[common skills]

solve scanf Eat back

getchar()

Input n Back carriage return , the second scanf Carriage return will be assigned to op. Use getchar() Get the carriage return first .

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

char n;
char op;
scanf("%c",&n);
getchar();
scanf("%c",&op);

char op[2]

char op[2];
char num;
scanf("%s",op);
scanf("%c",num);

cout<<op<<' '<<num;

Keep the specified decimal places

1. C++ cout Output specified decimal places

#include<iomanip>

cout<<setiosflags(ios::fixed)<<setprecision(3);
cout<<2.1<<endl; // Output 2.100

2. C++ printf Output specified decimal places

#include<cstdio>

printf("%.3f",2.1); // Output 2.100

character string 、 Array

1. int Array Allocate space size

int num[10]={
    0,1};
cout<<sizeof(num)/sizeof(num[0]);
// Output  10

2. String length

string str="123";
cout<<str.length();
// Output  3

3. character string turn char Array

#include<cstring>

char a[3];
string str="123";
strcpy(a,str.c_str());// Be careful  .c_str()
for(int i=0;i<3;i++)
    cout<<a[i]<<' ';
// Output  1 2 3

4. Count and character string Interturn

#include<sstream>

string tostr(int n)
{
    
    stringstream ss;
    ss<<n;
    string str;
    ss>>str;
    return str;
}
int toint(string str)
{
    
	stringstream ss;
    ss<<str;
    int num;
    ss>>num;
    return num;
}

5. Enter a line of string containing spaces

getline(cin,str);

gcd&lcm

1. Two numbers of gcd

int gcd(int a,int b)
{
    
    if(b==0) return a;
    return gcd(b,a%b);
}

2. More than one number of gcd

for(int i=0;i<n-1;i++)
{
    
     a[0]=gcd(a[i],a[0]);
}
// More than one number of gcd There is a[0]

3. Minimum common multiple

int lcm(int a,int b)
{
    
    return a*b/gcd(a,b);
    return a/gcd(a,b)*b; // More reasonable , prevent a*b overflow 
}

sort Sort

1. Default from small to large

sort(num,num+8);

2. Custom functions from large to small

int cmp(int a,int b)
{
    
	return a>b;
}

sort(num,num+8,cmp);