标签：strong hat sqrt cal lin ota fabs pair open
A - Alien Sunset
1 #include <bits/stdc++.h> 2 using namespace std; 3 typedef long long ll; 4 const int N=30; 5 int day[N],dx,rise[N],night[N]; 6 bool judge(int h, int i) 7 { 8 h%=day[i]; 9 if(rise[i] <= night[i] && (h<=rise[i] || h>=night[i])) return 1; 10 if(rise[i] > night[i] && (h<=rise[i] && h>=night[i])) return 1; 11 return 0; 12 } 13 int main() 14 { 15 int n; 16 scanf("%d",&n); 17 for(int i=1; i<=n; i++) 18 { 19 scanf("%d%d%d",&day[i],&rise[i],&night[i]); 20 dx=max(dx, day[i]); 21 } 22 for(int i=0; i<dx * 1825; i++) 23 { 24 bool f=0; 25 for(int j=1; j<=n; j++) if(!judge(i, j)) {f=1; break;} 26 if(!f){printf("%d\n",i); return 0;} 27 } 28 puts("impossible"); 29 return 0; 30 }
B - Breaking Biscuits （来自队友B
1 #include <bits/stdc++.h> 2 using namespace std; 3 const double eps = 1e-16; 4 int sgn(double x) 5 { 6 if(fabs(x) < eps)return 0; 7 if(x < 0)return -1; 8 else return 1; 9 } 10 struct point 11 { 12 double x,y; 13 point(int _x = 0,int _y = 0) 14 { 15 x = _x; 16 y = _y; 17 } point operator -(const point &b)const 18 { 19 return point(x - b.x, y - b.y); 20 } double operator ^(const point &b)const 21 { 22 return x*b.y - y*b.x; 23 } double operator *(const point &b)const 24 { 25 return x*b.x + y*b.y; 26 } void sc() 27 { 28 scanf("%lf%lf",&x,&y); 29 } 30 }; 31 struct Line 32 { 33 point s,e; 34 Line() {} Line(point _s,point _e) 35 { 36 s = _s; 37 e = _e; 38 } pair<int,point> operator &(const Line &b)const 39 { 40 point res = s; 41 if(sgn((s-e)^(b.s-b.e)) == 0) 42 { 43 if(sgn((s-b.e)^(b.s-b.e)) == 0) return make_pair(0,res); 44 else return make_pair(1,res); 45 } 46 double t = ((s-b.s)^(b.s-b.e))/((s-e)^(b.s-b.e)); 47 res.x += (e.x-s.x)*t; 48 res.y += (e.y-s.y)*t; 49 return make_pair(2,res); 50 } 51 }; 52 double dist(point a,point b) 53 { 54 return (double)sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y)); 55 } 56 point NearestpointToLineSeg(point P,Line L) 57 { 58 point result; 59 double t = ((P-L.s)*(L.e-L.s))/((L.e-L.s)*(L.e-L.s)); 60 if(t >= 0.000 && t <= 1.0000) 61 { 62 result.x = L.s.x + (L.e.x - L.s.x)*t; 63 result.y = L.s.y + (L.e.y - L.s.y)*t; 64 } 65 else 66 { 67 if(dist(P,L.s) < dist(P,L.e)) result = L.s; 68 else result = L.e; 69 } 70 return result; 71 } 72 double emmm(point P,Line L) 73 { 74 point zz=NearestpointToLineSeg(P,L); 75 return dist(zz,P); 76 } 77 const int MAXN = 2e5+7; 78 point List[MAXN]; 79 int Stack[MAXN],top; 80 bool _cmp(point p1,point p2) 81 { 82 double tmp = (p1-List[0])^(p2-List[0]); 83 if(tmp > 0)return true; 84 else if(tmp == 0 && dist(p1,List[0]) <= dist(p2,List[0])) return true; 85 else return false; 86 } 87 void Graham(int n) 88 { 89 point p0; 90 int k = 0; 91 p0 = List[0]; 92 for(int i = 1; i < n; i++) 93 { 94 if(p0.y > List[i].y || (p0.y == List[i].y && p0.x > List[i].x)) 95 { 96 p0 = List[i]; 97 k = i; 98 } 99 } 100 swap(List[k],List[0]); 101 sort(List+1,List+n,_cmp); 102 if(n == 1) 103 { 104 top = 1; 105 Stack[0] = 0; 106 return; 107 } 108 if(n == 2) 109 { 110 top = 2; 111 Stack[0] = 0; 112 Stack[1] = 1; 113 return; 114 } 115 Stack[0] = 0; 116 Stack[1] = 1; 117 top = 2; 118 for(int i = 2; i < n; i++) 119 { 120 while(top > 1 && ((List[Stack[top-1]]-List[Stack[top-2]])^(List[i]-List[Stack[top-2]])) <= 0) top--; 121 Stack[top++] = i; 122 } 123 } 124 double rotating_calipers(point p[],int n) 125 { 126 double ans = 999999999999999999999.0; 127 point v; 128 int cur = 1; 129 for(int i = 0; i < n; i++) 130 { 131 v = p[i]-p[(i+1)%n]; 132 while((v^(p[(cur+1)%n]-p[cur])) < 0) 133 { 134 cur = (cur+1)%n; 135 } 136 Line qw=Line(p[(i)%n],p[(i+1)%n]); 137 // point zz=NearestpointToLineSeg(p[cur],qw); 138 double as=emmm(p[cur],qw); //as=max(as,zx); 139 ans=min(ans,as); 140 // ans=min(ans,max(dist(p[i],p[cur]),dist(p[(i+1)%n],p[(cur+1)%n]))); 141 } 142 return ans; 143 } 144 point p[MAXN]; 145 int main() 146 { 147 int n,r; 148 scanf("%d",&n); 149 for(int i = 0; i < n; i++)List[i].sc(); 150 Graham(n); 151 for(int i = 0; i < top; i++)p[i] = List[Stack[i]]; 152 printf("%.7f\n",rotating_calipers(p,top)); 153 return 0; 154 }
C - Cued In
1 #include <bits/stdc++.h> 2 using namespace std; 3 typedef long long ll; 4 const int N=1009; 5 int a[N]; 6 string s; 7 map<string,int> psi; 8 void init() 9 { 10 psi["red"]=1; 11 psi["yellow"]=2; 12 psi["green"]=3; 13 psi["brown"]=4; 14 psi["blue"]=5; 15 psi["pink"]=6; 16 psi["black"]=7; 17 } 18 int main() 19 { 20 init(); 21 int n; 22 scanf("%d",&n); 23 for(int i=0; i<n; i++) 24 { 25 cin>>s; 26 a[psi[s]]++; 27 } 28 if (a[1]==n) puts("1"); 29 else 30 { 31 ll ans=0; 32 if (a[1]) 33 { 34 int p=7; 35 while(!a[p]) --p; 36 ans+=p*a[1]+a[1]; 37 } 38 for(int i=2; i<=7; i++) ans+=i*a[i]; 39 cout<<ans<<endl; 40 } 41 return 0; 42 }
D - Deranging Hat
1 #include <bits/stdc++.h> 2 using namespace std; 3 typedef long long ll; 4 const int N=2e3+10; 5 char s1[N],s2[N]; 6 int main() 7 { 8 cin>>s2; 9 strcpy(s1+1,s2); 10 strcpy(s2+1,s1+1); 11 int n=strlen(s2+1); 12 sort(s2+1,s2+1+n); 13 // cout<<(s2+1)<<endl; 14 for(int i=1; i<=n; i++) 15 { 16 if(s1[i]!=s2[i]) 17 { 18 for(int j=i+1; j<=n; j++) 19 { 20 if(s2[j]==s1[i]) 21 { 22 if(s2[i]>s2[j]) printf("%d %d\n",i,j); 23 else printf("%d %d\n",j,i); 24 swap(s2[j],s2[i]); 25 break; 26 } 27 } 28 } 29 } 30 return 0; 31 }
E - Education （来自队友A
1 #include <bits/stdc++.h> 2 typedef long long ll; 3 const int maxn = 1e5 + 100; 4 const int INF = 1e9 + 7; 5 using namespace std; 6 int a[maxn],b[maxn]; 7 bool cmp(int x,int y) 8 { 9 return x>y; 10 } 11 int main() { 12 int n, m; 13 scanf("%d%d", &n, &m); 14 for(int i=1; i<=n; i++){ 15 scanf("%d",&a[i]); 16 } 17 for(int i=1; i<=m; i++) 18 scanf("%d",&b[i]); 19 sort(a+1,a+n+1,cmp); 20 sort(b+1,b+m+1,cmp); 21 int v; 22 for(int i=1; i<=m; i++) 23 scanf("%d",&v); 24 int k=1; 25 for(int i=1; i<=m; i++){ 26 if(b[i]>=a[k]){ 27 k++; 28 if(k==n+1){ 29 printf("possible\n"); 30 return 0; 31 } 32 } 33 } 34 printf("impossible\n"); 35 return 0; 36 }
F - Flipping Coins （来自队友B
#include <iostream> #include <cstdio> #include <algorithm> #include <bits/stdc++.h> const int N=1e6+10; typedef long long ll; using namespace std; int n,m,k; double dp[450][450],c[450][450],p[450]; void init(){ c[0][0]=1; p[0]=1; for(int i=1;i<=420;i++){ for(int j=0;j<=i;j++){ if(!j||j==i) c[i][j]=1; else c[i][j]=c[i-1][j-1]+c[i-1][j]; } p[i]=p[i-1]/2.0; } } int main() { init(); scanf("%d%d",&n,&m); int k=1; double ans=0.0; memset(dp,0,sizeof(dp)); dp[0][0]=1; for(int i=0;i<m;i++){ for(int j=0;j<=n;j++){ if(!dp[i][j]) continue;//剪枝 for(int t=0;t<=k;t++){ if(j+k<=n){ dp[i+1][j+t]+=1.0*dp[i][j]*c[k][t]*p[k]; } else{//j+k>n int u=j+k-n; dp[i+1][j-u+t]+=1.0*dp[i][j]*c[k][t]*p[k]; } } } } for(int i=1;i<=n;i++){ ans+=i*dp[m][i]; } printf("%.7lf\n",ans); return 0; }
I - I Work All Day （队友口述思路之后扔给我写
1 #include <bits/stdc++.h> 2 using namespace std; 3 const int N=19; 4 int a[N]; 5 int main() 6 { 7 int n,m,f=0,ans=1; 8 scanf("%d",&n); 9 for(int i=1; i<=n; i++) scanf("%d",&a[i]); 10 scanf("%d",&m); 11 for(int i=2;i<=n;i++) 12 if(m%a[i]<m%a[ans]) ans=i; 13 printf("%d\n",a[ans]); 14 return 0; 15 }
J - Just A Minim （来自队友B
#include <bits/stdc++.h> using namespace std; typedef long long ll; const int N=2e3+10; int mp[20]; int n,x; int main() { scanf("%d",&n); double ans=0.0; for(int i=1;i<=n;i++){ scanf("%d",&x); mp[x]++; } ans=mp[0]*2.0+mp[1]+mp[2]*0.5+mp[4]*0.25+mp[8]*0.125+mp[16]*0.0625; printf("%0.7lf\n",ans); return 0; }
K - Knightsbridge Rises （来自队友A
1 #include <bits/stdc++.h> 2 3 typedef long long ll; 4 const int maxn = 200; 5 const int INF = 1e9 + 7; 6 using namespace std; 7 struct node { 8 int f; 9 int l, r; 10 } s[maxn]; 11 int x[maxn]; 12 int a[maxn][maxn]; 13 14 int main() { 15 int n; 16 scanf("%d", &n); 17 int zero = 0; 18 for (int i = 1; i <= n; i++) { 19 scanf("%d%d", &s[i].l, &s[i].r); 20 if (s[i].l == 0) 21 zero++; 22 } 23 int m; 24 scanf("%d", &m); 25 int flag = 1; 26 for (int i = 1; i <= m; i++) 27 scanf("%d", &x[i]); 28 if (zero < m) { 29 printf("impossible\n"); 30 return 0; 31 } 32 for (int i = 1; i <= m; i++) { 33 int v = x[i], pos = -1, val = INF; 34 a[i][0] = 1; 35 while (v != 0) { 36 pos = -1; 37 val = INF; 38 for (int j = 1; j <= n; j++) { 39 if (s[j].r >= v && s[j].f == 0) { 40 if (s[j].r < val) { 41 val = s[j].r; 42 pos = j; 43 } 44 } 45 } 46 if (pos == -1) { 47 flag = 0; 48 break; 49 } 50 v = s[pos].l; 51 a[i][a[i][0]++] = pos; 52 s[pos].f = 1; 53 } 54 if (flag == 0) { 55 break; 56 } 57 } 58 if (flag == 0) { 59 printf("impossible\n"); 60 } else { 61 for (int i = 1; i <= m; i++) { 62 for (int j = a[i][0] - 1; j > 0; j--) { 63 if (j == 1) 64 printf("%d", a[i][j]); 65 else 66 printf("%d ", a[i][j]); 67 } 68 printf("\n"); 69 } 70 } 71 return 0; 72 } 73 //6 74 //0 0 75 //0 1 76 //1 2 77 //3 2 78 //0 3 79 //0 3 80 //3 81 //1 2 3
L - Lounge Lizards
1 #include <bits/stdc++.h> 2 using namespace std; 3 typedef long long ll; 4 typedef pair<int,int> pii; 5 typedef pair<ll,int>pli; 6 const int inf=0x3f3f3f3f,N=1e6+7; 7 map<pii,int>m; 8 int n,res,cnt,a[N],lis[N]; 9 vector<pli>g[N]; 10 inline ll read() 11 { 12 int x=0,f=1; 13 char ch=getchar(); 14 while(ch<‘0‘||ch>‘9‘) 15 { 16 if(ch==‘-‘)f=-1; 17 ch=getchar(); 18 } 19 while(ch>=‘0‘&&ch<=‘9‘) 20 { 21 x=x*10+ch-‘0‘; 22 ch=getchar(); 23 } 24 return x*f; 25 } 26 inline int gcd(int a,int b) 27 { 28 return b?gcd(b,a%b):a; 29 } 30 inline ll dis(int x,int y) 31 { 32 return (ll)x*x+(ll)y*y; 33 } 34 inline int LIS(int n) 35 { 36 for(int i=1; i<n; i++) lis[i]=inf; 37 lis[0]=a[0]; 38 int len=1; 39 for(int i=0; i<n; i++) 40 { 41 if(a[i]>lis[len-1]) lis[len++]=a[i]; 42 else 43 { 44 int p=lower_bound(lis, lis+len, a[i])-lis; 45 lis[p]=a[i]; 46 } 47 } 48 return len; 49 } 50 inline void init(int x,int y,int h) 51 { 52 ll d=dis(x,y); 53 if(x==0) y/=abs(y); 54 else if(y==0) x/=abs(x); 55 else 56 { 57 int z=gcd(abs(x),abs(y)); 58 x/=z,y/=z; 59 } 60 if(m.find(pii(x,y))==m.end()) m[pii(x,y)]=++res; 61 g[m[pii(x,y)]].push_back(pli(d,h)); 62 } 63 int main() 64 { 65 int tx,ty,x,y,h; 66 tx=read(),ty=read(),n=read(); 67 for(int i=1; i<=n; i++) 68 { 69 x=read(),y=read(),h=read(); 70 x-=tx,y-=ty; 71 init(x,y,h); 72 } 73 int ans=0; 74 for(int i=1; i<=res; i++) 75 { 76 sort(g[i].begin(),g[i].end()); 77 cnt=g[i].size(); 78 for(int j=0; j<cnt; j++) a[j]=g[i][j].second; 79 ans+=LIS(cnt); 80 } 81 printf("%d\n",ans); 82 return 0; 83 }
标签：strong hat sqrt cal lin ota fabs pair open
原文地址：https://www.cnblogs.com/lorelei/p/9550176.html
