uva 465(高精度)

题目：

Write a program that reads an expression consisting of two non-negative integer and an operator. Determine if either integer or the result of the expression is too large to be represented as a ``normal'' signed integer (type integer if you are working Pascal, type int

input

An unspecified number of lines. Each line will contain an integer, one of the two operators + or *, and another integer.

output

For each line of input, print the input followed by 0-3 lines containing as many of these three messages as are appropriate: ``first number too big'', ``second number too big'', ``result too big''.

sample input

300 + 3
9999999999999999999999 + 11

sample output

300 + 3
9999999999999999999999 + 11
first number too big
result too big

题解：用正常做法，将两个数提取，将他们的结果也提取，判断是否大于int，，注意 0 * 这个陷阱。巧妙的方法有将两个数所在的字符串用atof()函数转成double类型，和int比大小。。。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
using namespace std;

const char c[10] = {'2', '1', '4', '7', '4', '8', '3', '6', '4', '7'};

int k, flag;
int judge(string s) {
	int len = s.size(), i;
	if (len > 10) { 
		for (i = 0; i < len; i++)
			if (s[i] != '0' && (len - i - 1) >= 10)
				return 1;
	}
	else if (len == 10 ) 
		for (i = 0; i < len; i++) {
			if (s[i] < c[i])
				break;
			if (s[i] > c[i])
				return 1;
		}
	return 0;
}
int result(string s1, string s2) {
	int ans[500] = {0}, a[500] = {0}, b[500] = {0};
	string anss = "";
	int i, j, q;
	int len1 = s1.size();
	int len2 = s2.size();
	if (flag == 1) {
		for (i = 0; i < len1; i++) 
			a[i] = s1[len1 - 1 - i] - '0';
		for (i = 0; i < len2; i++) 
			b[i] = s2[len2 - 1 - i] - '0';
		int lenx = len1 > len2 ? len1 : len2;
		for (i = 0; i < lenx; i++) {
			ans[i] = a[i] + b[i] + ans[i];
			ans[i + 1] = ans[i] / 10;
			ans[i] = ans[i] % 10;
		}
		if (ans[lenx])
			lenx++;
		for (j = lenx - 1; j >= 0; j--) 
			anss += ans[j] + '0';
//		cout << anss << endl;
		int a = judge(anss);
		if (a == 1)
			return 1;
		else
		 	return 0;
	}
	if (flag == 2) {
		int ans[500] = {0};
		for (i = len1 - 1; i >= 0; i--)
			for (j = len2 - 1, q = len1 - 1 - i; j >= 0; j--)
				ans[q++] += (s1[i] - '0') * (s2[j] - '0');
		for (i = 0; i < 500; i++) {
			ans[i + 1] += ans[i] / 10;
			ans[i] %= 10;
		}
		for (i = 499; i >=0; i--)
			if (ans[i] != 0)
				break;
		if (i == -1)
			return 0;
		else
			for (q = i; q >= 0; q--)
				anss += ans[q] + '0';
		int a = judge(anss);
		if (a == 1)
			return 1;
		else
			return 0;
	}
}

int main(){
	string str, str1, str2;
	while(getline(cin, str)) {
		k = 0;
		int length = str.size();
		str1 = "";
		str2 = "";
		flag = 0;
		for (int i = 0; i < length; i++)
			if(str[i] >= '0' && str[i] <= '9' && flag == 0)
				str1 += str[i];
			else if (str[i] == '+')
				flag = 1;
			else if (str[i] == '*')
				flag = 2;
			else if (str[i] >= '0' && str[i] <= '9' && flag != 0)
				str2 += str[i];
		cout << str << endl;
		int k1 = judge(str1);
		int k2 = judge(str2);
		if (k1 == 1) 
			cout << "first number too big" << endl;
		if (k2 == 1) 
			cout << "second number too big" << endl;
		if (flag == 2) {  //判断0 * ？ 结果不会大于int
			int i, j;
			if (k1 == 1) {
				int len2 = str2.size();
				for (i = 0; i < len2; i++)
					if (str2[i] != '0')
						break;
				if (i == len2)
					continue;
			}
			else if (k2 == 1) {
				int len1 = str1.size();
				for (j = 0; j < len1; j++)
					if (str1[j] != '0')
						break;
				if (j == len1)
					continue;
			}
		}
		if (k1 == 1 || k2 == 1)
			cout << "result too big" << endl;
		else
			if (result(str1, str2) == 1)
				cout << "result too big" << endl;
	}
	return 0;
}