【openjudge 计算概论（A）】[基础编程练习1]

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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-5using 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;}

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