http://codeforces.com/gym/100495
K题 草地的面积减去相交的面积，计算几何，垃圾题，避免不必要的计算损失精度（能约分的约分）
卡了老子一个星期了 再加前几天的一道题 这一星期真的是难受 什么都没干
AC代码
1 #include <stdio.h> 2 #include <math.h> 3 #include <string.h> 4 #include <stdlib.h> 5 #include <iostream> 6 #include <sstream> 7 #include <algorithm> 8 #include <string> 9 #include <queue> 10 #include <map> 11 #include <vector> 12 using namespace std; 13 const int maxn = 1e3+50; 14 const int maxm = 1e4+10; 15 const int inf = 0x3f3f3f3f; 16 const int mod = 998244353; 17 const double epx = 1e-10; 18 typedef long long ll; 19 const ll INF = 1e18; 20 const double pi = acos(-1.0); 21 int main() 22 { 23 int t,kase=1; 24 double x1,y1,r1,x2,y2,r2; 25 cin>>t; 26 while(t--) 27 { 28 scanf("%lf %lf %lf %lf %lf %lf",&x1,&y1,&r1,&x2,&y2,&r2); 29 double len=sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2)); 30 double ans; 31 if(len>=r1+r2) 32 { 33 ans=r1*r1*pi; 34 } 35 else if(fabs(r1-r2)>=len) 36 { 37 if(r1>r2) 38 ans=pi*r1*r1-pi*r2*r2; 39 else 40 ans=0; 41 } 42 else 43 { 44 double a1=acos((len*len+r1*r1-r2*r2)/(2*len*r1)); 45 double a2=acos((len*len+r2*r2-r1*r1)/(2*len*r2)); 46 double area1=a1*r1*r1; 47 double area2=a2*r2*r2; 48 double area3=len*r1*sin(a1); 49 ans=pi*r1*r1-(area1+area2-area3); 50 } 51 printf("Case #%d: %.8lf ",kase++,ans); 52 } 53 } 54 close
给出被卡一万年的K的代码
1 #include <stdio.h> 2 #include <math.h> 3 #include <string.h> 4 #include <stdlib.h> 5 #include <iostream> 6 #include <sstream> 7 #include <algorithm> 8 #include <string> 9 #include <queue> 10 #include <map> 11 #include <vector> 12 using namespace std; 13 const int maxn = 1e3+50; 14 const int maxm = 1e4+10; 15 const int inf = 0x3f3f3f3f; 16 const int mod = 998244353; 17 const double epx = 1e-10; 18 typedef long long ll; 19 const ll INF = 1e18; 20 const double pi = acos(-1.0); 21 int main() 22 { 23 int t,kase=1; 24 double x1,y1,r1,x2,y2,r2; 25 cin>>t; 26 while(t--) 27 { 28 scanf("%lf %lf %lf %lf %lf %lf",&x1,&y1,&r1,&x2,&y2,&r2); 29 double len=sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2)); 30 double ans; 31 if(len>=r1+r2) 32 { 33 ans=r1*r1*pi; 34 } 35 else if(fabs(r1-r2)>=len) 36 { 37 if(r1>r2) 38 ans=pi*r1*r1-pi*r2*r2; 39 else 40 ans=0; 41 } 42 else 43 { 44 double a1=2.0*acos((len*len+r1*r1-r2*r2)/(2*len*r1)); 45 double a2=2.0*acos((len*len+r2*r2-r1*r1)/(2*len*r2)); 46 double area1=0.5*a1*r1*r1; 47 double area2=0.5*a2*r2*r2; 48 double area3=len*r1*sin(a1); 49 ans=pi*r1*r1-(area1+area2-area3); 50 } 51 printf("Case #%d: %.11lf ",kase++,ans); 52 } 53 }
B题 把每个字符串开头和结尾中间的字符排序，用map标记一下 然后在判断是否出现过就好了
1 #include <stdio.h> 2 #include <math.h> 3 #include <string.h> 4 #include <stdlib.h> 5 #include <iostream> 6 #include <sstream> 7 #include <algorithm> 8 #include <string> 9 #include <queue> 10 #include <map> 11 #include <vector> 12 using namespace std; 13 const int maxn = 1e3+50; 14 const int maxm = 1e4+10; 15 const int inf = 0x3f3f3f3f; 16 const int mod = 998244353; 17 const double epx = 1e-10; 18 typedef long long ll; 19 const ll INF = 1e18; 20 const double pi = acos(-1.0); 21 string s; 22 int ans[maxn]; 23 int main() 24 { 25 int n,m,t,kase=1; 26 cin>>t; 27 while(t--) 28 { 29 map<string,int> mp; 30 cin>>n>>m; 31 for(int i=0;i<n;i++) 32 { 33 cin>>s; 34 int len=s.length(); 35 if(len>2) 36 sort(s.begin()+1,s.end()-1); 37 //cout<<len<<" "<<s<<endl; 38 mp[s]++; 39 } 40 for(int i=0;i<m;i++) 41 { 42 cin>>s; 43 int len=s.length(); 44 if(len>2) 45 sort(s.begin()+1,s.end()-1); 46 //cout<<len<<" "<<s<<endl; 47 if(mp[s]>0) 48 ans[i]=1; 49 else 50 ans[i]=0; 51 } 52 printf("Case #%d: ",kase++); 53 for(int i=0;i<m;i++) 54 cout<<ans[i]; 55 cout<<endl; 56 } 57 }
E题 可以推出要求子串满足min+a>=max 的最大长度 尺取法 右端点从左端点开始推进 直到右端点不符合条件
再从左端点开始到右端点找最值 并且不断推进左端点 直到满足条件 再向右推进
1 #include <stdio.h> 2 #include <math.h> 3 #include <string.h> 4 #include <stdlib.h> 5 #include <iostream> 6 #include <sstream> 7 #include <algorithm> 8 #include <string> 9 #include <queue> 10 #include <map> 11 #include <vector> 12 using namespace std; 13 const int maxn = 1e5+50; 14 const int maxm = 1e4+10; 15 const int inf = 0x3f3f3f3f; 16 const int mod = 998244353; 17 const double epx = 1e-10; 18 typedef long long ll; 19 const ll INF = 1e18; 20 int s[maxn]; 21 int main() 22 { 23 int t,a,n,kase=1; 24 scanf("%d",&t); 25 while(t--) 26 { 27 scanf("%d %d",&n,&a); 28 memset(s,0,sizeof(s)); 29 for(int i=1;i<=n;i++) 30 scanf("%d",&s[i]); 31 int len=1; 32 int l=1,maxx,minn; 33 maxx=minn=s[l]; 34 for(int r=1;r<=n;) 35 { 36 if(minn+a>=maxx) 37 { 38 len=max(len,r-l+1); 39 r++; 40 if(r>n) 41 break; 42 maxx=max(maxx,s[r]); 43 minn=min(minn,s[r]); 44 } 45 else 46 { 47 if(s[r]==maxx) 48 { 49 while(minn+a<maxx) 50 { 51 int tempmin=s[r]; 52 for(int i=l+1;i<=r;i++) 53 tempmin=min(tempmin,s[i]); 54 minn=tempmin; 55 l++; 56 } 57 } 58 else if(s[r]==minn) 59 { 60 while(minn+a<maxx) 61 { 62 int tempmaxx=s[r]; 63 for(int i=l+1;i<=r;i++) 64 tempmaxx=max(tempmaxx,s[i]); 65 maxx=tempmaxx; 66 l++; 67 } 68 } 69 } 70 //cout<<"l="<<l<<" r="<<r<<" min="<<minn<<" max="<<maxx<<endl; 71 } 72 printf("Case #%d: %d ",kase++,len); 73 } 74 } 75 close
D题 水题 边乘边取模
1 #include <stdio.h> 2 #include <math.h> 3 #include <string.h> 4 #include <stdlib.h> 5 #include <iostream> 6 #include <sstream> 7 #include <algorithm> 8 #include <string> 9 #include <queue> 10 #include <map> 11 #include <vector> 12 using namespace std; 13 const int maxn = 1e6+10; 14 const int maxm = 1e4+10; 15 const int inf = 0x3f3f3f3f; 16 const int mod = 998244353; 17 const double epx = 1e-10; 18 typedef long long ll; 19 const ll INF = 1e18; 20 ll n,m; 21 char a[maxn]; 22 int sum[maxn]; 23 int sushu(int x) 24 { 25 for(int i=2;i<n;i++) 26 if(x%i==0) 27 return 0; 28 return 1; 29 } 30 int poww(int x,int y,int z) 31 { 32 int ans=1; 33 for(int i=1;i<=y;i++) 34 ans=(ans*x)%n; 35 return ans; 36 } 37 int main() 38 { 39 int t; 40 cin>>t; 41 int kase=1; 42 while(t--) 43 { 44 cin>>n; 45 if(n==1) 46 printf("Case #%d: %d ",kase,n); 47 else if(sushu(n)) 48 { 49 printf("Case #%d: %d ",kase,poww(2,n-1,n)%n); 50 } 51 else 52 printf("Case #%d: %d ",kase,poww(n-1,2,n)%n); 53 kase++; 54 } 55 } 56 close
F题 一遍dfs回溯，记录最大深度就好了
1 #include <stdio.h> 2 #include <math.h> 3 #include <string.h> 4 #include <stdlib.h> 5 #include <iostream> 6 #include <sstream> 7 #include <algorithm> 8 #include <string> 9 #include <queue> 10 #include <map> 11 #include <vector> 12 using namespace std; 13 const int maxn = 1e3+50; 14 const int maxm = 1e4+10; 15 const int inf = 0x3f3f3f3f; 16 const int mod = 998244353; 17 const double epx = 1e-10; 18 typedef long long ll; 19 const ll INF = 1e18; 20 char a[maxn][maxn]; 21 int visit[maxn][maxn]; 22 int n,m,k; 23 int sum,flag; 24 int dire[4][2]= {{1,0},{-1,0},{0,-1},{0,1}}; 25 void dfs(int x,int y) 26 { 27 visit[x][y]=1; 28 for(int i=0; i<4; i++) 29 { 30 int x1=(x+dire[i][0]+n)%n; 31 int y1=(y+dire[i][1]+m)%m; 32 //printf("x=%d y=%d x1=%d y1=%d sum=%d c=%c ",x,y,x1,y1,sum,a[x1][y1]); 33 if(visit[x1][y1]==0&&a[x1][y1]!='#'&&a[x1][y1]!='x') 34 { 35 if(a[x1][y1]=='o') 36 { 37 dfs(x1,y1); 38 visit[x1][y1]=0; 39 } 40 else if(a[x1][y1]=='.') 41 { 42 sum++; 43 if(sum>=k) 44 flag=1; 45 dfs(x1,y1); 46 visit[x1][y1]=0; 47 sum--; 48 } 49 } 50 } 51 } 52 int main() 53 { 54 int t,kase=1; 55 cin>>t; 56 while(t--) 57 { 58 cin>>n>>m>>k; 59 for(int i=0; i<n; i++) 60 cin>>a[i]; 61 int x,y; 62 for(int i=0; i<n; i++) 63 for(int j=0; j<m; j++) 64 if(a[i][j]=='x') 65 x=i,y=j; 66 // cout<<x<<" "<<y<<endl; 67 // for(int i=0;i<4;i++) 68 // cout<<dire[i][0]<<" "<<dire[i][1]<<endl; 69 memset(visit,0,sizeof(visit)); 70 flag=0;sum=1;dfs(x,y); 71 // for(int i=0;i<n;i++) 72 // { 73 // for(int j=0;j<m;j++) 74 // { 75 // cout<<visit[i][j]<<" "; 76 // } 77 // cout<<endl; 78 // } 79 if(flag==1) 80 printf("Case #%d: Fits perfectly! ",kase++); 81 else 82 printf("Case #%d: Oh no, snake's too fat! ",kase++); 83 } 84 }