Codeforces Beta Round #1
A:大水题;
代码:
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 #include<iostream> 2 #define ll long long 3 using namespace std; 4 5 int main() 6 { 7 ll n,m,a; 8 cin>>n>>m>>a; 9 cout<<(((n+a-1)/a)*((m+a-1)/a)); 10 return 0; 11 }
B:灵活的运用sscanf
代码:
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 #include<cstdio> 2 #include<cstring> 3 #define maxn 1000005 4 using namespace std; 5 int n; 6 char s[50]; 7 void a_to_b() 8 { 9 int a,b; 10 int p=maxn-1; 11 char t[maxn]; 12 sscanf(s,"R%dC%d",&a,&b); 13 t[p--]=0; 14 while(a) 15 { 16 t[p--]=a%10+'0'; 17 a/=10; 18 } 19 while(b) 20 { 21 t[p--]=(b-1)%26+'A'; 22 b=(b-1)/26; 23 } 24 printf("%s\n",&t[p+1]); 25 } 26 27 void b_to_a() 28 { 29 char t[maxn]; 30 int c = 0; 31 int r; 32 sscanf(s,"%[A-Z]%d",t,&r); 33 int len = strlen(t); 34 int p = 0; 35 while (p<len) 36 { 37 c+=t[p++]-'A'+1; 38 c*=26; 39 } 40 c/=26; 41 printf("R%dC%d\n",r,c); 42 } 43 44 int main() 45 { 46 int a,b; 47 scanf("%d",&n); 48 while(n--) 49 { 50 scanf("%s",s); 51 if(sscanf(s,"R%dC%d",&a,&b)==2)a_to_b(); 52 else b_to_a(); 53 } 54 return 0; 55 }
C:
很水的一个计算几何,关键是枚举,因为最多100个角;
稍微要用到一点圆的知识,因为它是正的多边形,所以所有的点都在外接圆上,判断三角形的内角是否是单位圆心角一半的倍数;
代码:
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 #include<cstdio> 2 #include<cmath> 3 #define eps 0.0001 4 #define pi acos(-1) 5 using namespace std; 6 7 struct point 8 { 9 double x,y; 10 point(double x=0,double y=0):x(x),y(y){ } 11 }p[3]; 12 point operator - (point a,point b){return point(a.x-b.x,a.y-b.y);} 13 double dis(point a,point b){return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));} 14 double cross(point a,point b){return a.x*b.y-a.y*b.x;} 15 16 bool isok(double a,int n) 17 { 18 double alph=a*n/pi; 19 double x=floor(alph+eps); 20 if(alph-x<eps)return 1; 21 return 0; 22 } 23 24 int main() 25 { 26 for(int i=0;i<3;i++)scanf("%lf%lf",&p[i].x,&p[i].y); 27 double d1=dis(p[0],p[2]); 28 double d2=dis(p[1],p[0]); 29 double d3=dis(p[2],p[1]); 30 double a=asin(cross(p[1]-p[0],p[2]-p[0])/d1/d2); 31 double b=asin(cross(p[2]-p[1],p[0]-p[1])/d2/d3); 32 double c=asin(cross(p[0]-p[2],p[1]-p[2])/d3/d1); 33 double r=d1*d2*d3/2/cross(p[1]-p[0],p[2]-p[0]); 34 int i; 35 for(i=3;i<=100;i++) 36 if(isok(a,i)&&isok(b,i)&&isok(c,i))break; 37 double ans=i*r*r*sin(2*pi/i)*0.5; 38 printf("%lf",ans); 39 }