Are They Equal (25)
难度: ⭐⭐⭐
题目连接
题目描述
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*105 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.
输入描述:
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 A and 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.
输出描述:
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.d1…dN*10^k” (d1>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.
输入例子:
3 12300 12358.9
输出例子:
YES 0.123*10^5
大意
给出两个浮点数f1, f2 和一个整数n,判断这两个浮点数在类似于科学记数法的形式下是否相等,并输出这两个浮点数用这种计数法表示的形式。这种计数法只保存前面n位有效数字和指数,形式为0.xxxxx * 10 ^ k。
分析
首先分析规律,然后用字符串处理的方法解决。要表示这种计数法只要得到一个浮点数的前面n为有效数字和指数即可。然而这样的方法对程序的健壮性有较高的要求,因为问题的输入案例往往有很多你意料不到的输入,有时不看失败样例的话很难猜到是什么情况导致答案不对。另外比较坑的是不足n位要在后面补0题目没有说明。
MYCODE
#include<iostream>
#include<stdio.h>
#include<string>
#include<string.h>
using namespace std;
//得到指数值
int getExp(string s){
int common = -1;
for (int i = 0; i < s.length(); i++){ //寻找逗号位置
if (s[i] == '.') {
common = i;
break;
}
}
int isNum = -1;
for (int i = 0; i < s.length(); i++){ //寻找第一个有效数字位置
if (s[i]>'0' && s[i] <= '9'){
isNum = i;
break;
}
}
if (isNum == -1) return 0; //有可能为 00000.0
if (common == -1 && isNum != 0) return 1; //如00001这样的数
if (common == -1) return s.length();
int ans = common - isNum;
if (ans >= 0) return ans;
return ans + 1;
}
//得到前n为有效数字
string getUse(string s, int n){
int sss = n;
string temp = "";
int i = 0;
for (; i < s.length() && (s[i] == '0' || s[i] == '.'); i++);
for (; i < s.length() && n; i++){
if (s[i] != '.'){ temp += s[i]; n--; }
}
if (temp == "") temp = "0";
temp = "0." + temp;
while (temp.length() <= sss+1) temp += "0";
return temp;
}
struct mynum {
string use;
int exp;
void init(string s, int n){
this->use = getUse(s, n);
this->exp = getExp(s);
}
void printf(){
cout << use << "*10^" << exp;
}
};
int main(){
string s1,s2;
mynum t1, t2;
int n;
cin >> n >> s1 >> s2;
t1.init(s1,n);
t2.init(s2,n);
if (t1.exp == t2.exp && t1.use == t2.use){
cout << "YES ";
t1.printf();
cout << endl;
}
else{
cout << "NO ";
t1.printf();
cout << " ";
t2.printf();
cout << endl;
}
}
// 99 0 0.000
// 5 00000001 1