1060. Are They Equal (25)

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.

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 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.

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.d₁...d_N*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.

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

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判断两个数以题目给定的模式表示是否一样，并且输出结果。要考虑的情况比较多，比如123，0.002，0002.2，02.200003，0.，.023等形式的数字都可能有。我这里主要分成两大类，就是大于等于1的和小于1的。对于一个以字符串形式输入的数，首先先把前面的0排除掉。如果当前字符为‘.'且后面还有数，说明小于1，就遍历字符串，找出n位有效数字，如果不够n位就补零补足n位，同时也求出指数ex。如果当前字符不是‘.'，说明大于等于1，遍历字符串直到找到‘.'，同时添加有效数字，和求出指数ex，然后跳过‘.'，继续补全有效数字，如果最后也不足n个，就补零。最后比较有效数字就能判断两个数是否“相等“。最后按题目指定形式输出结果。

代码：

#include <iostream>#include <cstring>#include <cstdlib>#include <cstdio>#include <vector>using namespace std;int get_num(char *s,int n,string &num){int m=strlen(s);int i=0,ex;while(i<m&&s[i]=='0') i++;if(i<m&&s[i]=='.'){i++;ex=0;while(i<m&&s[i]=='0'){i++;ex--;}while(i<m&&num.size()<n){num+=s[i++];}while(num.size()<n){num+='0';}return ex;}else{ex=0;while(i<m&&s[i]!='.'){ex++;if(num.size()<n){num+=s[i];}i++;}i++;while(i<m&&num.size()<n){num+=s[i++];}while(num.size()<n){num+='0';}return ex;}}int main(){int n;char s1[105],s2[105];scanf("%d %s %s",&n,s1,s2);string res1,res2;int ex1=get_num(s1,n,res1);int ex2=get_num(s2,n,res2);if(res1[0]=='0') ex1=0;if(res2[0]=='0') ex2=0;if(res1==res2&&ex1==ex2){printf("YES 0.%s*10^%d",res1.c_str(),ex1);}else{printf("NO 0.%s*10^%d 0.%s*10^%d",res1.c_str(),ex1,res2.c_str(),ex2);}}