PAT甲级 1060 Are They Equal （25 分）科学计算表示

1060 Are They Equal （25 分）

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 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

题目大意：给出两个数，问将它们写成保留N位小数的科学计数法后是否相等。如果相等，输出YES，输出他们的科学记数法表示方法。如果不相等输出NO，分别输出他们的科学计数法
分析：
1. cnta 和 cntb 通过扫描字符串得到小数点所在的下标（初始化cnta cntb为字符串长度，即下标为strlen(str））
2. 考虑到可能前面有多余的零，用 p 和 q 通过扫描字符串使 p q 开始于第一个非0（且非小数点）处的下标
3. 如果cnta >= p ，说明小数点在第一个开始的非0数的下标的右边，那么科学计数法的指数为cnta – p ; 否则应该为cnta – p + 1; 字符串b同理。
4. 如果 p 和 q 等于字符串长度，说明字符串是 0，此时直接把 cnta（或者cntb）置为0，因为对于0来说乘以几次方都是相等的，如果不置为0可能会出现两个0比较导致判断为它们不相等
5. indexa = 0开始给新的A数组赋值，共赋值n位除去小数点外的正常数字，从p的下标开始。如果p大于等于strlen，说明字符串遍历完毕后依旧没能满足需要的位数，此时需要在A数组后面补上0直到满足n位数字。indexb同理，产生新的B数组
6. 判断A和B是否相等，且cnta和cntb是否相等。如果相等，说明他们用科学计数法表示后是相同的，输出YES，否则输出NO，同时输出正确的科学计数法
注意：
– 10的0次方和1次方都要写。
– 题目中说，无需四舍五入。

#include<iostream>
#include<cstdio>
using namespace std;
int main(){
	int n;
	string a,b,ta,tb;
	cin>>n>>a>>b;
	//cnta是字符串a的下标，p是第一个非0字符下标,exa记录a字符串指数 
	int p = 0,q = 0,exa = 0,exb = 0;
	int cnta = a.length(),cntb = b.length();
	for(int i=0;i<a.length();i++)
		if(a[i] == '.'){
			cnta = i;
			break;
		}	
	for(int i=0;i<b.length();i++)
		if(b[i] == '.'){
			cntb = i;
			break;
		}
	while(a[p]=='0' || a[p] == '.') p++;
	while(b[q]=='0' || b[q] == '.') q++;
	if(cnta >= p)
		exa = cnta - p; 
	else
		exa = cnta - p + 1;
	if(cntb >= q)
		exb = cntb - q; 
	else
		exb = cntb - q + 1;
	if(p == a.length())//若字符串全是0 
		exa = 0;
	if(q == b.length())
		exb = 0;
		
	int index = 0;
	while(index < n){
		if(a[p] != '.' && p < a.length()){//把所有有效数字加入ta 
			ta += a[p];
			index++;	
		}
		else if(p >= a.length()){
			ta += '0';//不足补0
			index++;
		}	
		p++;
	}
	index = 0;
	while(index < n){
		if(b[q] != '.' && q < b.length()){
			tb += b[q];
			index++;
		}	
		else if(q >= b.length()){
			tb += '0';
			index++;
		}	
		q++;
	}
	if(ta == tb && exa == exb)
		cout<<"YES 0."<<ta<<"*10^"<<exa;
	else{
		cout<<"NO 0."<<ta<<"*10^"<<exa;
		cout<<" 0."<<tb<<"*10^"<<exb;
	}		
	return 0;
}