P1a+b

#include<iostream>
using namespace std;

int main(){
    int a,b; cin>>a>>b;
    cout<<a+b;
    return 0;
}

P2求圆环的面积

#include<iostream>
#include<iomanip>
using namespace std;
const double pi = 3.14;

int main(){
    int r1,r2; cin>>r1>>r2; 
    cout<<fixed<<setprecision(2)<< pi * r1 *r1 - pi * r2*r2;
    return 0;
}

P3种蔬菜

#include<iostream>
#include<iomanip>
using namespace std;
const double pi = 3.14;

int main(){
  int a,b,c,d; char p;
  cin>>a>>p>>b>>p>>c>>p;
  
  d = 100 - a - b - c;
  double x = 80.0 * a/100;
  double y = 80.0 * b/100;
  double z = 80.0 * c/100;
  double q = 80.0 * d/100;
  cout<<fixed<<setprecision(1)<<x<<" "<<y<<" "<<z<<" "<<q;
  return 0;
}

P4求三位数数字和

#include<iostream>
#include<iomanip>
using namespace std;
const double pi = 3.14;

int main(){
    int n; cin>>n;
    int a = n/100;
    int b = n/10%10; 
    int c = n%10; 
    cout<<a+b+c;
    return 0;
}

P5数字反转

#include<iostream>
#include<iomanip>
using namespace std;
const double pi = 3.14;

int main(){
	int n; cin>>n;
	
	int a = n/100;
	int b = n/10%10;
	int c = n%10;
	cout<<a+b+c;
	
  return 0;
}

P6求倒数和

#include<iostream>
using namespace std;

int main(){
    int n; cin>>n;
    int a = n/100;
    int b = n/10%10;
    int c = n%10;
    cout<<n + c*100+b*10+a;
    return 0;
}

P7加密四位数

#include<iostream>
using namespace std;

int main(){
    int n; cin>>n;
    int a = n/1000;
    int b = n/100%10;
    int c = n/10%10;
    int d = n%10;
    
    a+=5;
    b+=5;
    c+=5;
    d+=5;
    
    a%=10;
    b%=10;
    c%=10;
    d%=10;
    cout<<d<<c<<b<<a;
    return 0;
}

P8纸箱/carton

#include<iostream>
using namespace std;
typedef long long LL; // 以后使用 LL 就等效于 long long

int main(){
   int a,b,c; cin>>a>>b>>c;
    
//    if(a < b) {
//    	int t = a; a=b; b=t;
//	}
	if(a<c){
		int t=a; a=c; c=t;
	}
	if(b<c){
		int t=b; b=c; c=t;
	}
	cout<<1LL * 2*(a*b + a*c+b*c) - a*b;
    return 0;
}

P9糖果共享

#include<iostream>
using namespace std;

int main(){
    int n; cin>>n;
    cout<<n/2;
    return 0;
}

P10小鱼的游泳时间

#include<iostream>
using namespace std;

int main(){
    int a,b,c,d; cin>>a>>b>>c>>d;
    int start = a*60 + b;
    int end = c *60 +d;
    int time = end-start;
    int e = time/60, f = time %60;
    cout<<e<<" "<<f; 
    return 0;
}