If a machine can save only 3 significant digits, the float numbers 12300 and 12358.9 are considered equal since they are both saved as 0.123×10^5 with simple chopping. Now given the number of significant digits on a machine and two float numbers, you are supposed to tell if they are treated equal in that machine.
Input Specification:
Each input file contains one test case which gives three numbers N, A and B, where N (<100) is the number of significant digits, and Aand B are the two float numbers to be compared. Each float number is non-negative, no greater than 10100, and that its total digit number is less than 100.
Output Specification:
For each test case, print in a line YES
if the two numbers are treated equal, and then the number in the standard form 0.d[1]...d[N]*10^k
(d[1]
>0 unless the number is 0); or NO
if they are not treated equal, and then the two numbers in their standard form. All the terms must be separated by a space, with no extra space at the end of a line.
Note: Simple chopping is assumed without rounding.
Sample Input 1:
3 12300 12358.9
Sample Output 1:
YES 0.123*10^5
Sample Input 2:
3 120 128
Sample Output 2:
NO 0.120*10^3 0.128*10^3
题意:
给出两个数,问:将他们写成保留N (<100)位小数的科学计数法 0.d[1]...d[N]*10^k
(d[1]
>0 unless the number is 0); 后,是否相同。若相等,输出YES,并转换结果; 如果不相等,输出NO,并给出两个数的转换结果。
思路:
本题的思路难想而且情况判断非常复杂。
题目要求科学计数法时,两个数是否相等。因此只要判断科学技术法时的本体部分以及指数部分是否相等即可。
对于数据来说,要分为大于 1 与小于 1 来判断,要考虑各种情况下的数字,比如:0000, 000.00, 00123.4, 0.012, 0.00 等
知识点:
string
- 添加 #include <string> 以及 using namespace std;
- 定义:string str = "abcd";
- 下标直接访问 string, str.length() 长度
- 读入和输出整个字符串,只能用 cin 和 cout
- 迭代器访问:一般采用 3 即可访问, 但 insert() 和 erase() 函数要求以迭代器为参数,定义为:string::iterator it; 得到迭代器 it, 可以采用 *it 来访问 string 里的每一位
- erase() 删除单个元素:erase(it),it 为需要删除的元素的迭代器
#include <iostream>
#include <string>
using namespace std;
int n; //精度
string deal(string s, int& e){ //s为输入的字符串,e 为指数
int k = 0;
while(s.length() > 0 && s[0] == '0'){
s.erase(s.begin()); //取出s的前导零
}
if(s[0] == '.'){ //若去掉 0 之后,遇到小数点,则说明s是小于1的小数
s.erase(s.begin()); //去掉小数点
while(s.length() > 0 && s[0] == '0'){
s.erase(s.begin()); //去掉小数点后非零位之前的所有零
e--; //每去掉一个零,指数e减一
}
}
else{ //若去掉前导0之后,没有小数点,则找后面的小数点删除
while(k < s.length() && s[k] != '.'){ //寻找小数点
k++;
e++; //只要不遇到小数点,就让指数++
}
if(k < s.length()){ //while结束后,若k < s.length(),说明遇到了小数点
s.erase(s.begin() + k); //把小数点删除
}
}
if(s.length() == 0){
e = 0; //如果去除前导0之后s的长度变为0,则说明这个数是0
}
//此时,需要处理的就是大于1的数字
int num = 0;
k = 0;
string res; //返回值
while(num < n){ //只要精度没有到达n
if(k < s.length()) //只要还有数字,就添加到res末尾
res += s[k++];
else
res += '0'; //否则res末尾添加0
num++; //精度加1
}
return res;
}
int main(){
string s1, s2, s3, s4;
cin >> n >> s1 >> s2;
int e1 = 0, e2 = 0; //指数
s3 = deal(s1, e1);
s4 = deal(s2, e2);
if(s3 == s4 && e1 == e2){
cout << "YES 0." << s3 << "*10^" << e1 << endl;
}
else{
cout << "NO 0." << s3 << "*10^" << e1 << " 0." << s4 << "*10^" << e2 << endl;
}
return 0;
}