2014西安網絡賽

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A. Post Robot

把每種單詞都kmp跑一遍，順序輸出即可

#include <cstdio> #include <iostream> #include <algorithm> #include <map> #include <cstring> #include <cmath> #include <queue> #include <set> #include <vector> #include <iterator> #include <string> #include <deque> #define rep(i,a,b) for(int i=a;i<=b;++i) #define per(i,a,b) for(int i=a;i>=b;--i) #define pb push_back #define MP make_pair #define fr first #define sc second #define PII pair<int,int> #define VI vector<int> typedef long long ll; typedef unsigned long long ull;inline int read() {char c=getchar();int x=0,f=1;while(!isdigit(c)){if(f=='-')f=-1;c=getchar();}while(isdigit(c)){x=x*10+c-'0';c=getchar();}return x*f; } const int N = 1000500; using namespace std;int n; string s,t,s1="Apple",s2="iPhone",s3="iPod",s4="iPad",s5="Sony"; int nxt[N],vis[N]; void kmp_pre(string s) {int i,j,m=s.size();j=nxt[0]=-1;i=0;while(i<m){while(-1!=j&&s[i]!=s[j])j=nxt[j];nxt[++i]=++j;} } void kmp(string s,string t,int f) {int i,j,n=s.size(),m=t.size();i=j=0;while(i<n){while(-1!=j&&s[i]!=t[j])j=nxt[j];++i;++j;if(j>=m) {//printf("%d ",i);vis[i] = f;j=nxt[j];}} } int main() { // freopen("in.txt","r",stdin); // freopen("out.txt","w",stdout);memset(vis,-1,sizeof(vis));int f=0;while(getline(cin,t)){if(f) s+=" ";f=1;s+=t;}kmp_pre(s1); kmp(s,s1,0);kmp_pre(s2); kmp(s,s2,0);kmp_pre(s3); kmp(s,s3,0);kmp_pre(s4); kmp(s,s4,0);kmp_pre(s5); kmp(s,s5,1);rep(i,1,s.size())if(vis[i]==0) puts("MAI MAI MAI!");else if(vis[i]==1) puts("SONY DAFA IS GOOD!");return 0; }

B. Boring String Problem

后綴數組的幫我們排好序了，然后取后綴的前綴即可。嚴格第k小，就處理一下減去Height[i]即可去重，取前綴和，二分就可以找到嚴格第k小的串，然后問題就是，求一個串它在整個串中出現的最左邊的位置。以當前這個串的左端點為中心，直接在Height數組里二分最左或最右的公共前綴長度恰好大于等于串長的位置，然后在這個區間詢問最小的SA值即可。講道理正解挺好想的。。。（用了兩天發現自己的后綴數組板子有錯，然后又發現不會二分。。。網上又找了一份大佬板子

#include <cstdio> #include <iostream> #include <algorithm> #include <map> #include <cstring> #include <cmath> #include <queue> #include <set> #include <vector> #include <iterator> #include <string> #include <deque> #define rep(i,a,b) for(int i=a;i<=b;++i) #define per(i,a,b) for(int i=a;i>=b;--i) #define pb push_back #define MP make_pair #define fr first #define sc second #define PII pair<int,int> #define VI vector<int> typedef long long ll; typedef unsigned long long ull; const int N = 100005; inline int readint() {char c=getchar();int x=0,f=1;while(!isdigit(c)){if(f=='-')f=-1;c=getchar();}while(isdigit(c)){x=x*10+c-'0';c=getchar();}return x*f; } using namespace std; int n; char c[N],str[N]; int rnk[N] , SA[N] , Height[N]; int X[N] , Y[N] , sum[N]; int f[101000][20] , fm[101000][20]; bool cmp(int *r,int a,int b,int l) {return ( r[a] == r[b] && r[a+l] == r[b+l] ); } void calc() {int l , p , *x = X , *y = Y , m = 128;rep(i,0,m) sum[i] = 0;rep(i,1,n) sum[ x[i] = c[i] ] ++;rep(i,1,m) sum[i] += sum[i-1];per(i,n,1) SA[ sum[ x[i] ]-- ] = i;for ( l = 1 , p = 1 ; l <= n ; m = p , l *= 2 ) {p = 0;rep(i,n-l+1,n) y[++p] = i;rep(i,1,n) if ( SA[i] > l ) y[++p] = SA[i] - l;rep(i,0,m) sum[i] = 0;rep(i,1,n) sum[ x[y[i]] ] ++;rep(i,1,m) sum[i] += sum[i-1];per(i,n,1) SA[ sum[ x[y[i]] ]-- ] = y[i];swap( x , y );x[SA[1]] = 1; p = 2;rep(i,2,n)x[ SA[i] ] = cmp(y,SA[i-1],SA[i],l) ? p - 1 : p++;}rep(i,1,n) rnk[SA[i]] = i;p = 0;rep(i,1,n) {if ( rnk[i] == 1 ) continue;while ( c[i+p] == c[SA[rnk[i]-1]+p] ) p ++;Height[rnk[i]] = p;if ( p ) p --;} } void init() {n = strlen(str);for(int i=0;i<n;++i) c[i+1] = str[i];c[n+1]=0; } int ST_h[N][22], ST_sa[N][22],Log[N]; void init_rmq() {Log[1] = 0;rep(i,2,n) Log[i]=Log[i>>1]+1LL;rep(i,1,n) ST_h[i][0]=Height[i], ST_sa[i][0]=SA[i];rep(j,1,20)for(int i=1;i+(1<<j-1)<=n;++i) {ST_h[i][j] = min(ST_h[i][j-1],ST_h[i+(1<<(j-1))][j-1]);ST_sa[i][j] = min(ST_sa[i][j-1],ST_sa[i+(1<<(j-1))][j-1]);} } int ask_h(int l,int r) {if(l==r)return ST_h[l][0];if(l>r)swap(l,r);++l;int p = Log[r-l+1];return min(ST_h[l][p],ST_h[r-(1<<p)+1][p]); } int ask_sa(int l,int r) {if(l>r)swap(l,r);int p = Log[r-l+1];return min(ST_sa[l][p],ST_sa[r-(1<<p)+1][p]); } ll s[N];int fd(int L,int len) {int l1=1,r1=L-1,ansl=L;while(l1<r1) {int mid = (l1+r1)>>1;if(ask_h(mid,L)>=len) r1=mid;else l1=mid+1;}if(ask_h(l1,L)>=len)ansl = min(ansl,l1);int l2=L+1,r2=n,ansr=L,tmp=-1;while(l2<=r2) {int mid = (l2+r2)>>1;if(ask_h(L,mid)<len) tmp=mid,r2=mid-1;else l2=mid+1;}if(tmp!=-1&&ask_h(L,tmp)<len)--tmp;else if(tmp==-1&&ask_h(L,n)>=len) tmp=n;if(tmp>=L+1&&tmp<=n&&ask_h(L,r2)>=len)ansr = max(r2,ansr);int x = ask_sa(ansl,ansr);return x; } void solve(ll k,int &l,int &r) {int p = lower_bound(s+1,s+1+n,k) - s;l = SA[p];int len = k - s[p-1] + Height[p];l = fd(p,len);r = l+len-1;return; } int main() {while(scanf(" %s",str)!=EOF) {init();calc();init_rmq();sum[0]=0;rep(i,1,n) s[i] = s[i-1] + (n-SA[i]-Height[i]+1);int q=readint();int l=0,r=0;while(q--) {ll x;scanf("%I64d",&x);x = (ll)(l^r^x) + 1LL;if(x>(ll)s[n]) {l=r=0;puts("0 0");continue;}solve(x,l,r);printf("%d %d\n",l,r);}}return 0; }

C. Paint Pearls

先想暴力dp，\(dp[i] = min(dp[j-1]+D(j,i)^2), D(l,r)：表示[l,r]區間內不同數的個數\)。長得想斜率優化，可是搞不出來。然后，發現只有區間的長度大于區間種類的平方時，才會有貢獻，否則不如一個個加起來。于是枚舉以i為右端點的區間包含k種數字，顯然\(k^2\)不能大過i，現在就是要求以i為右端點包含k種數字，的最左邊的位置，一開始寫了主席樹，然后二分左端點的兩個log做法。tle了之后，發現可以雙指針預處理pre[i][j]表示以i為右端點，包含j種元素的最遠左端點，因為有個離散化，是一個log。所以預處理復雜度是\(O(n \sqrt{n} logn)\)，dp是\(O(n \sqrt{n})\)的，然而還是t了。。。可能這個復雜度還是有點問題。。。還是寫搓了？？我再努力補。。

不優化確實會tle，學習了一些優化，換成刷表法dp，最重要優化的是當計算出的dp[i]大于等于dp[n]時break。

#include <cstdio> #include <iostream> #include <algorithm> #include <map> #include <cstring> #include <cmath> #include <queue> #include <set> #include <vector> #include <iterator> #include <string> #include <deque> #define rep(i,a,b) for(int i=a;i<=b;++i) #define per(i,a,b) for(int i=a;i>=b;--i) #define mem(W) memset(W,0,sizeof(W)) #define pb push_back #define MP make_pair #define fr first #define sc second #define PII pair<int,int> #define VI vector<int> typedef long long ll; typedef unsigned long long ull; const int N = 50010; const ll inf = 1000000000000000LL; inline int read() {char c=getchar();int x=0,f=1;while(!isdigit(c)){if(f=='-')f=-1;c=getchar();}while(isdigit(c)){x=x*10+c-'0';c=getchar();}return x*f; } using namespace std; vector<ll> v; int getid(ll x) {return lower_bound(v.begin(),v.end(),x) - v.begin() + 1; }int n; int a[N],b[N],A[N]; ll dp[N]; int vis[N]; int main() {while(scanf("%d",&n)!=EOF) {v.clear();rep(i,1,n)a[i]=read(),dp[i]=inf,v.pb(a[i]);sort(v.begin(),v.end());v.erase(unique(v.begin(),v.end()),v.end());for(int i=1;i<=n;++i) b[i]=getid(a[i]);dp[0]=0;rep(i,0,n-1) {int sz = 0;rep(j,i+1,n) {if(!vis[b[j]]) vis[b[j]]=1, A[++sz]=b[j];if(dp[i]+sz*sz>=dp[n]) break; //***if(sz*sz>n) break;dp[j] = min(dp[j],dp[i]+sz*sz);}rep(j,1,sz) vis[A[j]]=0;}printf("%I64d\n",dp[n]);}return 0; }

E. Game

手推了幾組，找規律的。把每一堆的大小都異或起來就星了。

#include <cstdio> #include <algorithm> #include <map> #include <cstring> #include <cmath> #include <queue> #include <set> #include <vector> #include <iterator> #include <string> #include <deque> #define rep(i,a,b) for(int i=a;i<=b;++i) #define per(i,a,b) for(int i=a;i>=b;--i) #define pb push_back #define MP make_pair #define fr first #define sc second #define PII pair<int,int> #define VI vector<int> typedef long long ll; typedef unsigned long long ull;inline int read() {char c=getchar();int x=0,f=1;while(!isdigit(c)){if(f=='-')f=-1;c=getchar();}while(isdigit(c)){x=x*10+c-'0';c=getchar();}return x*f; } using namespace std;int n,x;int main() {while(scanf("%d",&n)!=EOF){int ans=0;rep(i,1,n){int x=read();ans^=x;}if(!ans)puts("Lose");else puts("Win");}return 0; }

F. Dice

直接從123456開始bfs出到每種狀態的距離即可。查詢時，替換下標即可。

#include <cstdio> #include <algorithm> #include <map> #include <cstring> #include <cmath> #include <iostream> #include <queue> #include <set> #include <vector> #include <iterator> #include <string> #include <deque> #define rep(i,a,b) for(int i=a;i<=b;++i) #define per(i,a,b) for(int i=a;i>=b;--i) #define pb push_back #define MP make_pair #define fr first #define sc second #define PII pair<int,int> #define VI vector<int> typedef long long ll; typedef unsigned long long ull; const int N = 654321 + 10; inline int read() {char c=getchar();int x=0,f=1;while(!isdigit(c)){if(f=='-')f=-1;c=getchar();}while(isdigit(c)){x=x*10+c-'0';c=getchar();}return x*f; } using namespace std; int hs(string s) {int hs = 0;for(int i=0;i<6;++i) hs = hs*10 + (s[i]-'0');return hs; } string t; string fhs(int x) {t.clear();while(x) {t += (char)(x%10+'0');x/=10;}reverse(t.begin(),t.end());return t; } string tx,tx2; int rt1(int x) {tx = fhs(x);tx2 = tx;tx2[2-1] = tx[3-1];tx2[4-1] = tx[2-1];tx2[1-1] = tx[4-1];tx2[3-1] = tx[1-1];return hs(tx2); } int rt2(int x) {tx = fhs(x);tx2 = tx;tx2[2-1] = tx[4-1];tx2[3-1] = tx[2-1];tx2[1-1] = tx[3-1];tx2[4-1] = tx[1-1];return hs(tx2); } int rt3(int x) {tx = fhs(x);tx2 = tx;tx2[2-1] = tx[5-1];tx2[6-1] = tx[2-1];tx2[1-1] = tx[6-1];tx2[5-1] = tx[1-1];return hs(tx2); } int rt4(int x) {tx = fhs(x);tx2 = tx;tx2[2-1] = tx[6-1];tx2[5-1] = tx[2-1];tx2[1-1] = tx[5-1];tx2[6-1] = tx[1-1];return hs(tx2); } int dis[N],vis[N]; void bfs(string s) {memset(dis,-1,sizeof(dis));memset(vis,0,sizeof(vis));int u = hs(s);dis[u] = 0;queue<int> q;q.push(u);vis[u]=1;while(!q.empty()) {int u = q.front(); q.pop();int tu = u;tu = rt1(u);if(!vis[tu]) {vis[tu] = 1;q.push(tu);dis[tu] = dis[u] + 1;}tu = rt2(u);if(!vis[tu]) {vis[tu] = 1;q.push(tu);dis[tu] = dis[u] + 1;}tu = rt3(u);if(!vis[tu]) {vis[tu] = 1;q.push(tu);dis[tu] = dis[u] + 1;}tu = rt4(u);if(!vis[tu]) {vis[tu] = 1;q.push(tu);dis[tu] = dis[u] + 1;}} }int n,a[10],b[10]; map<int,int> c; int main() {string s = "123456";bfs(s);while(scanf("%d%d%d%d%d%d",&a[1],&a[2],&a[3],&a[4],&a[5],&a[6])!=EOF) {scanf("%d%d%d%d%d%d",&b[1],&b[2],&b[3],&b[4],&b[5],&b[6]);c.clear();rep(i,1,6) c[a[i]] = i;int hs = 0;rep(i,1,6)hs=hs*10+c[b[i]];printf("%d\n",dis[hs]);}return 0; }

H. Number Sequence

一開始想起一道題，覺得要從高位到低位分治貪心。然后發現過的人有點多啊。就開始找規律。。。于是順利浪費了大量時間。。。首先發現最大值就是\(n*(n+1)\) ，而每一個值都可以用一些\(2^i-1\)組合成，于是我們從大到小貪心的把每個數都異或成\(2^i-1\)中的盡可能大的值即可

#include <cstdio> #include <iostream> #include <algorithm> #include <map> #include <cstring> #include <cmath> #include <queue> #include <set> #include <vector> #include <iterator> #include <string> #include <deque> #define rep(i,a,b) for(int i=a;i<=b;++i) #define per(i,a,b) for(int i=a;i>=b;--i) #define pb push_back #define MP make_pair #define fr first #define sc second #define PII pair<int,int> #define VI vector<int> typedef long long ll; typedef unsigned long long ull; const int N = 1e6 +100000; inline int read() {char c=getchar();int x=0,f=1;while(!isdigit(c)){if(f=='-')f=-1;c=getchar();}while(isdigit(c)){x=x*10+c-'0';c=getchar();}return x*f; } using namespace std; ll n; ll a[N],b[N]; int vis[N]; void solve() {memset(vis,0,sizeof(vis));for(int i=n;i>=0;--i) {int e = 0;for(ll j=20;j>=0;--j) {e=(1<<j)-1;if(!vis[(e^i)]&&(e^i)<=n) {vis[(e^i)]=1;b[i] = (e^i);break;}}} } int main() {while(scanf("%lld",&n)!=EOF){rep(i,0,n)a[i]=read();solve();int f=0;printf("%lld\n",(n+1ll)*n);rep(i,0,n) {if(f)printf(" ");f=1;printf("%lld",b[a[i]]);}puts("");}return 0; }

I. 233 Matrix

發現n非常小，m很大，于是想到矩陣快速冪，推一下就ok了

#include <cstdio> #include <algorithm> #include <map> #include <cstring> #include <cmath> #include <queue> #include <set> #include <vector> #include <iterator> #include <string> #include <deque> #define rep(i,a,b) for(int i=a;i<=b;++i) #define per(i,a,b) for(int i=a;i>=b;--i) #define pb push_back #define MP make_pair #define fr first #define sc second #define PII pair<int,int> #define VI vector<int> #define rg register typedef long long ll; typedef unsigned long long ull; const ll mod = 10000007; inline int read() {char c=getchar();int x=0,f=1;while(!isdigit(c)){if(f=='-')f=-1;c=getchar();}while(isdigit(c)){x=x*10+c-'0';c=getchar();}return x*f; } using namespace std; ll ans[20][20],c[20][20],d[20][20]; inline void mul(ll a[][20], ll b[][20], int n){for(rg int i=1;i<=n;++i)for(int j=1;j<=n;++j)ans[i][j]=0,c[i][j]=a[i][j],d[i][j]=b[i][j];for(rg int i=1;i<=n;++i)for(rg int k=1;k<=n;++k)if(c[i][k])for(rg int j=1;j<=n;++j)if(d[k][j]){ans[i][j] = (ans[i][j] + (c[i][k]*d[k][j]))%mod;}for(rg int i=1;i<=n;++i)for(rg int j=1;j<=n;++j)a[i][j]=ans[i][j]; } ll E[20][20],A[20][20]; inline void mx_pow(ll A[][20],ll b,int n) {while(b) {if(b&1) mul(E,A,n);mul(A,A,n);b>>=1LL;} }int n; ll m,a[20]; int main() {while(scanf("%d%lld",&n,&m)!=EOF) {memset(A,0,sizeof(A));memset(E,0,sizeof(E));rep(i,1,n) a[i] = read();a[n+1] = 233; a[n+2] = 1;rep(i,1,n+2)E[i][i] = 1;rep(i,1,n)rep(j,1,i) A[i][j] = 1;rep(i,1,n)A[i][n+1]=1;A[n+1][n+1] = 10; A[n+1][n+2] = 3;A[n+2][n+2]=1;mx_pow(A,m,n+2);ll res = 0;rep(i,1,n+2) res = (res + (E[n][i]*a[i])%mod)%mod;printf("%lld\n",res);}return 0; }

轉載于:https://www.cnblogs.com/RRRR-wys/p/9347399.html

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