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The Preliminary Contest for ICPC China Nanchang National Invitational

發布時間：2024/10/5 编程问答 26 豆豆

生活随笔收集整理的這篇文章主要介紹了 The Preliminary Contest for ICPC China Nanchang National Invitational 小編覺得挺不錯的,現在分享給大家,幫大家做個參考.

Problem A?PERFECT NUMBER PROBLEM

https://nanti.jisuanke.com/t/38220

題解：打表

Problem B?Greedy HOUHOU

https://nanti.jisuanke.com/t/38221

Problem C?Angry FFF Party

https://nanti.jisuanke.com/t/38222

題意：f(x)是斐波那契數的第x項，g(x)=f(f(x))，對于給定的n，將其表示成若干個g數的和，要求字典序最小，n<10^100000

題解：矩陣快速冪求一下f(x)，預處理前29項g(x)。

可以發現g(x)的增長速度非常快，其實在第29項就已經超出10^100000，由于后面的項差過大，所以方法基本唯一，所以就把貪心選一下，然后前幾項在特判一下取個最小字典序就ok了。
JAVA版本一

import java.io.*; import java.math.*; import java.util.*; public class Main {static int maxn = 29;static BigInteger []w = new BigInteger[maxn];static int []v = new int[maxn];public static BigInteger f(BigInteger n){n = n.subtract(BigInteger.ONE);BigInteger [][]a = new BigInteger[2][2];BigInteger [][]b = new BigInteger[2][2];BigInteger [][]c = new BigInteger[2][2];for(int i = 0; i < 2; i++) for(int j = 0; j < 2; j++) a[i][j] = BigInteger.ZERO;a[0][0] = a[1][0] = a[0][1] = BigInteger.ONE; for(int i = 0; i < 2; i++) for(int j = 0; j < 2; j++) b[i][j] = BigInteger.ZERO;for(int i = 0; i < 2; i++) b[i][i] = BigInteger.ONE;while(n.compareTo(BigInteger.ZERO) > 0){if(n.mod(new BigInteger("2")).equals(BigInteger.ONE)){for(int i = 0; i < 2; i++){for(int j = 0; j < 2; j++){c[i][j] = BigInteger.ZERO;for(int k = 0; k < 2; k++){c[i][j] = c[i][j].add(b[i][k].multiply(a[k][j]));}}}for(int i = 0; i < 2; i++){for(int j = 0; j < 2; j++){b[i][j] = c[i][j];}}}for(int i = 0; i < 2; i++){for(int j = 0; j < 2; j++){c[i][j] = BigInteger.ZERO;for(int k = 0; k < 2; k++){c[i][j] = c[i][j].add(a[i][k].multiply(a[k][j]));}}}for(int i = 0; i < 2; i++){for(int j = 0; j < 2; j++){a[i][j] = c[i][j];}}n = n.divide(new BigInteger("2"));}//System.out.println("len " + b[0][0].toString().length());return b[0][0];}public static void main(String[] args){Scanner cin = new Scanner (System.in);for(int i = 1; i < maxn; i++){w[i] = f(f(BigInteger.valueOf(i)));}// System.out.println(w[6] + " " + w[7]);int T = cin.nextInt();while(T > 0){T--;BigInteger n = cin.nextBigInteger();int tot = 0;for(int i = maxn - 1; i > 0; i--){if(n.equals(new BigInteger("5")) && i >= 4){v[tot++] = 4;v[tot++] = 3;v[tot++] = 2;v[tot++] = 1;n = BigInteger.ZERO;break;}if(n.equals(new BigInteger("4")) && i >= 4){v[tot++] = 4;v[tot++] = 2;v[tot++] = 1;n = BigInteger.ZERO;break;}if(n.equals(new BigInteger("3")) && i >= 3){v[tot++] = 3;v[tot++] = 2;v[tot++] = 1;n = BigInteger.ZERO;break;}if(n.equals(new BigInteger("2")) && i >= 2){v[tot++] = 2;v[tot++] = 1;n = BigInteger.ZERO;break;}if(n.equals(new BigInteger("1")) && i >= 1){v[tot++] = 1;n = BigInteger.ZERO;break;}if(n.compareTo(w[i]) >= 0){v[tot++] = i;n = n.subtract(w[i]);}}if(!n.equals(BigInteger.ZERO)) System.out.println(-1);else{for(int i = tot - 1; i >= 0; i--) System.out.printf("%d%c", v[i], i == 0 ? '\n' : ' ');}}cin.close();} }

Problem D?Match Stick Game

https://nanti.jisuanke.com/t/38223

題解：

先將表達式表示成用最少火柴棍能表達的形式，然后在此基礎上添加火柴棍。

先預處理出i位數添加j個火柴棍能夠得到的最大值，然后背包一下就ok了。

舉個例子

1+1+222

最簡表示1-1-111，剩余11根棒子。

一共三項，1，-1，-111，然后dp[i][j]表示前i項加j根棒子的和的最大值，轉移顯然。

關鍵是預處理比較麻煩。

#include <bits/stdc++.h> #define my_max(a, b) ((a) > (b) ? (a) : (b)) #define my_min(a, b) ((a) < (b) ? (a) : (b)) #define fi first #define se second #define pb push_back #define eb emplace_back #define rep(i, s, t) for(int i = (int)(s); i <= (int)(t); i++) #define rev(i, t, s) for(int i = (int)(t); i >= (int)(s); i--) #define lson rt << 1 #define rson rt << 1 | 1 #define sz(x) (int)(x).size()typedef long long ll; typedef long double lb; typedef std::pair<int, int> pii; typedef std::pair<ll, ll> pll; typedef std::vector<int> VI; typedef std::vector<ll> VL; #include <ext/pb_ds/tree_policy.hpp> #include <ext/pb_ds/assoc_container.hpp> using namespace __gnu_pbds; using namespace std; typedef tree<int,null_type,less<int>,rb_tree_tag,tree_order_statistics_node_update> rbtree; ll gcd(ll x, ll y){ return y % x == 0 ? x : gcd(y % x, x); } template<class T>T my_abs(T a){ if(a < 0) a = -a; return a; } inline ll read(){ll ret = 0, sign = 1;char c = getchar();while(c < '0' || c > '9'){ if(c == '-') sign = -1; c = getchar();}while(c >= '0' && c <= '9'){ ret = ret * 10 + c - '0'; c = getchar(); }return ret * sign; } inline void write(ll x){if(x < 0){putchar('-');x = -x;}if(x > 9) write(x / 10);putchar(x % 10 + '0'); }//void expand_stack_space(){ // int size = 256 << 20; // 256MB // char *p = (char*)malloc(size) + size; // __asm__("movl %0, %%esp\n" :: "r"(p)); //} const int maxn = 100 + 7; const ll inf = 1e18; ll dp[maxn][maxn * 10], sa[maxn]; ll a[maxn][maxn], b[maxn][maxn]; int n; char str[maxn]; int count(char c){if(c == '-' || c == '1') return 0;if(c == '7' || c == '+') return 1;if(c == '4') return 2;if(c == '2' || c == '3' || c == '5') return 3;if(c == '0' || c == '6' || c == '9') return 4;if(c == '8') return 5; } char icount(int c){if(c == 1) return '7';if(c == 2) return '4';if(c == 3) return '5';if(c == 4) return '9';if(c == 5) return '8'; } void init(){for(int i = 1; i <= 10; i++){for(int j = 0; j <= 5 * i; j++){int c = j;string s = "";for(int k = 1; k <= i; k++){if(c >= count('9')) s += '9', c -= count('9');else if(c >= count('7')) s += '7', c -= count('7');else s += '1';}for(int k = i; k >= 1; k--){if(c == 0) break;c += count(s[k - 1]);if(c >= count('8')) s[k - 1] = '8', c -= count('8');else{s[k - 1] = icount(c);c = 0;}}ll res = 0;//cout<<"s " + s<<endl;for(char ss : s) res = res * 10 + ss - '0';a[i][j] = res;b[i][j + 1] = res;}b[i][0] = -b[i][1];} } int main(){ #ifndef ONLINE_JUDGE // freopen("in.txt", "r", stdin); // freopen("out.txt", "w", stdout); #endif // ONLINE_JUDGEint T;init();scanf("%d", &T);while(T--){scanf("%d", &n);scanf("%s", str);int cnt = 0;for(int i = 0; i < n; i++) cnt += count(str[i]);int num = 0, tot = 0;for(int i = 0; i < n; i++){if(str[i] == '-' || str[i] == '+'){sa[++tot] = num;num = 0;}else num++;}//printf("cnt %d\n", cnt);sa[++tot] = num;for(int i = 0; i <= tot; i++) for(int j = 0; j <= cnt; j++) dp[i][j] = -inf;//比賽時初始化寫錯了，還過了。emmmmdp[0][0] = 0;for(int i = 1; i <= tot; i++){int len = sa[i];//printf("len %d\n", len);if(i == 1){for(int k = 0; k <= 5 * len; k++){for(int j = cnt; j >= k; j--){dp[i][j] = max(dp[i][j], dp[i - 1][j - k] + a[len][k]);}}}else{for(int k = 0; k <= 5 * len; k++){for(int j = cnt; j >= k; j--){dp[i][j] = max(dp[i][j], dp[i - 1][j - k] + b[len][k]);}}}}printf("%lld\n", dp[tot][cnt]);}return 0; }

Problem E?Card Game

https://nanti.jisuanke.com/t/38224

Problem F?Information Transmitting

https://nanti.jisuanke.com/t/A2254

Problem G tsy's number

https://nanti.jisuanke.com/t/38226

題解：

#include <cstdio> #include <algorithm> #include <cstring> #include <iostream> using namespace std;typedef unsigned int ll;const int N = 1e7 + 10;int prime[N], phi[N], tot; ll f[N], gg[N], fu[N], res[N]; bool is_prime[N];void init() {memset(is_prime, true, sizeof(is_prime));for (ll i = 1; i < N; i++)f[i] = f[i - 1] + 1;for (ll i = 1; i < N; i++) {gg[i] = gg[i - 1] + i;f[i] = f[i] * gg[i];}for (ll i = 1; i < N; i++) {gg[i] = gg[i - 1] + i * i;f[i] = f[i] * gg[i];}phi[1] = gg[1] = fu[1] = 1;for (int i = 2; i < N; i++) {if (is_prime[i]) {prime[tot++] = i;fu[i] = i;phi[i] = i - 1;gg[i] = i - 2;}for (int j = 0; j < tot && i * prime[j] < N; j++) {is_prime[i * prime[j]] = false;if (i % prime[j] == 0) {phi[i * prime[j]] = phi[i] * prime[j];fu[i * prime[j]] = fu[i] * prime[j];if (fu[i] == i) {gg[i * prime[j]] = phi[i * prime[j]] - phi[i];}else {gg[i * prime[j]] = gg[i / fu[i]] * gg[fu[i * prime[j]]];}break;}else {fu[i * prime[j]] = prime[j];phi[i * prime[j]] = phi[i] * phi[prime[j]];gg[i * prime[j]] = gg[i] * gg[prime[j]];}}}for (ll i = 1; i < N; i++) {res[i] = i * i * i * gg[i] + res[i - 1];} }int n;void solve() {ll ans = 0;for (ll l = 1, r; l <= n; l = r + 1) {r = n / (n / l);ans += (res[r] - res[l - 1]) * f[n / l];}printf("%d\n", ans % (1 << 30)); }int main() {//freopen("0in.txt", "r", stdin);init();int T;scanf("%d", &T);while (T--) {scanf("%d", &n);solve();}return 0; }

Problem H?Coloring Game

https://nanti.jisuanke.com/t/38227

題解：快速冪+規律

/* *@Author: STZG *@Language: C++ */ #include <bits/stdc++.h> #include<iostream> #include<algorithm> #include<cstdlib> #include<cstring> #include<cstdio> #include<string> #include<vector> #include<bitset> #include<queue> #include<deque> #include<stack> #include<cmath> #include<list> #include<map> #include<set> //#define DEBUG #define RI register int #define endl "\n" using namespace std; typedef long long ll; //typedef __int128 lll; const int N=100000+10; const int M=100000+10; const int MOD=1e9+7; const double PI = acos(-1.0); const double EXP = 1E-8; const int INF = 0x3f3f3f3f; int t,n,m,k,p,l,r,u,v; int ans,cnt,flag,temp,sum; int a[N]; char str; struct node{};ll POW(ll a,ll b,ll c){ll res=1;ll base=a%c;while(b){if(b&1)res=(res*base)%c;base=(base*base)%c;b>>=1;}return res; } int main() { #ifdef DEBUGfreopen("input.in", "r", stdin);//freopen("output.out", "w", stdout); #endif//ios::sync_with_stdio(false);//cin.tie(0);//cout.tie(0);//scanf("%d",&t);//while(t--){scanf("%d",&n);if(n==1){cout<<1<<endl;return 0;}cout<<(4*POW(3,n-2,MOD))%MOD<<endl;//}#ifdef DEBUGprintf("Time cost : %lf s\n",(double)clock()/CLOCKS_PER_SEC); #endif//cout << "Hello world!" << endl;return 0; }

Problem I?Max answer

https://nanti.jisuanke.com/t/38228

題意：求區間最小值×區間和的最大值。

題解：用單調棧維護以每個數為最小值能拓展到最左邊和最右邊。然后用線段樹維護一個前綴和。

對于一個數如果是正數查詢當前位置到最右端的前綴和最大值，然后查詢最左端到當前點的最小值做減法。

對于一個數如果是負數查詢正好和上述情況相反。

C++版本一

#include<bits/stdc++.h> using namespace std; #define LL long long #define sca(x) scanf("%d",&x) #define N 500005 const LL inf = 1e18;int a[N],L[N],R[N]; LL sum[N]; stack<int>S; struct node {LL mmax,mmin; }T[N*4];void build(int rt,int l,int r) {if(l==r){T[rt].mmax=sum[l];T[rt].mmin=sum[l];return ;}int m=(l+r)>>1;build(rt<<1,l,m);build(rt<<1|1,m+1,r);T[rt].mmax=max(T[rt<<1].mmax,T[rt<<1|1].mmax);T[rt].mmin=min(T[rt<<1].mmin,T[rt<<1|1].mmin); }LL query_max(int rt,int l,int r,int ql,int qr) {if(l>=ql && r<=qr){return T[rt].mmax;}int m=(l+r)>>1;LL ans1=-inf,ans2=-inf;if(ql<=m) ans1=query_max(rt<<1,l,m,ql,qr);if(qr>m) ans2=query_max(rt<<1|1,m+1,r,ql,qr);return max(ans1,ans2); }LL query_min(int rt,int l,int r,int ql,int qr) {if(l>=ql && r<=qr){return T[rt].mmin;}int m=(l+r)>>1;LL ans1=inf,ans2=inf;if(ql<=m) ans1=query_min(rt<<1,l,m,ql,qr);if(qr>m) ans2=query_min(rt<<1|1,m+1,r,ql,qr);return min(ans1,ans2); }int main() {int n;sca(n);sum[0]=0;for(int i=1;i<=n;i++){sca(a[i]);sum[i]=sum[i-1]+a[i];}for(int i=1;i<=n;i++){while(!S.empty()&&a[S.top()]>=a[i])S.pop();if(S.size()==0)L[i]=0;else L[i]=S.top();S.push(i);}while(!S.empty())S.pop();for(int i=n;i>=1;i--){while(!S.empty()&&a[S.top()]>=a[i])S.pop();if(S.size()==0)R[i]=n;else R[i]=S.top()-1;S.push(i);}build(1,0,n);LL ans=-inf;for(int i=1;i<=n;i++){if(a[i]>=0){LL ans1=query_max(1,0,n,i,R[i]);LL ans2=query_min(1,0,n,L[i],i);ans=max(ans,a[i]*(ans1-ans2));}else{LL ans1=query_min(1,0,n,i,R[i]);LL ans2=query_max(1,0,n,L[i],i);ans=max(ans,a[i]*(ans1-ans2));}}cout<<ans<<endl; }

Problem J?Distance on the tree

https://nanti.jisuanke.com/t/38229

題意：給你一棵樹，每條邊都有一個權值，m個詢問，問你從x到y的路徑上有多少條邊的值<=w。

題解：LCA
一邊dfs一邊建樹，每一個點的狀態是從它父親的狀態轉移過來的。那么兩個點之間的所有可能就是一個點到他們的lca+另一個點到他們的lca。注意存邊權的話，是要存在兒子節點上的。
?

#include<bits/stdc++.h> using namespace std; #define maxn 100005 int ls[maxn*40],rs[maxn*40],sum[maxn*40],rt[maxn*40]; int tot,a[maxn],b[maxn]; int n,m; struct node {int to,next,wei; }e[maxn<<1]; int head[maxn],fa[maxn][25],cnt,dep[maxn]; void add(int x,int y,int z) {e[cnt].to=y;e[cnt].next=head[x];e[cnt].wei=z;head[x]=cnt++; } void update(int l,int r,int &root,int last,int val) {root=++tot;ls[root]=ls[last];rs[root]=rs[last];sum[root]=sum[last]+1;if(l==r)return ;int mid=l+r>>1;if(mid>=val)update(l,mid,ls[root],ls[last],val);elseupdate(mid+1,r,rs[root],rs[last],val); } int query(int l,int r,int ql,int qr,int num) {if(l==r)return sum[qr]-sum[ql];int mid=l+r>>1;if(num<=mid)return query(l,mid,ls[ql],ls[qr],num);else{int ans=sum[ls[qr]]-sum[ls[ql]];ans+=query(mid+1,r,rs[ql],rs[qr],num);return ans;} } int all; void dfs(int son,int ffa,int val) {dep[son]=dep[ffa]+1;fa[son][0]=ffa;if(son!=1)update(1,maxn,rt[son],rt[ffa],val);for(int i=head[son];~i;i=e[i].next){int ne=e[i].to;if(ne==ffa)continue;dfs(ne,son,lower_bound(b+1,b+1+all,e[i].wei)-b);} } void deal(){for(int j=1;(1<<j)<=n;j++)for(int i=1;i<=n;i++)fa[i][j]=fa[fa[i][j-1]][j-1];}int lca(int x,int y){if(dep[x]<dep[y])swap(x,y);for(int i=20;i>=0;i--)if(dep[fa[x][i]]>=dep[y])x=fa[x][i];if(x==y)return x;for(int i=20;i>=0;i--)if(fa[x][i]!=fa[y][i])x=fa[x][i],y=fa[y][i];return fa[x][0]; }int main() {memset(head,-1,sizeof(head));scanf("%d%d",&n,&m);int x,y,z;for(int i=1;i<n;i++){scanf("%d%d%d",&x,&y,&z),b[i]=z;add(x,y,z),add(y,x,z);}sort(b+1,b+n);all=unique(b+1,b+n)-(b+1);tot=0;dfs(1,1,0);deal();while(m--){scanf("%d%d%d",&x,&y,&z);z=upper_bound(b+1,b+1+all,z)-b-1;if(z==0)printf("0\n");else{int l=lca(x,y),ans=0;if(x!=l)ans=query(1,maxn,rt[l],rt[x],z);if(y!=l)ans+=query(1,maxn,rt[l],rt[y],z);printf("%d\n",ans);}}return 0; }

Problem K?MORE XOR

https://nanti.jisuanke.com/t/38230

題解：規律

C++版本一

#include<bits/stdc++.h> #define ll long long #define pb push_back #define INF 0x3f3f3f3f #define FI first #define SE second #define MP make_pair #define PI pair<int,int> #define lson l,m,rt<<1,ls,rs #define rson m+1,r,rt<<1|1,ls,rs #define test printf("here!!!\n") using namespace std; const int mx=1e5+10; int n,m,p,q; int qz[4][4][mx]; int hz[4][4][mx]; int a[mx]; int main() {int t;scanf("%d",&t);while (t--){memset(qz,0,sizeof(qz));memset(hz,0,sizeof(hz));scanf("%d",&n);for (int i=1;i<=n;++i){scanf("%d",&a[i]);}for (int i=1;i<=n;++i){for (int j=0;j<4;++j){int id=(i-1)%4;int hgf=(1+j)%4;if (id==j){qz[1][j][i]=qz[1][j][i-1]^a[i];}else qz[1][j][i]=qz[1][j][i-1];if (id==j||id==hgf){qz[2][j][i]=qz[2][j][i-1]^a[i];}else qz[2][j][i]=qz[2][j][i-1];if (id==hgf){qz[3][j][i]=qz[3][j][i-1]^a[i];}else qz[3][j][i]=qz[3][j][i-1];}}for (int i=n;i>=1;--i){for (int j=0;j<4;++j){int id=(i-1)%4;int hgf=(1+j)%4;if (id==j){hz[1][j][i]=hz[1][j][i+1]^a[i];}else hz[1][j][i]=hz[1][j][i+1];if (id==j||id==hgf){hz[2][j][i]=hz[2][j][i+1]^a[i];}else hz[2][j][i]=hz[2][j][i+1];if (id==hgf){hz[3][j][i]=hz[3][j][i+1]^a[i];}else hz[3][j][i]=hz[3][j][i+1];}}scanf("%d",&q);int l,r;for (int i=1;i<=q;++i){scanf("%d%d",&l,&r);int dis=(r-l+1)%4;int cxk=(l-1)%4;if (dis!=0) printf("%d\n",qz[dis][cxk][n]^qz[dis][cxk][l-1]^hz[dis][cxk][r+1]);else printf("0\n");}} }

C++版本二

題解：規律

打表代碼：

while(1){scanf("%d",&n);for(int l=1;l<=n;l++){for(int r=l;r<=n;r++){memset(vis,0,sizeof(vis));for(int l1=l;l1<=r;l1++){for(int r1=l1;r1<=r;r1++){for(int l2=l1;l2<=r1;l2++){for(int r2=l2;r2<=r1;r2++){for(int i=l2;i<=r2;i++){vis[i]=!vis[i];}}}}}cout<<l<<" "<<r<<endl;for(int i=1;i<=12;i++){cout<<vis[i];}cout<<endl;}}}

1、可以發現當區間確定時，是否取用成一個固定的規律；

2、同長度的區間，取法相同；

3、對于一定的區間取法循環節長度為4，例如【1,9】循環節1000，整個區間取法100010001，并且區間長度循環節也為4，例如長度為1的區間和長度為5的區間取法循環節相同；

/* *@Author: STZG *@Language: C++ */ #include <bits/stdc++.h> #include<iostream> #include<algorithm> #include<cstdlib> #include<cstring> #include<cstdio> #include<string> #include<vector> #include<bitset> #include<queue> #include<deque> #include<stack> #include<cmath> #include<list> #include<map> #include<set> //#define DEBUG #define RI register int #define endl "\n" using namespace std; typedef long long ll; //typedef __int128 lll; const int N=100000+10; const int M=100000+10; const int MOD=1e9+7; const double PI = acos(-1.0); const double EXP = 1E-8; const int INF = 0x3f3f3f3f; int t,n,m,k,p,l,r,u,v; int ans,cnt,flag,temp,sum; int a[N]; int vis[N]; char str; int qz[4][4][N];//第一維:取法，第二維:開始位置 int main() { #ifdef DEBUGfreopen("input.in", "r", stdin);//freopen("output.out", "w", stdout); #endif//ios::sync_with_stdio(false);//cin.tie(0);//cout.tie(0);scanf("%d",&t);while(t--){scanf("%d",&n);for(int i=1;i<=n;i++){scanf("%d",&a[i]);}memset(qz,0,sizeof(qz));for(int i=1;i<=n;i++){for(int j=0;j<4;j++){int pos=(i-1)%4;int dis=(j+1)%4;if(pos==j){qz[1][j][i]=qz[1][j][i-1]^a[i];}else{qz[1][j][i]=qz[1][j][i-1];}if(pos==j||dis==pos){qz[2][j][i]=qz[2][j][i-1]^a[i];}else{qz[2][j][i]=qz[2][j][i-1];}if(dis==pos){qz[3][j][i]=qz[3][j][i-1]^a[i];}else{qz[3][j][i]=qz[3][j][i-1];}}}scanf("%d",&m);while(m--){scanf("%d%d",&l,&r);int pos=(r-l+1)%4;int dis=(l-1)%4;int res=qz[pos][dis][r]^qz[pos][dis][l-1];cout<<res<<endl;}}#ifdef DEBUGprintf("Time cost : %lf s\n",(double)clock()/CLOCKS_PER_SEC); #endif//cout << "Hello world!" << endl;return 0; }

C++版本三

Problem?L qiqi'tree

https://nanti.jisuanke.com/t/38231

Problem M?Subsequence

https://nanti.jisuanke.com/t/38232

題解：預處理+貪心

/* *@Author: STZG *@Language: C++ */ #include <bits/stdc++.h> #include<iostream> #include<algorithm> #include<cstdlib> #include<cstring> #include<cstdio> #include<string> #include<vector> #include<bitset> #include<queue> #include<deque> #include<stack> #include<cmath> #include<list> #include<map> #include<set> //#define DEBUG #define RI register int #define endl "\n" using namespace std; typedef long long ll; //typedef __int128 lll; const int N=100000+10; const int M=100000+10; const int MOD=1e9+7; const double PI = acos(-1.0); const double EXP = 1E-8; const int INF = 0x3f3f3f3f; int t,n,m,k,p,l,r,u,v; int ans,cnt,flag,temp,sum; int a[N]; char str[N],s[N]; int pos[N][130]; struct node{}; int main() { #ifdef DEBUGfreopen("input.in", "r", stdin);//freopen("output.out", "w", stdout); #endif//ios::sync_with_stdio(false);//cin.tie(0);//cout.tie(0);//scanf("%d",&t);//while(t--){cin>>str+1;scanf("%d",&n);int len=strlen(str+1);for(int i=0;i<128;i++){pos[len][i]=len+1;}for(int i=len-1;i>=0;i--){for(int j=0;j<128;j++){pos[i][j]=pos[i+1][j];}pos[i][str[i+1]]=i+1;}while(n--){scanf("%s",s);int l=strlen(s);int id=pos[0][s[0]];int r=0;while(id<=len&&r<l){r++;id=pos[id][s[r]];}//cout<<l<<" "<<r<<endl;cout<<((r==l)?"YES":"NO")<<endl;}//}#ifdef DEBUGprintf("Time cost : %lf s\n",(double)clock()/CLOCKS_PER_SEC); #endif//cout << "Hello world!" << endl;return 0; }

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