autumn is a Genius 大数加减法

时间:2016-01-28 19:20:26      阅读:251      评论:0      收藏:0      [点我收藏+]

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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; }





POJ 2756 Autumn is a Genius 大数加减法

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原文地址:http://www.cnblogs.com/mengfanrong/p/5167118.html

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