csps模拟测试707172部分题解myc

题面：https://www.cnblogs.com/Juve/articles/11678524.html

骆驼：构造题，留坑

根据5×5的矩形构造成大矩形

毛一琛：

mid in the middle思想，先dfs一半，然后dfs后一半

 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #include<unordered_map>
 #include<set>
 #define int long long
 #define re register
 using namespace std;
 const int MAXN=;
 int n,m[MAXN],ans=,tot=;
 unordered_map<int,int>mp;
 set<int>s[<<];
 bool vis[<<];
 void DFS(re int now,re int num,re int st){
     if(now>n){
         if(mp.find(num)==mp.end()) return ;
         int t=mp[num];
         for(set<int>::iterator it=s[t].begin();it!=s[t].end();++it){
             if(vis[(*it)|st]==) ++ans;
             vis[(*it)|st]=;
         }
         return ;
     }
     DFS(now+,num,st);
     DFS(now+,num+m[now],st|(<<(now-)));
     DFS(now+,num-m[now],st|(<<(now-)));
 }
 void dfs(re int now,re int num,re int st){
     if(now>n/){
         int t;
         if(mp.find(num)==mp.end()) t=mp[num]=++tot;
         else t=mp[num];
         s[t].insert(st);
         return ;
     }
     dfs(now+,num,st);
     dfs(now+,num+m[now],st|(<<(now-)));
     dfs(now+,num-m[now],st|(<<(now-)));
 }
 signed main(){
     scanf("%lld",&n);
     for(re int i=;i<=n;++i) scanf("%lld",&m[i]);
     vis[]=;
     dfs(,,);DFS(n/+,,);
     printf("%lld\n",ans);
     return ;
 }

毛二琛：dp，不会

毛三琛：

二分思路考场上想到了，但是没有想到优化骗分

把x随机排个序，二分前先check一下是否比当先的ans优，不优就不二分

再加个clock

 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #define int long long
 #define re register
 using namespace std;
 const int MAXN=1e4+;
 int n,k,p,a[MAXN],res,l=0x3f3f3f3f,r=,ans=,x[MAXN],b[MAXN];
 inline bool check(re int lim){
     re int tot=,sum=;
     for(re int i=;i<=n;++i){
         if(b[i]>lim) return ;
         if(sum+b[i]>lim){
             ++tot;
             sum=b[i];
             if(tot>=k) return ;
         }else sum+=b[i];
     }
     return ;
 }
 signed main(){
     re double ck=clock();
     scanf("%lld%lld%lld",&n,&p,&k);
     for(re int i=;i<=n;++i){
         scanf("%lld",&a[i]);
         r+=a[i],l=min(l,a[i]);
     }
     for(re int i=;i<=p;++i) x[i]=i-;
     srand(time());
     random_shuffle(x+,x+p+);
     ans=r;
     for(re int i=;i<=p;++i){
         if(clock()-ck>) break;
         l=,r=ans;
         for(re int j=;j<=n;++j){
             b[j]=(a[j]+x[i])%p;
             l=max(l,b[j]);
         }
         if(!check(r)) continue;
         while(l<r){
             re int mid=(l+r)>>;
             if(check(mid)) r=mid;
             else l=mid+;
         }
         ans=min(ans,l);
     }
     printf("%lld\n",ans);
     return ;
 }

简单的序列：

推式子，dp也可以

 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #define int long long
 using namespace std;
 const int MAXN=1e5+,mod=1e9+;
 int n,m,tot=,ans=,cnt=;
 char s[MAXN];
 int max(int a,int b){
     return a>b?a:b;
 }
 int q_pow(int a,int b,int p){
     int res=;
     while(b){
         if(b&) res=res*a%p;
         a=a*a%p;
         b>>=;
     }
     return res;
 }
 int fac(int N){
     int res=;
     for(int i=;i<=N;++i) res=res*i%mod;
     return res%mod;
 }
 int C(int n,int m){
     if(m>n) return ;
     if(m==n) return ;
     return fac(n)*q_pow(fac(m)%mod,mod-,mod)%mod*q_pow(fac(n-m)%mod,mod-,mod)%mod;
 }
 int sta[MAXN],top=;
 signed main(){
     scanf("%lld%lld%s",&n,&m,s+);
     for(int i=;i<=m;++i){
         if(s[i]=='(') sta[++top]=;
         else if(top) ++cnt,--top;
     }
     if(n&){
         puts("");
         return ;
     }
     ans=C(n-m+,n/-m+cnt);
     printf("%lld\n",ans);
     return ;
 }

简单的期望

dp[i][j][k][0/1]表示操作了i次，最后8位是j，右移8位后最后有连续k位是0/1。

 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #define int long long
 #define re register
 using namespace std;
 const int MAXN=;
 int x,n,pre[];
 double p,ans,f[MAXN][(<<)+][][];
 signed main(){
     scanf("%lld%lld%lf",&x,&n,&p);
     for(int i=;i<=(<<);++i) pre[i*]=pre[i]+;
     int d=x&,b=x&,c=;
     x>>=;p/=100.0;
     while(x&&(x&)==d){
         x>>=;
         ++c;
     }
     f[][b][c][d]=1.0;
     for(int i=;i<=n;++i){
         for(int j=;j<(<<);++j){
             for(int k=;k<=;++k){
                 if(j==((<<)-)){
                     f[i][][k][]+=f[i-][j][k][]*(1.0-p);
                     f[i][][][]+=f[i-][j][k][]*(1.0-p);
                 }else{
                     f[i][j+][k][]+=f[i-][j][k][]*(1.0-p);
                     f[i][j+][k][]+=f[i-][j][k][]*(1.0-p);
                 }
                 if(j>=(<<)){
                     int b=(j<<)&((<<)-);
                     int tmp=((j<<)&(<<))>>;
                     if(tmp!=){
                         f[i][b][k+][]+=f[i-][j][k][]*p;
                         f[i][b][][tmp]+=f[i-][j][k][]*p;
                     }else{
                         f[i][b][][tmp]+=f[i-][j][k][]*p;
                         f[i][b][k+][]+=f[i-][j][k][]*p;
                     }
                 }else{
                     f[i][j<<][][]+=f[i-][j][k][]*p;
                     f[i][j<<][k+][]+=f[i-][j][k][]*p;
                 }
             }
         }
     }
     for(int j=;j<=(<<);j++){
         for(int k=;k<=;k++)
             ans+=pre[j]*(f[n][j][k][]+f[n][j][k][]);
     }
     for(int k=;k<=;k++)
         ans+=((f[n][][k][]+f[n][][k][])*)+f[n][][k][]*k;
     printf("%0.6lf\n",ans);
     return ;
 }