//T1014 # 空格分隔输出

#这题其实很简单

#思路 1.导入头文件 2.定义一个变量n 3.输入数字n 4.输出

#include <iostream>
#include <iomanip>
using namespace std;
int main()
{
    char a;
    int b;
    float c;
    double d;
    cin >> a;
    cin >> b;
    cin >> c;
    cin >> d;
    cout << a << " ";
    cout << b << " ";
    cout << fixed << setprecision(6) << c << " ";
    cout << fixed << setprecision(6) << d << " ";
    return 0;
}

#include<bits/stdc++.h>
using namespace std;
int main()
{
    char a;
    int b;
    float c;
    double d;
    cin>>a>>b>>c>>d;
    cout<<a<<" "<<b<<" "<<fixed<<setprecision(6)<<c<<" "<<fixed<<setprecision(6)<<d;
    return 0;
}

👆🏼依题意就可编码

#include <iostream>
#include <cstdio>
using namespace std;

char a;
int b;
float c;
double d;

int main()
{
    scanf("%c%d%f%lf", &a, &b, &c, &d);
    printf("%c %d %.6f %.6lf", a, b, c, d);
    return 0;
}

#include <iostream>
#include <cstdio>
#include <iomanip>
using namespace std;
int main()
{
    char a;
    long long  b;
    float c;
    double d;
    cin >> a >> b >> c >> d;
    cout << a<<" ";
    cout << b<<" ";
    cout << fixed << setprecision(6)<<c<<" ";
    cout << fixed << setprecision(6)<<d<<" ";
    return 0;
用printf("%.nf",参数)的形式会更简单

//最简单方法（不是最短)
#include <bits/stdc++.h>
using namespace std;
int main()
{
    //分类输入
    char a;
    int b;
    float c;
    double d;
    cin >> a;
    cin >> b;
    cin >> c;
    cin >> d;
    //输出
    cout << a << ' ' ;
    cout << b << ' ' ;
    //浮点数保留6位
    cout << fixed << setprecision(6) << c << ' ';
    cout << fixed << setprecision(6) << d;
    return 0;
}

太简单了,没的分析

#include<iostream>
using namespace std;
#include<iomanip>
 
int main()
{
	char a1;
	int a2;
	float a3;
	double a4;
	cin >> a1>>a2>>a3>>a4;
	cout <<a1<<" "<<a2<<" "<< fixed << setprecision(6) << a3<<" "<<a4;
 
}