Codeforces 691C Exponential notation(模拟)
Exponential notation
time limit per test
2 secondsmemory limit per test
256 megabytesinput
standard inputoutput
standard outputYou are given a positive decimal number x.
Your task is to convert it to the "simple exponential notation".
Let x = a·10b, where 1 ≤ a < 10, then in general case the "simple exponential notation" looks like "aEb". If b equals to zero, the part "Eb" should be skipped. If a is an integer, it should be written without decimal point. Also there should not be extra zeroes in a and b.
Input
The only line contains the positive decimal number x. The length of the line will not exceed 106. Note that you are given too large number, so you can't use standard built-in data types "float", "double" and other.
Output
Print the only line — the "simple exponential notation" of the given number x.
Examples
input
16
output
1.6E1
input
01.23400
output
1.234
input
.100
output
1E-1
input
100.
output
1E2
#include<iostream> #include<algorithm> #include<cstring> using namespace std; const int maxn=1e6+5; char s[maxn]; int main() { while(cin>>s) { int len=strlen(s); int h=-1,t=len,p=len;//h第一个不为0的位置,t为最后一个不为0的位置,p是小数点的位置。 for(int i=0;i<len;i++) { if(s[i]=='.') p=i; if(s[i]!='.'&&s[i]!='0'&&h==-1) h=i; } for(int i=len-1;i>=0;i--) { if(s[i]!='.'&&s[i]!='0') { t=i; break; } } if(h==-1) cout<<'0'<<endl;//说明都是0000.0000形式的 else { cout<<s[h];//输出第一个数字 if(h!=t)//后面存在有效数字 cout<<'.'; for(int i=h+1;i<=t;i++) if(s[i]!='.') cout<<s[i]; int e;//计算幂次 if(h>p) e=p-h; else e=p-h-1; if(e) cout<<'E'<<e<<endl; else cout<<endl; } } return 0; }