A.小迟的比赛
最优策略永远是努力应战，dp[i][j]表示前i轮赢了j局的概率，dp[i][j]=dp[i-1][j]*(1-p[i][j])+dp[i-1][j-1]*p[i-1][j-1]
#include <iostream> #include<cstdio> #include<cstring> #include<cmath> #include<algorithm> using namespace std; double p[1005][1005]; double dp[1005][1005]; int main() { int n;scanf("%d",&n); for(int i=1;i<=n;i++){ for(int j=0;j<=i-1;j++){ scanf("%lf",&p[i][j]); } } dp[1][1]=p[1][0]; dp[1][0]=1-p[1][0]; for(int i=2;i<=n;i++){ for(int j=0;j<=i;j++){ if(j==0) dp[i][j]=dp[i-1][j]*(1-p[i][j]); else if(j==i) dp[i][j]=dp[i-1][j-1]*p[i][j-1]; else dp[i][j]=dp[i-1][j]*(1-p[i][j])+dp[i-1][j-1]*p[i][j-1]; //printf("%.2f ",dp[i][j]); } } double ans=0; for(int i=0;i<=n;i++){ ans=ans+i*dp[n][i]; } printf("%.2f ",ans); return 0; }
meet in middle 搜索
#include <iostream> #include<cstdio> #include<cstring> #include<cmath> #include<algorithm> typedef long long ll; using namespace std; ll n,v; ll p[45]; ll all[2000000],cnt=0; ll ans=0; void dfs1(ll x,ll tot){ if(tot>v) return; if(x>=n/2){ all[++cnt]=tot; return; } dfs1(x+1,tot); dfs1(x+1,tot+p[x+1]); } void dfs2(ll x,ll tot){ if(tot>v) return; if(x>=n){ ll tmp=v-tot; ll p=upper_bound(all+1,all+1+cnt,tmp)-all; ans=max(ans,tot+all[p-1]); return; } dfs2(x+1,tot); dfs2(x+1,tot+p[x+1]); } int main() { scanf("%lld%lld",&n,&v); for(ll i=1;i<=n;i++) scanf("%lld",&p[i]); dfs1(1,0); dfs1(1,p[1]); sort(all+1,all+1+cnt); //for(ll i=1;i<=cnt;i++) printf("%lld ",all[i]); dfs2(n/2+1,0); dfs2(n/2+1,p[n/2+1]); printf("%lld ",v-ans); return 0; }
A.阿瓦的手套加强版
并查集
#include <iostream> #include<cstdio> #include<cstring> typedef long long ll; #define mod 998244353 using namespace std; ll fa[100005]; ll findd(ll x){ if(x==fa[x]) return x; else return fa[x]=findd(fa[x]); } ll qpow(ll a,ll b){ ll ret=1; while(b){ if(b&1) ret=ret*a%mod; a=a*a%mod; b>>=1; } return ret; } int main() { ll n,m,T;scanf("%lld%lld%lld",&n,&m,&T); for(ll i=1;i<=n;i++) fa[i]=i; ll x,y; for(ll i=1;i<=T;i++){ scanf("%lld%lld",&x,&y); ll fax=findd(x),fay=findd(y); if(fax==fay) continue; else fa[fax]=fay; } ll cnt=0; for(ll i=1;i<=n;i++){ if(fa[i]==i) cnt++; } ll ans=qpow(m,cnt); printf("%lld ",ans); return 0; }
B.阿卡吃百奇
规律
import java.math.BigInteger; import java.util.Scanner; public class Main { public static void main(String[] args) { Scanner cin=new Scanner(System.in); int t=cin.nextInt(); for(int i=1;i<=t;i++){ int r=cin.nextInt(); if(r==1){ System.out.println(1); continue; } BigInteger ans; int tmp=(r-1)%3; if(tmp==0){ int b=(r-1)/3-1; ans=BigInteger.valueOf(1); BigInteger a=BigInteger.valueOf(3); while(b!=0){ if((b&1)!=0) ans=ans.multiply(a); a=a.multiply(a); b=b>>1; } ans=ans.multiply(BigInteger.valueOf(4)); } else{ int b=(r-1)/3; ans=BigInteger.valueOf(1); BigInteger a=BigInteger.valueOf(3); while(b!=0){ if((b&1)!=0) ans=ans.multiply(a); a=a.multiply(a); b=b>>1; } if(tmp==1) ans=ans.multiply(BigInteger.valueOf(2)); else ans=ans.multiply(BigInteger.valueOf(3)); } System.out.println(ans); } } }
H.Jollo
枚举
#include <iostream> #include<cstdio> #include<cstring> #include<algorithm> using namespace std; int a[5],b[5],c[5]; int vis[55]; int main() { int t;scanf("%d",&t); while(t--){ memset(vis,0,sizeof(vis)); scanf("%d%d%d%d%d",&a[1],&a[2],&a[3],&b[1],&b[2]); vis[a[1]]=vis[a[2]]=vis[a[3]]=vis[b[1]]=vis[b[2]]=1; int i; for(i=1;i<=52;i++){ if(vis[i]) continue; c[1]=b[1],c[2]=b[2],c[3]=i; int cnt=0; sort(a+1,a+4);sort(c+1,c+4); for(int p1=3,p2=3;p1>=1&&p2>=1;){ if(a[p1]>c[p2]){ cnt++; p1--,p2--; } else p2--; } if(cnt>=2) continue; else break; } if(i<=52) printf("%d ",i); else printf("-1 "); } return 0; }
F.小Z的省选
模拟
#include <iostream> #include<cstdio> #include<cstring> #include<algorithm> typedef long long ll; using namespace std; ll id; char res[10]; char times[35]; char first_sub[100005][35]; char first_ac[100005][35]; ll days1[13]={366,31,29,31,30,31,30,31,31,30,31,30,31}; ll days2[13]={365,31,28,31,30,31,30,31,31,30,31,30,31}; bool run(ll y){ if((y%4==0&&y%100!=0)||y%400==0) return 1; return 0; } struct Node{ ll y,m,d; bool operator <(const Node& b)const{ if(y!=b.y) return y<b.y; else if(m!=b.m) return m<b.m; else return d<b.d; } }; Node get_times(char s[]){ ll y1=0,m1=0,d1=0; ll flag=0; ll len=(ll)strlen(s); for(ll k=0;k<len;k++){ if(s[k]=='/') {flag++;continue;} if(flag==0) y1=y1*10+(s[k]-'0'); else if(flag==1) d1=d1*10+(s[k]-'0'); else m1=m1*10+(s[k]-'0'); } return Node{y1,m1,d1}; } int main() { ll n,m;scanf("%lld%lld",&n,&m); for(ll i=1;i<=n;i++){ first_sub[i][0]='n'; first_ac[i][0]='n'; } for(ll i=1;i<=m;i++){ scanf("%d",&id); scanf("%s%s",res,times); if(res[0]!='C'){ if(first_sub[id][0]=='n'||get_times(times)<get_times(first_sub[id])) strcpy(first_sub[id],times); } if(res[0]=='A'){ if(first_ac[id][0]=='n'||get_times(times)<get_times(first_ac[id])) strcpy(first_ac[id],times); } } ll ans=0; for(ll i=1;i<=n;i++){ if(first_ac[i][0]=='n') continue; ll y1=get_times(first_sub[i]).y,m1=get_times(first_sub[i]).m,d1=get_times(first_sub[i]).d; ll y2=get_times(first_ac[i]).y,m2=get_times(first_ac[i]).m,d2=get_times(first_ac[i]).d; ll tot=0; if(y1==y2){ if(m1==m2) tot=(d2-d1+1); else{ ll tmp1; if(run(y1)) tmp1=days1[m1]-d1+1; else tmp1=days2[m1]-d1+1; ll tmp2=d2; ll tmp3=0; for(ll k=m1+1;k<=m2-1;k++){ if(run(y1)) tmp3+=days1[k]; else tmp3+=days2[k]; } tot=tmp1+tmp2+tmp3; } } else{ ll tmp1=0; for(ll k=1;k<=m1-1;k++){ if(run(y1)) tmp1+=days1[k]; else tmp1+=days2[k]; } tmp1+=d1; if(run(y1)) tmp1=days1[0]-tmp1+1; else tmp1=days2[0]-tmp1+1; ll tmp2=0; for(ll k=1;k<=m2-1;k++){ if(run(y2)) tmp2+=days1[k]; else tmp2+=days2[k]; } tmp2+=d2; ll tmp3=0; if(y2-1>=y1+1){ ll num=(y2-1)/4-(y1)/4-((y2-1)/100-(y1)/100)+((y2-1)/400-(y1)/400); tmp3=(y2-1-(y1+1)+1)*365+num; } tot=tmp1+tmp2+tmp3; } ans+=tot; } printf("%lld ",ans); return 0; }