信息学奥赛一本通---第二章---顺序结构程序设计--- 1024-1038 (第四节 至 第五节)题解

注:此为c++代码

    虽然我的代码无注释,但如想复制,必须完全搞懂代码所对应的题!

Dev c++这东西点进来的都有吧,详情见我的另一篇文章。

正式进入题解部分:          (本人小白,大佬们请勿吐槽)

1024:

#include<bits/stdc++.h> 
using namespace std;
float n;  
int main() 
{ 
    cin>>n; 
    cout<<setiosflags(ios::fixed)<<setprecision(3); 
    cout<<n<<endl; 
    return 0; 
}

1025:

#include<bits/stdc++.h> 
using namespace std;
float n;  
int main() 
{ 
    cin>>n; 
    cout<<setiosflags(ios::fixed)<<setprecision(12); 
    cout<<n<<endl; 
    return 0; 
}

1026:

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

1027:

#include<bits/stdc++.h>
using namespace std;
double a;
int main()
{
	scanf("%lf",&a);
	printf("%f\n%.5f\n%e\n%g\n",a,a,a,a);
	return 0;
 }

1028:

#include<bits/stdc++.h>
using namespace std;
char a;
int main()
{
	cin>>a;
	cout<<"  "<<a<<"  "<<endl;
	cout<<" "<<a<<a<<a<<" "<<endl;
	cout<<a<<a<<a<<a<<a<<endl;
	cout<<" "<<a<<a<<a<<" "<<endl;
	cout<<"  "<<a<<"  "<<endl;
	return 0;
}

1029:

#include<bits/stdc++.h>
using namespace std;
double a, b;
int main()
{
    cin>>a>>b;
    double r=a-int(a/b)*b;
    if(r<0)
    {
       r+=b;
    }
    cout<<r<<endl;
}

1030:

#include<bits/stdc++.h>
using namespace std;
double r,v;
int main()
{ 
    cin>>r;
    v=(4*3.14*r*r*r)/3.0;
    cout<<setiosflags(ios::fixed)<<setprecision(2);
    cout<<v<<endl;
    return 0;
}

1031:

#include<bits/stdc++.h>
using namespace std;
int n,a,b,c;
int main()
{
    cin>>n;
    a=n/100;
    b=n%100/10;
    c=n%10;
    cout<<c<<b<<a<<endl;
    return 0;
}

1032:

#include<bits/stdc++.h>
using namespace std;
double r,c,h,pi,v,water;
int main()
{
    pi=3.14159;
    cin>>h>>r;
    v=pi*r*r*h/1000.0;
    water=20.0/v;
    c=ceil(water);
    cout<<c<<endl;
    return 0;
}

1033:

#include<bits/stdc++.h>
double Xa,Xb,Ya,Yb,len;
using namespace std;
int main()
{
    cin>>Xa>>Ya;
    cin>>Xb>>Yb;
    len=sqrt( (Xa-Xb)*(Xa-Xb) + (Ya-Yb)*(Ya-Yb) );
    cout<<setiosflags(ios::fixed)<<setprecision(3);
    cout<<len<<endl;
    return 0;
}

1034:

#include<iostream>
double x1,x2,x3,y1,y2,y3,a,b,c,area,p;
using namespace std;
int main()
{ 
	cin>>x1>>y1>>x2>>y2>>x3>>y3;
    a=sqrt((x2-x1)*(x2-x1)+(y2-y1)*(y2-y1));
	b=sqrt((x3-x1)*(x3-x1)+(y3-y1)*(y3-y1));
	c=sqrt((x2-x3)*(x2-x3)+(y2-y3)*(y2-y3));
	p=(a+b+c)/2.0;
	area=sqrt(p*(p-a)*(p-b)*(p-c));
	cout<<setiosflags(ios::fixed)<<setprecision(2)<<area<<endl;  
	return 0;
}

1035:

#include<bits/stdc++.h>
using namespace std;
int a_1,a_2,n,a_n,d;
int main()
{
    cin>>a_1>>a_2>>n;
    d=a_2-a_1;
    a_n=a_1+(n-1)*d;
    cout<<a_n<<endl;
    return 0;
}

1036:

#include<bits/stdc++.h>
long long int a,b,c;
using namespace std;
int main()
{
	scanf("%lld %lld",&a,&b);
	c=a*b;
	printf("%lld",c);
	return 0;
}

1037:

#include<bits/stdc++.h>
using namespace std;
int n,result;
int main()
{
    cin>>n;
    result=pow(2,n);
    cout<<result<<endl;
    return 0;
}

1038:

#include<bits/stdc++.h>
using namespace std;
double n,x,y,c;
int main()
{
    cin>>n>>x>>y;
    c=n-ceil(y*1.0/x);
    if(c<0)
    {
        c=0;
    }
    cout<<c<<endl;
    return 0;
}

个人码风，不喜勿喷

明天发:

信息学奥赛一本通---第三章---程序的控制结构--- 1039-1048 (第一节)题解