noip模擬71(待補)
阿新 • • 發佈：2021-10-09
「T1 出了個大陰間題·T2 最簡單辣快來做·T3 是我的你不要搶·T4 顯然也是我整的」
T1 出了個大陰間題
狀壓 \(dp\)，設 \(f_{s,k}\) 表示 \(s\) 狀態下，當前的 \(max_a -\) 最大值等於 \(j\) 的所有代價.
有一個不會對分數有什麼影響的規律，就是 \(k\) 永遠 \(\le 1\).
A_code
#include<bits/stdc++.h> using namespace std; namespace BSS { #define ll long long #define ull unsigned ll #define lf long double #define lbt(x) (x&(-x)) #define mp(x,y) make_pair(x,y) #define lb lower_bound #define ub upper_bound #define Fill(x,y) memset(x,y,sizeof x) #define Copy(x,y) memcpy(x,y,sizeof x) #define File(x) freopen(#x".in","r",stdin),freopen(#x".out","w",stdout) inline ll read() { ll w=0; bool cit=1; char ch; while(!isdigit(ch=getchar())) if(ch=='-') cit=0; while(isdigit(ch)) w=(w<<3)+(w<<1)+(ch^48),ch=getchar(); return cit?w:-w; } } using namespace BSS; const ll mod=1e9+7; ll m,n,U,maxval; ll val[25],b[25]; ll num[1<<18]; map<ll,ll> f[1<<18],g[1<<18]; signed main(){ File(repair); n=read(),m=read()%mod,U=(1<<n)-1; for(ll i=1;i<=n;i++) val[i]=read(); for(ll i=2;i<=n;i++) b[i]=b[i-1]<<1|1; for(ll i=1;i<=n;i++) f[1<<i-1][val[i]]=0,g[1<<i-1][val[i]]++; for(ll i=1;i<U;i++){ num[i]=num[i&(i-1)]+1; for(ll j=1,k;j<=n;j++){ if((i>>j-1)&1) continue; k=i|(1<<j-1); for(auto c : f[i]){ ll to=(c.first==val[j] ? val[j]+1 : max(c.first,val[j])); f[k][to]=(f[k][to]+f[i][c.first]+(to*m%mod+b[num[i]])%mod*g[i][c.first]%mod)%mod; g[k][to]=(g[k][to]+g[i][c.first])%mod; } } } for(auto i : f[U]) maxval=max(maxval,i.first); printf("%lld %lld\n",maxval,f[U][maxval]%mod),exit(0); }
T2 最簡單辣快來做
自以為是打的是正解，其實連別人的暴力都不如.
自以為把簡單的想複雜就可能是正解，像個笑話.
正解是利用分塊思想，考慮把每個橫縱座標都有當成一條線，於是變成了 \(n^2\) 個矩形.
然後字首和即可.
學習了光速冪.
B_code
#include<bits/stdc++.h> using namespace std; namespace BSS { #define ll long long #define ull unsigned ll #define lf long double #define lbt(x) (x&(-x)) #define mp(x,y) make_pair(x,y) #define lb lower_bound #define ub upper_bound #define Fill(x,y) memset(x,y,sizeof x) #define Copy(x,y) memcpy(x,y,sizeof x) #define File(x) freopen(#x".in","r",stdin),freopen(#x".out","w",stdout) inline ll read() { ll w=0; bool cit=1; char ch; while(!isdigit(ch=getchar())) if(ch=='-') cit=0; while(isdigit(ch)) w=(w<<3)+(w<<1)+(ch^48),ch=getchar(); return cit?w:-w; } } using namespace BSS; const ll N=2e3+21,W=1e5; ll m,n,s,ops,mod,cntx,cnty; ll lshx[N],lshy[N]; ll rk[N][N]; ll pre[5][N][N]; struct I { ll x,y,h,id; } p[N*100]; struct Flash_Pow{ ll val; ll sm[W+21],bg[W+21],po[N]; inline void Pre(){ sm[0]=1; for(ll i=1;i<=W;i++) sm[i]=sm[i-1]*val%mod; bg[0]=1; for(ll i=1;i<=W;i++) bg[i]=bg[i-1]*sm[W]%mod; } inline ll qpow(ll x){ return sm[x%W]*bg[x/W]%mod; } } A,B; signed main(){ File(satellite); s=read(),ops=read(),n=read(),m=read(); ll x,y,res; mod=read(),A.val=read(),B.val=read(),A.Pre(),B.Pre(); for(ll i=1;i<=s;i++){ p[i].h=read()%mod,p[i].id=i; lshx[++cntx]=(p[i].x=read()),lshy[++cnty]=(p[i].y=read()); } sort(lshx+1,lshx+1+cntx),cntx=unique(lshx+1,lshx+1+cntx)-lshx-1; sort(lshy+1,lshy+1+cnty),cnty=unique(lshy+1,lshy+1+cnty)-lshy-1; for(ll i=1;i<=cntx;i++) A.po[i]=A.qpow(lshx[i]-lshx[i-1]); for(ll i=1;i<=cnty;i++) B.po[i]=B.qpow(lshy[i]-lshy[i-1]); for(ll i=1;i<=s;i++){ p[i].x=lb(lshx+1,lshx+1+cntx,p[i].x)-lshx,p[i].y=lb(lshy+1,lshy+1+cnty,p[i].y)-lshy; rk[p[i].x][p[i].y]=p[i].id; pre[1][p[i].x][p[i].y]+=p[i].h,pre[2][p[i].x][p[i].y]+=p[i].h; pre[3][p[i].x][p[i].y]+=p[i].h,pre[4][p[i].x][p[i].y]+=p[i].h; } for(ll i=1;i<=cntx;i++){ for(ll j=1;j<=cnty;j++){ y=pre[1][i][j]%mod; y=(y+pre[1][i-1][j]*A.po[i]%mod+pre[1][i][j-1]*B.po[j]%mod)%mod; y=(y-pre[1][i-1][j-1]*A.po[i]%mod*B.po[j]%mod+mod)%mod,pre[1][i][j]=y; } for(ll j=cnty;j>=1;j--){ y=pre[2][i][j]%mod; y=(y+pre[2][i-1][j]*A.po[i]%mod+pre[2][i][j+1]*B.po[j+1]%mod)%mod; y=(y-pre[2][i-1][j+1]*A.po[i]%mod*B.po[j+1]%mod+mod)%mod,pre[2][i][j]=y; } } for(ll i=cntx;i>=1;i--){ for(ll j=1;j<=cnty;j++){ y=pre[3][i][j]%mod; y=(y+pre[3][i+1][j]*A.po[i+1]%mod+pre[3][i][j-1]*B.po[j]%mod)%mod; y=(y-pre[3][i+1][j-1]*A.po[i+1]%mod*B.po[j]%mod+mod)%mod,pre[3][i][j]=y; } for(ll j=cnty;j>=1;j--){ y=pre[4][i][j]%mod; y=(y+pre[4][i+1][j]*A.po[i+1]%mod+pre[4][i][j+1]*B.po[j+1]%mod)%mod; y=(y-pre[4][i+1][j+1]*A.po[i+1]%mod*B.po[j+1]%mod+mod)%mod,pre[4][i][j]=y; } } while(ops--){ x=read(),y=read(),res=0; ll i=ub(lshx+1,lshx+1+cntx,x)-lshx-1,j=ub(lshy+1,lshy+1+cnty,y)-lshy-1; res=(res+pre[1][i][j]*A.qpow(abs(x-lshx[i]))%mod*B.qpow(abs(y-lshy[j]))%mod)%mod; res=(res+pre[2][i][j+1]*A.qpow(abs(x-lshx[i]))%mod*B.qpow(abs(y-lshy[j+1]))%mod)%mod; res=(res+pre[3][i+1][j]*A.qpow(abs(x-lshx[i+1]))%mod*B.qpow(abs(y-lshy[j]))%mod)%mod; res=(res+pre[4][i+1][j+1]*A.qpow(abs(x-lshx[i+1]))%mod*B.qpow(abs(y-lshy[j+1]))%mod)%mod; printf("%lld\n",res); } exit(0); }
T3 是我的你不要搶
\(Hash\) 能水過.
正解還沒打.
C_code
#include<bits/stdc++.h> using namespace std; namespace BSS { #define ll int #define ull unsigned ll #define lf long double #define lbt(x) (x&(-x)) #define mp(x,y) make_pair(x,y) #define lb lower_bound #define ub upper_bound #define Fill(x,y) memset(x,y,sizeof x) #define Copy(x,y) memcpy(x,y,sizeof x) #define File(x) freopen(#x".in","r",stdin),freopen(#x".out","w",stdout) inline ll read() { ll w=0; bool cit=1; char ch; while(!isdigit(ch=getchar())) if(ch=='-') cit=0; while(isdigit(ch)) w=(w<<3)+(w<<1)+(ch^48),ch=getchar(); return cit?w:-w; } } using namespace BSS; const ll N=6e5+21; const ull P=131; ll m,n,ops,maxlen,ans; ll len[N]; ull po[N]; vector<ull> c[N],pre[N]; unordered_map<ll,ll> mp1[N]; inline ull getsh(ll x,ll l,ll r){ return pre[x][r]-pre[x][l-1]*po[r-l+1]; } signed main(){ File(string); n=read(),ops=read(); char s[N]; ll x,y; for(ll i=1;i<=n;i++){ scanf("%s",s+1); c[i].push_back(0),pre[i].push_back(0); len[i]=strlen(s+1),maxlen=max(maxlen,len[i]); for(ll j=1;j<=len[i];j++){ c[i].push_back(s[j]-'a'+1); pre[i].push_back(pre[i][j-1]*P+c[i][j]); } } po[0]=1; for(ll i=1;i<=maxlen;i++) po[i]=po[i-1]*P; while(ops--){ y=read(),x=read(); maxlen=min(len[x],len[y]); if(mp1[x].find(y)!=mp1[x].end()){ printf("%lld\n",mp1[x][y]); continue; } for(ll i=maxlen;i>=0;i--){ if(getsh(x,1,i)==getsh(y,len[y]-i+1,len[y])){ ans=i; break; } } mp1[x][y]=ans; printf("%d\n",ans); } exit(0); }
T4 顯然也是我整的
還沒會，先鴿.
這幾場考得都不是很讓自己滿意.
每次考試的策略都很差.
這次考試上來先入手碼了碼量最長的題，甚至連 \(T1\) 和 \(T3\) 的暴力都沒打.
\(T1\) 絕對在自己的能力範圍內，自己知道是狀壓，可能推個規律就出來了.
可是自己偏偏就是看著 \(T2\) 去碼較難的.
\(T4\) 連部分分都沒拿全，這樣的策略怎麼可能行..
一定要規劃好.