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 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 (<) 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 1, 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题意：
给出两个数，问：将他们写成保留N (<100)位小数的科学计数法 0.d[1]...d[N]*10^k (d[1]>0 unless the number is 0); 后，是否相同。若相等，输出YES，并转换结果； 如果不相等，输出NO，并给出两个数的转换结果。
题解:
本题的思路难想而且情况判断非常复杂。
题目要求科学计数法时，两个数是否相等。因此只要判断科学技术法时的本体部分以及指数部分是否相等即可。
对于数据来说，要分为大于 1 与小于 1 来判断，要考虑各种情况下的数字，比如：0000， 000.00， 00123.4， 0.012, 0.00 等
一开始没有考虑太多的异常情况，拿了19分，第二天看了题解的注意点重做，还是只有21分，
第三天理了理思路，其实，只要关注，格式化好后的小数点要点在哪里，次数是多少就可以了，在此基础上，特判0，去掉前导零。比较的时候注意不要超限，输出不够0补齐。
自己编的测试样例：
0.0015 0000.001520000
YES 0.15*^-
010.25 0010.23
YES 0.102*^
0.1 00.100
YES 0.10000*^
000.0012 0000000.0012000000000
YES 0.1200000000*^-
0.000
YES 0.000*^
AC代码：
#include<iostream>
#include<string>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
int n;
char a[];
char b[];
int main(){
cin>>n>>a>>b;
int la=strlen(a);
int lb=strlen(b);
int sta=-,stb=-;
int isa0=,isb0=;//特判是不是0 //从第一个不是0的数开始
for(int i=;i<la;i++){
if(a[i]!='.'&&a[i]!=''){
sta=i;
break;
}
}
for(int i=;i<lb;i++){
if(b[i]!='.'&&b[i]!=''){
stb=i;
break;
}
}
if(sta==-) isa0=;
if(stb==-) isb0=; //小数点前有几位
int ia=,ib=;
int f=;
for(int i=;i<la;i++){
if(a[i]=='.') break;
if(a[i]!='') f=;
if(f){
ia++;
}
}
f=;
for(int i=;i<lb;i++){
if(b[i]=='.') break;
if(b[i]!='') f=;
if(f){
ib++;
}
}
//cout<<"sta:"<<sta<<" ia:"<<ia<<" ha:"<<ha<<endl;
//cout<<"stb:"<<stb<<" ib:"<<ib<<" hb:"<<hb<<endl; //比较
f=;
int pa=sta,pb=stb;
if(isa0!=isb0 || ia!=ib) f=;
for(int i=;i<n;i++){
if(pa+i>=la||pb+i>=lb) break;
if(a[pa]=='.') pa++;
if(b[pb]=='.') pb++;
//cout<<"pa "<<pa+i<<" a[pa] "<<a[pa+i]<<endl;
//cout<<"pb "<<pa+i<<" b[pb] "<<b[pa+i]<<endl;
if(a[pa+i]!=b[pb+i]){
f=;
break;
}
}
if(isa0==&&isa0==isb0) f=;//如果两个都是0 //输出
if(f){
cout<<"YES 0.";
if(isa0==){//0的特判
for(int i=;i<=n;i++) cout<<"";
cout<<"*10^0";
}else{
pa=sta;
for(int i=;i<n;i++){
if(pa+i>=la) {
cout<<"";
continue;
}
if(a[pa+i]=='.') pa++;
if(pa+i>=la) cout<<"";
else cout<<a[pa+i];
}
cout<<"*10^";
if(ia!=) cout<<ia;//次数>=于0
else{//次数<0
int k=-;
for(int i=;i<la;i++){
if(a[i]=='.'){
k=i;
break;
}
}
cout<<k-sta+;//.的位置-开始的位置+1
}
}
}else{//同上
cout<<"NO 0.";
if(isa0==){
for(int i=;i<=n;i++) cout<<"";
cout<<"*10^0 ";
}else{
for(int i=;i<n;i++){
if(pa+i>=la) {
cout<<"";
continue;
}
if(a[pa+i]=='.') pa++;
if(pa+i>=la) cout<<"";
else cout<<a[pa+i];
}
cout<<"*10^";
if(ia!=) cout<<ia<<" 0.";
else{
int k=-;
for(int i=;i<la;i++){
if(a[i]=='.'){
k=i;
break;
}
}
cout<<k-sta+<<" 0.";
}
}
if(isb0==){
for(int i=;i<=n;i++) cout<<"";
cout<<"*10^0";
}else{
for(int i=;i<n;i++){
if(pb+i>=lb) {
cout<<"";
continue;
}
if(b[pb+i]=='.') pb++;
if(pb+i>=lb) cout<<"";
else cout<<b[pb+i];
}
cout<<"*10^";
if(ib!=) cout<<ib;
else{
int k=-;
for(int i=;i<lb;i++){
if(b[i]=='.'){
k=i;
break;
}
}
cout<<k-stb+;
}
}
}
return ;
}