Light OJ Basic Geometry
2013-02-13 03:24
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Basic
Geometry
一共十八道题,在前面的beginners problems里面做了8道:1022 1072 1107 1211 1216 1305 1331 1433 详见:beginners problems
剩下的10道题,解决了6道,剩下的4道等什么时候有思路了再做吧,先放放。
1043 Triangle
Partitioning
相似
1058 Parallelogram Counting
求平行四边形个数
1118 Incredible
Molecules
1178 Trapezium
1366 Pair
of Touching Circles
1385 Kingdom Division
Remained:
1378 The Falling Circle
1383 Underwater
Snipers
1388 Trapezium
Drawing
1391 Speed Zones
Geometry
一共十八道题,在前面的beginners problems里面做了8道:1022 1072 1107 1211 1216 1305 1331 1433 详见:beginners problems
剩下的10道题,解决了6道,剩下的4道等什么时候有思路了再做吧,先放放。
1043 Triangle
Partitioning
相似
#include <cstdio> #include <cmath> int main() { int t; scanf("%d",&t); for(int i=1;i<=t;i++) { double a,b,c,ratio; scanf("%lf%lf%lf%lf",&a,&b,&c,&ratio); printf("Case %d: %lf\n",i,a/sqrt((ratio+1)/ratio)); } return 0; }
1058 Parallelogram Counting
求平行四边形个数
// 求给定点能构成平行四边形的个数 // 由于平行四边形对角线中点相重,所以我们先把这种点找出来 // 满足中点的k条线段用C(k,2)得到以某点为中心的平行四边形 #include <cstdio> #include <iostream> #include <algorithm> using namespace std; struct point { double x,y; }; point p[1001],mid[1001010]; bool cmp(point a,point b) { if(a.x==b.x) return a.y<b.y; return a.x<b.x; } int main() { int t; scanf("%d",&t); for(int ii=1;ii<=t;ii++) { int n; scanf("%d",&n); int total=0; for(int i=1;i<=n;i++) scanf("%lf%lf",&p[i].x,&p[i].y); for(int i=1;i<n;i++) for(int j=i+1;j<=n;j++) { mid[++total].x=p[i].x+p[j].x; mid[total].y=p[i].y+p[j].y; } sort(mid+1,mid+total+1,cmp); int sum=1; int ans=0; for(int i=1;i<total;i++) { if(mid[i].x==mid[i+1].x&&mid[i].y==mid[i+1].y) sum++; else { ans+=sum*(sum-1)/2; sum=1; } } printf("Case %d: %d\n",ii,ans); } return 0; }
1118 Incredible
Molecules
#include <cstdio> #include <cstring> #include <iostream> #include <cmath> #define PI (2*acos(0.0)) using namespace std; struct point { double x,y; }; double Distance(point a,point b) { return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y)); } double triangle(double a,double b,double c) { double p=(a+b+c)/2; return sqrt(p*(p-a)*(p-b)*(p-c)); } double jiao(double a,double b,double c) { return acos((a*a+b*b-c*c)/(2*a*b)); } int main() { int t; scanf("%d",&t); for(int i=1;i<=t;i++) { point p1,p2; double r1,r2; scanf("%lf%lf%lf%lf%lf%lf",&p1.x,&p1.y,&r1,&p2.x,&p2.y,&r2); double d=Distance(p1,p2); double ans; if(d<=fabs(r1-r2))// 等于号掉了会WA呃 { double r=r1<r2?r1:r2; ans=r*r*PI; } else if(d>=r1+r2) { ans=0; } else { double a0=jiao(r1,r2,d); double a1=jiao(r1,d,r2); double a2=jiao(r2,d,r1); ans=r1*r1*a1; ans+=r2*r2*a2; ans-=sin(a0)*r1*r2; } printf("Case %d: %.8lf\n",i,ans); } return 0; }
1178 Trapezium
#include <cstdio> #include <cmath> #include <iostream> #include <algorithm> using namespace std; double triangle(double a,double b,double c) { double p=(a+b+c)/2; return sqrt(p*(p-a)*(p-b)*(p-c)); } int main() { int t; scanf("%d",&t); for(int i=1;i<=t;i++) { double a,b,c,d; scanf("%lf%lf%lf%lf",&a,&b,&c,&d); if(a>c) swap(a,c); double ans; double tmp=c-a; double cnt=acos((b*b+tmp*tmp-d*d)/(2*b*tmp)); ans=a*b*sin(cnt); ans+=triangle(c-a,b,d); printf("Case %d: ",i); printf("%lf\n",ans); } return 0; }
1366 Pair
of Touching Circles
// 参考别人的uva5815 #include <iostream> #include <cstdio> #include <algorithm> #include <cmath> using namespace std; int total; int x[501],y[501],z[501]; void solve() //预处理,先把勾股数放在数组 { total=0; for(int i=1;i<=500;i++) for(int j=i+1;j<=500;j++) { int k=(int)sqrt(i*i+j*j); if(i*i+j*j==k*k) { total++; x[total]=i; y[total]=j; z[total]=k; } } } int main() { int t; scanf("%d",&t); solve(); for(int cases=1;cases<=t;cases++) { int H,W; scanf("%d%d",&H,&W); printf("Case %d: ",cases); long long sum=0,tmp; int mx=max(H,W); for(int i=2;i<mx;i+=2)//垂直或水平相切 for(int j=i;i+j<=mx;j+=2) { int length=i+j; int broad=max(i,j); tmp=0; if(broad<=H&&length<=W) tmp+=(H+1-broad)*(W+1-length); if(broad<=W&&length<=H) tmp+=(W+1-broad)*(H+1-length); if(i!=j) tmp*=2; sum+=tmp; } for(int i=1;i<=total;i++)//两圆斜切 for(int j=1;j<z[i];j++) { tmp=0; int length=max(z[i]+x[i],2*max(j,z[i]-j)); int broad=max(z[i]+y[i],2*max(j,z[i]-j)); if(length<=H&&broad<=W) tmp+=(H+1-length)*(W+1-broad); if(broad<=H&&length<=W) tmp+=(H+1-broad)*(W+1-length); tmp*=2; sum+=tmp; } printf("%lld\n",sum); } return 0; }
1385 Kingdom Division
// 将EF连起来,区域d划分为x和y // 由比例关系知 // a/b=x/c // (a+x+y)/(b+c)=y/(x+c) // 最终求d=x+y的值 // b=x的时候A点就不存在了 #include <cstdio> #include <cstring> #include <cmath> #include <iostream> #define eps 10-9 using namespace std; int main() { int t; scanf("%d",&t); for(int i=1;i<=t;i++) { double a,b,c,x,y; scanf("%lf%lf%lf",&a,&b,&c); x=a*c/b; y=(a+x)*(c+x)/(b-x); if(b-x<=eps) printf("Case %d: -1\n",i); else printf("Case %d: %lf\n",i,x+y); } return 0; }
Remained:
1378 The Falling Circle
1383 Underwater
Snipers
1388 Trapezium
Drawing
1391 Speed Zones
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