1060. Are They Equal (25)-PAT

1060. Are They Equal (25)

時間限制 50 ms

記憶體限制 32000 kB

代碼長度限制 16000 B

判題程式 Standard 作者 CHEN, Yue

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...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.

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      

推薦指數：※※      

來源：http://pat.zju.edu.cn/contests/pat-a-practise/1060      

這道題本身并沒有複雜的算法，隻要考慮好幾個case就好了。      

1.數的前面有0，例如 5 003 03      

2.零的情況，例如 3 0.0000 00.0      

3.指數的計算要考慮周全.      

#include<iostream>
#include<string.h>
using namespace std;
#define  N 200
int formatstr(char *str,char *sa,int len,int strsize){
	int i;
	int exp_num=0;
	int isbig=0,first_use=0;//first_use is the first use number,because of the case may have such 003 00.03
	for(i=0;i<strsize;i++){//check num is large than 1 or not
		if(str[i]>='1'&&str[i]<='9'){
			first_use=i;
			isbig=1;
			break;
		}
		else if(str[i]=='.'){
			first_use=i;
			isbig=0;
			break;
		}

	}
	int t,is_point_done,has_num;
	if(isbig==1){// num is large than 1
		is_point_done=0;
		for(i=first_use,t=0;t<len;i++){
			if(i<strsize){
				if(str[i]>='0'&&str[i]<='9'){
					sa[t++]=str[i];
					if(0==is_point_done)
						exp_num++;
				}
				else if(str[i]=='.')
					is_point_done=1;
			}else
				sa[t++]='0';
		}
		for(;i<strsize&&is_point_done==0;i++){//check number whether arrive point '.' 
			if(str[i]=='.')
				break;
			exp_num++;
		}
	}
	else{
		has_num=0;
		for(i=first_use+1;i<strsize&&str[i]=='0';i++)//skip 0
			exp_num--;
		for(t=0;t<len;i++){
			if(i<strsize){
				sa[t++]=str[i];
				if(str[i]>='1'&&str[i]<='9')
					has_num=1;
			}
			else
				sa[t++]='0';
		}
		if(0==has_num)//check whether the case of 00.000000
			exp_num=0;
	}
	sa[t]='\0';
	return exp_num;
}
void print_num(char *a,int len,int exp_num){
	int i;
	cout<<"0.";
	for(i=0;i<len;i++)
		cout<<a[i];
	cout<<"*10^"<<exp_num;
}
int main(){
	int i,n,exp_num_a,exp_num_b;
	char a[N],b[N];
	char sa[N],sb[N];
	cin>>n>>a>>b;
	exp_num_a=formatstr(a,sa,n,strlen(a));//format for two number
	exp_num_b=formatstr(b,sb,n,strlen(b));
	if(exp_num_a==exp_num_b){
		for(i=0;i<n;i++){
			if(sa[i]!=sb[i]){
				cout<<"NO ";
				print_num(sa,n,exp_num_a);
				cout<<" ";
				print_num(sb,n,exp_num_b);
				break;		
			}
		}
		if(i==n){
			cout<<"YES ";
			print_num(sa,n,exp_num_a);
		}
	}
	else{
		cout<<"NO ";
		print_num(sa,n,exp_num_a);
		cout<<" ";
		print_num(sb,n,exp_num_b);
	}
	return 0;
}