【Codeforces Round 362 (Div 2)B】【模拟】Barnicle 科学计数法转普通表示法

B. Barnicle

time limit per test
1 second

memory limit per test
256 megabytes

input
standard input

output
standard output

Barney is standing in a bar and starring at a pretty girl. He wants to shoot her with his heart arrow but he needs to know the distance between him and the girl to make his shot accurate.



Barney asked the bar tender Carl about this distance value, but Carl was so busy talking to the customers so he wrote the distance value (it's a real number) on a napkin. The problem
is that he wrote it in scientific notation. The scientific notation of some real number x is
the notation of form AeB, where A is
a real number and B is an integer and x = A × 10B is
true. In our case A is between 0 and 9 and B is
non-negative.
Barney doesn't know anything about scientific notation (as well as anything scientific at all). So he asked you to tell him the distance value in usual decimal representation with
minimal number of digits after the decimal point (and no decimal point if it is an integer). See the output format for better understanding.

Input
The first and only line of input contains a single string of form a.deb where a, d and b are
integers and e is usual character 'e' (0 ≤ a ≤ 9, 0 ≤ d < 10100, 0 ≤ b ≤ 100) —
the scientific notation of the desired distance value.
a and b contain
no leading zeros and d contains no trailing zeros (but may be equal to 0).
Also, b can not be non-zero
if a is zero.

Output
Print the only real number x (the
desired distance value) in the only line in its decimal notation.
Thus if x is
an integer, print it's integer value without decimal part and decimal point and without leading zeroes.
Otherwise print x in
a form of p.q such that p is
an integer that have no leading zeroes (but may be equal to zero), and q is an
integer that have no trailing zeroes (and may not be equal to zero).

Examples

input

8.549e2

output

854.9

input

8.549e3

output

8549

input

0.33e0

output

0.33

#include<stdio.h>
#include<iostream>
#include<string.h>
#include<string>
#include<ctype.h>
#include<math.h>
#include<set>
#include<map>
#include<vector>
#include<queue>
#include<bitset>
#include<algorithm>
#include<time.h>
using namespace std;
void fre() { freopen("c://test//input.in", "r", stdin); freopen("c://test//output.out", "w", stdout); }
#define MS(x,y) memset(x,y,sizeof(x))
#define MC(x,y) memcpy(x,y,sizeof(x))
#define MP(x,y) make_pair(x,y)
#define ls o<<1
#define rs o<<1|1
typedef long long LL;
typedef unsigned long long UL;
typedef unsigned int UI;
template <class T1, class T2>inline void gmax(T1 &a, T2 b) { if (b>a)a = b; }
template <class T1, class T2>inline void gmin(T1 &a, T2 b) { if (b<a)a = b; }
const int N = 1010, M = 0, Z = 1e9 + 7, ms63 = 0x3f3f3f3f;
char s
;
char b
;
int main()
{
while (~scanf("%s", s))
{
int l = strlen(s);
int dot;
int pe;
for (int i = 0; i < l; ++i)
{
if (s[i] == '.')dot = i;
if (s[i] == 'e')pe = i;
}
int p; sscanf(s + pe + 1, "%d", &p);
s[l = pe] = 0;
int mid = pe - dot - 1;
int st = 0;
int n = 0;
MS(b, 0);
if (p >= mid)//没有小数点
{
for (int i = 0; i < l; ++i)if(s[i]!='.')b[n++] = s[i];
for (int i = mid + 1; i <= p; ++i)b[n++] = '0';
while (st < n - 1 && b[st] == '0')++st;
}
else//有小数点
{
for (int i = 0; i <= dot + p; ++i)if(s[i]!='.')b[n++] = s[i];
while (st < n - 1 && b[st] == '0')++st;
b[n++] = '.';
for (int i = dot + p + 1; i < l; ++i)b[n++] = s[i];
while (b[n - 1] == '0')b[--n] = 0;
if (b[n - 1] == '.')b[--n] = 0;
}
puts(b + st);
}
return 0;
}
/*
【题意】
给你一个科学计数法表示的数，让你把其转化为普通类型的数。

【类型】
模拟

【分析】
这道题是一个模拟。
科学计数法的数有什么特性呢？
可能会有小数点'.'，有特殊表示符号'e'（不妨一开始我们默认设置'.'为整个字符串的最后位置）
然后我们分别找到'.'和'e'的位置，
并得到'e'之后的数是多少。
然后我们对小数点的位置做一定的平移，去前导零，去后缀零。
注意细节就可以啦

【时间复杂度&&优化】
O(n)

【数据】
0.000000000010e1

*/