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 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.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：机器只能存储N位数，给出一个最多100位的浮点数，判断它们在机器中是否是相同的存储。
//一开始想用字符数组，但是在字符数组中查找小数点的位置还要遍历，不如使用string直接find方便。
#include <iostream>
#include<stdio.h>
#include<cmath>
#include<vector>
using namespace std; int main() {
int n;
string s1,s2;
cin>>n;
cin>>s1>>s2;
int len1,len2;
int pos=s1.find(".");
pos==string::npos?len1=s1.size():len1=pos;
pos=s2.find(".");
pos==string::npos?len2=s2.size():len2=pos; string temp1=s1.substr(,n);
string temp2=s2.substr(,n); if(len1!=len2){
cout<<"NO ";
cout<<"0."<<temp1<<"*10^"<<len1<<" ";
cout<<"0."<<temp2<<"*10^"<<len2;
}else{
if(temp1==temp2){
cout<<"YES ";
cout<<"0."<<temp1<<"*10^"<<len1;
}
else{
cout<<"NO ";
cout<<"0."<<temp1<<"*10^"<<len1<<" ";
cout<<"0."<<temp2<<"*10^"<<len2;
}
} return ;
}
//提交一次之后，没有通过，牛客网上报错：
测试用例:
99 0 0.000
对应输出应该为:
YES 0.000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000*10^0
你的输出为:
NO 0.0*10^1 0.0.000*10^1
//但是我发现我不知道怎么处理，我写的这个bug存在问题，
//看了大佬的代码之后发现，我对科学计数法并不太明白。
代码来自：https://www.liuchuo.net/archives/2293
#include <iostream>
#include <cstring>
#include<stdio.h>
using namespace std;
int main() {
int n, p = , q = ;
char a[], b[], A[], B[];
scanf("%d%s%s", &n, a, b);
int cnta = strlen(a), cntb = strlen(b);
for(int i = ; i < strlen(a); i++) {//居然真的是遍历，我还以为需要字符串那样呢。
if(a[i] == '.') {
cnta = i;
break;
}
}
for(int i = ; i < strlen(b); i++) {
if(b[i] == '.') {
cntb = i;
break;
}
} while(a[p] == '' || a[p] == '.') p++;//找到不为0的位置。
while(b[q] == '' || b[q] == '.') q++;
if(cnta >= p)//这个小数点出现在非0位之后，如123.45，指数就为cnta-p=3
cnta = cnta - p;
else//小数点后仍有0位，如0.0002，那么指数就是负值，
cnta = cnta - p + ;
if(cntb >= q)
cntb = cntb - q;
else
cntb = cntb - q + ;//
if(p == strlen(a))//如果是0.000这样的形式，将其赋值为0.
cnta = ;//不置为0，有可能两个0判断不相等。
if(q == strlen(b))
cntb = ;
int indexa = , indexb = ;
while(indexa < n) {
if(a[p] != '.' && p < strlen(a))//除了小数点以外都赋值，
A[indexa++] = a[p];
else if(p >= strlen(a))//当遍历完后，n较大时，仍然不够n位，那么后n位补0
A[indexa++] = '';
p++;
}
while(indexb < n) {
if(b[q] != '.' && q < strlen(b))
B[indexb++] = b[q];
else if(q >= strlen(b))
B[indexb++] = '';
q++;
}
if(strcmp(A, B) == && cnta == cntb)
printf("YES 0.%s*10^%d", A, cnta);//读入和输出字符数组都可以用%s.
else
printf("NO 0.%s*10^%d 0.%s*10^%d" , A, cnta, B, cntb);
return ;
}
//深刻地学习了科学记数法
1.科学记数法有两个重要的因素，就是小数点的位置，和首位非0位的位置，通过这两个可以确定指数，
2.小数点的位置，怎么办呢？先初始化位字符串长度，对应整数，没有小数位。
3.首位非0的元素也初始化为字符串的长度。
4.需要注意的是如果首位非零元长度=字符串长度，那么这个字符串表示的数就是0.需要特殊处理，就是为0.0..0*10^0。否则计算指数。
5.科学记数法如果不是0，那么处理后的结果，小数点后边一定要是非0元！。
学习了！