Educational Codeforces Round 14 C 好东西啊！！！

最新推荐文章于 2025-03-18 13:57:21 发布

原创最新推荐文章于 2025-03-18 13:57:21 发布 · 390 阅读

0 ·

CC 4.0 BY-SA版权

-----Codeforces----- 同时被 3 个专栏收录

149 篇文章

订阅专栏

---------暴力模拟---------

94 篇文章

订阅专栏

模拟

34 篇文章

订阅专栏

本文介绍了一种将大型正十进制数转换为简单指数表示的方法，并提供了一个C++实现示例。该方法遵循特定的格式规则，适用于无法使用标准浮点数据类型的场景。

链接：戳这里

C. Exponential notation

time limit per test2 seconds

memory limit per test256 megabytes

inputstandard input

outputstandard output

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.

Examples
input
16
output
1.6E1
input
01.23400
output
1.234
input
.100
output
1E-1
input
100.
output
1E2

题意：

给出一串只包含数字和一个或0个小数点的字符串，输出科学记数法表示这个数

思路：

这种东西慢慢磨吧以后肯定有用的

代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<string>
#include<vector>
#include <ctime>
#include<queue>
#include<set>
#include<map>
#include<stack>
#include<iomanip>
#include<cmath>
#include<bitset>
#define mst(ss,b) memset((ss),(b),sizeof(ss))
///#pragma comment(linker, "/STACK:102400000,102400000")
typedef long long ll;
typedef long double ld;
#define INF (1ll<<60)-1
#define Max 1e9
using namespace std;
string s;
int main(){
    cin>>s;
    int n=s.size();
    if(s[n-1]=='.') n--;
    int f=0;
    for(int i=0;i<n;i++){
        if(s[i]=='.'){
            f=1;
            break;
        }
    }
    if(f==0){
        int t=n-1;
        while(s[t]=='0' && t>=0) t--;
        int l=0;
        while(l<n && s[l]=='0') l++;
        if(t==-1) cout<<0<<endl;
        else if(n==1) cout<<s[0]<<endl;
        else if(l==n-1) cout<<s[l]<<endl;
        else {
            cout<<s[l];
            if(t!=0) {
                cout<<".";
                for(int i=l+1;i<=t;i++) cout<<s[i];
            }
            cout<<"E"<<n-1-l<<endl;
        }
    } else {
        int pos=-1;
        for(int i=0;i<n;i++){
            if(s[i]=='.') {
                pos=i;
                break;
            }
        }
        int l=0,r=n-1;
        while(s[l]=='0' && l<n) l++;
        while(s[r]=='0' && r>=0) r--;
        ///cout<<l<<" "<<r<<" "<<pos<<endl;
        if(l==r) cout<<0<<endl;
        else if(s[l]=='.'){
            int x=l+1;
            while(s[x]=='0' && x<n) x++;
            cout<<s[x];
            if(x!=r){
                cout<<".";
                for(int i=x+1;i<=r;i++) cout<<s[i];
            }
            cout<<"E"<<l-x;
        }
        else {
            if(pos!=r){
                cout<<s[l];
                cout<<".";
                for(int i=l+1;i<=r;i++) {
                    if(s[i]=='.') continue;
                    cout<<s[i];
                }
                if(pos-l-1!=0) {
                    cout<<"E";
                    cout<<pos-l-1<<endl;
                }
            } else {
                cout<<s[l];
                if(l+1!=pos){
                    int x=l+1;
                    while(s[x]=='0' && x<n) x++;
                    if(x==pos) cout<<"E"<<pos-l-1;
                    else {
                        cout<<".";
                        for(int i=l+1;i<=x;i++) cout<<s[i];
                        cout<<"E"<<pos-1-l<<endl;
                    }
                }
            }
        }
    }
    return 0;
}
/*
0001.001000
0000.000001
0001100
10.0
*/