【openjudge 计算概论（A）】[基础编程练习1]

1:大象喝水

#include<cmath>
#include<cstdio>
using namespace std;
const double pi=3.14159;
int main()
{
	int h,r;
	scanf("%d%d",&h,&r);
	double v=(double)h*r*r*pi;
	double sum=20000/v;
	int ans=ceil(sum);
	printf("%d\n",ans);
	return 0;
}

2:苹果和虫子

#include<cmath>
#include<cstdio>
using namespace std;
int main()
{
	int n,x,y;
	scanf("%d%d%d",&n,&x,&y);
	int sum=n-ceil((double)y/(double)x);
	printf("%d\n",sum);
	return 0;
}

3:晶晶赴约会

#include<cstdio>
using namespace std;
int main()
{
	int i;
	scanf("%d",&i);
	if(i!=1&&i!=3&&i!=5) printf("YES\n");
	 else printf("NO\n");
	return 0;
}

4:求一元二次方程的根(这道题的坑点在于要限制精度)

#include<cmath>
#include<cstdio>
#include<cstring>
#include<algorithm>
#define eps 1e-5
using namespace std;
int n;
double a,b,c;
inline double solve(double x)
{
	if (x<eps&&x>-eps) return 0;
    return x;
}
int main()
{
	scanf("%d",&n);
	while(n--)
	 {
	 	scanf("%lf%lf%lf",&a,&b,&c);
	 	double check=(b*b-4*a*c);
	 	if(check<eps&&check>-eps)
	 	 {
	 	 	double ans=(-b)/(2*a);
	 	 	printf("x1=x2=%.5lf\n",solve(ans));
		  }
		 else
		  if(check>eps)
		   {
		   	double ans1=(-b+sqrt(check))/(2*a);
		   	double ans2=(-b-sqrt(check))/(2*a);
		   	if(ans1<ans2) swap(ans1,ans2);
		   	printf("x1=%.5lf;x2=%.5lf\n",solve(ans1),solve(ans2));
		   }
		  else
		   {
		   	double shi=(-b)/(2*a);
		   	double ans=sqrt(-check)/(2*a);
		   	printf("x1=%.5lf+%.5lfi;x2=%.5lf-%.5lfi\n",solve(shi),solve(ans),solve(shi),solve(ans));
		   }
	 }
	return 0;
}

5:鸡兔同笼（别忘了考虑腿是奇数的情况）

#include<cstdio>
using namespace std;
int main()
{
	int n;
	scanf("%d",&n);
	if(n%2) {printf("0 0\n"); return 0;}
	int sm=n/4+(n%4)/2;
	int sx=n/2+(n%2)/4;
	printf("%d %d\n",sm,sx);
	return 0;
}

6:判断闰年（这个题，听说春天的时候有小伙伴忘记怎么判闰年。。。）

#include<cstdio>
using namespace std;
int main()
{
	int n;
	scanf("%d",&n);
	if(!(n%400)||!(n%4)&&(n%100)) printf("Y");
	 else printf("N");
	return 0;
}

7:奇数求和

#include<cstdio>
using namespace std;
int ans,m,n;
int main()
{
	scanf("%d%d",&n,&m);
	for(int i=n;i<=m;++i)
	 if(i%2) ans+=i;
	printf("%d\n",ans);
	return 0;
}

8:与7无关的数

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
int n;
int sum;
int main()
{
    scanf("%d",&n);
    for (int i=1;i<=n;i++)
    {
        int a=i/10,b=i%10;
        if (i%7==0)
            continue;
        if (a==7||b==7)
            continue;
        sum+=i*i;
    }
    printf("%d\n",sum);
    return 0;
}