cf 619c 科学记数法1
| Time Limit: 2000MS | Memory Limit: 262144KB | 64bit IO Format: %I64d & %I64u |
Description
You 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.
Sample Input
Input
16
Output
1.6E1
Input
01.23400
Output
1.234
Input
.100
Output
1E-1
Input
100.
Output
1E2
代码如下：
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<iostream>
#define M 1000005
char s[M],ss[M];
using namespace std;
int main()
{
int i,k,j,m,k1,k2;
while(~scanf("%s",&s))
{
m=strlen(s);
for(i=0; i<m; i++)
{
if(s[i]=='.')//标记小数点的位置
{
k1=i;
break;
}
else k1=m;
}
for(i=0; i<m; i++)
{
if(s[i]>'0'&&s[i]<='9')
{
ss[0]=s[i];
ss[1]='.';
k=1;
k2=i+1;
for(j=i+1; j<m; j++)
{if(s[j]>='0'&&s[j]<='9')
ss[++k]=s[j];
}
break;
}
}
for(i=k; i>=1; i--)//去掉多余的0和小数点
{
if(ss[i]=='0'||ss[i]=='.')
k--;
else break;
}
for(i=0; i<k+1; i++)
printf("%c",ss[i]);
if(k1>k2)
{
printf("E%d",k1-k2);
}
else if(k1<k2)
printf("E%d",k1-k2+1);
printf("\n");
}
return 0;
}
