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...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



#include<iostream>
#include<cstdio>
#include<cstring>
#include<vector>
using namespace std;
vector<char>va;
vector<char>vb;
int n;
vector<char> change(char s[],int &index){
	int point=-1;
	int firstchar=-1;
	int len=strlen(s);
	vector<char>tmp;
	int i, j;
	for(i=0;i<len;i++){
		if(s[i]=='.'){
			point=i;
			continue;
		}
		if(s[i]!='0'||firstchar!=-1){
			if(firstchar==-1){
				firstchar=i;
			}
			tmp.push_back(s[i]);
		}
	}
	if(point==-1){
		index=i;
	}else if(point<firstchar){
		index=-(firstchar-point-1);
	}else{
		index=point-firstchar;
	}
	if(point==-1&&firstchar!=-1){//字符串为s=0000000001的情况 
		index=len-firstchar;
	}
	len = tmp.size();
	if(len==0){//字符串为s=0的情况; 
		index=0;
	}
	for(i=len;i<n;i++){
		tmp.push_back('0'); 
	}
	return tmp;
}
int main(){
	char a[120],b[120];
	scanf("%d %s %s",&n,a,b);
	int aindex,bindex;
	va=change(a,aindex);
	vb=change(b,bindex);
	int flag=0;
	for(int i=0;i<n;i++){
		if(va[i]!=vb[i]){
			flag=1;
			break;
		}
	}
	if(flag==0){
		printf("YES 0.");
		for(int i=0;i<n;i++){
			printf("%c",va[i]);
		}
		printf("*10^%d\n",aindex);
	}else {
		printf("NO 0.");
		for(int i=0;i<n;i++){
			printf("%c",va[i]);
		}
		printf("*10^%d 0.",aindex);
		for(int i=0;i<n;i++){
			printf("%c",vb[i]);
		}
		printf("*10^%d\n",bindex);
	}
	return 0;
} 

 



posted @ 2017-11-11 12:37  gaoren  阅读(205)  评论(0编辑  收藏  举报