• The Preliminary Contest for ICPC China Nanchang National Invitational and International Silk-Road Programming Contest


    两个队友链接:

    A. PERFECT NUMBER PROBLEM 题库链接

    思路:2^1*(2^2-1), 2^2*(2^3-1), 2^4*(2^5-1), 2^6*(2^7-1), 2^12*(2^13-1)

    代码:

    #include<bits/stdc++.h>
    using namespace std;
    int main()
    {
       cout<<"6
    28
    496
    8128
    33550336
    ";
       return 0;
    }
    View Code

    B. Greedy HOUHOU 题库链接

    C. Angry FFF Party 题库链接

    思路:java大数算,贪心减

    代码:

    import java.util.*;
    import java.math.*;
    
    public class Main {
    
        /**
         * @param args
         */
        
        static class Matrix {
            BigInteger a[][] = new BigInteger [2][2];
            public void init() {
                for (int i = 0; i < 2; ++i) for (int j = 0; j < 2; ++j) a[i][j] = BigInteger.ZERO;
            }
            public void _init() {
                init();
                for (int i = 0; i < 2; ++i) a[i][i] = BigInteger.ONE;
            }
            public static Matrix mul(Matrix A, Matrix B) {
                Matrix res = new Matrix();
                res.init();
                for (int i = 0; i < 2; ++i) {
                    for (int j = 0; j < 2; ++j) {
                        for (int k = 0; k < 2; ++k) {
                            res.a[i][k] = res.a[i][k].add(A.a[i][j].multiply(B.a[j][k]));
                        }
                    }
                }
                return res;
            }
            public static Matrix q_pow(Matrix A, BigInteger k) {
                Matrix res = new Matrix();
                res._init();
                while(k.compareTo(BigInteger.ZERO) > 0) {
                    if(k.mod(BigInteger.valueOf(2)).compareTo(BigInteger.ZERO) > 0) res = mul(res, A);
                    A = mul(A, A);
                    k = k.shiftRight(1);
                }
                return res;
            }
            public static BigInteger get_fib(BigInteger n) {
                if(n.compareTo(BigInteger.ONE) == 0) return BigInteger.ONE;
                if(n.compareTo(BigInteger.valueOf(2)) == 0) return BigInteger.ONE;
                Matrix A = new Matrix();
                A.a[1][1] = BigInteger.ZERO;
                A.a[0][0] = A.a[1][0] = A.a[0][1] = BigInteger.ONE;
                A = q_pow(A, n.subtract(BigInteger.valueOf(2)));
                return A.a[0][0].add(A.a[0][1]);
            }
        }
        
        public static BigInteger f[] = new BigInteger[100];
        public static int ans[] = new int[100];
        public static void main(String[] args) {
            // TODO Auto-generated method stub
            Scanner in = new Scanner(System.in);
            int T = in.nextInt();
            for (int i = 1; i <= 28; ++i) {
                f[i] = Matrix.get_fib(Matrix.get_fib(BigInteger.valueOf(i)));
            }
            
            while(T > 0) {
                --T;
                BigInteger W = in.nextBigInteger();
                int cnt = 0;
                for (int i = 28; i >= 6; --i) {
                    if(f[i].compareTo(W) <= 0) {
                        ans[++cnt] = i;
                        W = W.subtract(f[i]);
                    }
                }
                if(W.compareTo(BigInteger.valueOf(1)) == 0) {
                    ans[++cnt] = 1;
                }
                else if(W.compareTo(BigInteger.valueOf(2)) == 0){
                    ans[++cnt] = 2;
                    ans[++cnt] = 1;
                }
                else if(W.compareTo(BigInteger.valueOf(3)) == 0) {
                    ans[++cnt] = 3;
                    ans[++cnt] = 2;
                    ans[++cnt] = 1;
                }
                else if(W.compareTo(BigInteger.valueOf(4)) == 0) {
                    ans[++cnt] = 4;
                    ans[++cnt] = 2;
                    ans[++cnt] = 1;
                }
                else if(W.compareTo(BigInteger.valueOf(5)) == 0) {
                    ans[++cnt] = 4;
                    ans[++cnt] = 3;
                    ans[++cnt] = 2;
                    ans[++cnt] = 1;
                }
                else if(W.compareTo(BigInteger.valueOf(6)) == 0) {
                    ans[++cnt] = 5;
                    ans[++cnt] = 1;
                }
                else if(W.compareTo(BigInteger.valueOf(7)) == 0) {
                    ans[++cnt] = 5;
                    ans[++cnt] = 2;
                    ans[++cnt] = 1;
                }
                else if(W.compareTo(BigInteger.valueOf(8)) == 0) {
                    ans[++cnt] = 5;
                    ans[++cnt] = 3;
                    ans[++cnt] = 2;
                    ans[++cnt] = 1;
                }
                else if(W.compareTo(BigInteger.valueOf(9)) == 0) {
                    ans[++cnt] = 5;
                    ans[++cnt] = 4;
                    ans[++cnt] = 2;
                    ans[++cnt] = 1;
                }
                else if(W.compareTo(BigInteger.valueOf(10)) == 0) {
                    ans[++cnt] = 5;
                    ans[++cnt] = 4;
                    ans[++cnt] = 3;
                    ans[++cnt] = 2;
                    ans[++cnt] = 1;
                }
                else if(W.compareTo(BigInteger.valueOf(0)) != 0){
                    System.out.println(-1);
                    continue;
                }
                for (int i = cnt; i >= 1; --i) {
                    System.out.print(ans[i]);
                    if(i == 1)System.out.println("");
                    else System.out.print(" ");
                }
            }
        }
    
    }
    View Code

    D. Match Stick Game 题库链接

    思路:mx[i][j]:用i根火柴组成一个位数为j的数这个数的最大值 mn[i][j]:用i根火柴组成一个位数为j的数这个数的最小值

    dp[i][j]:处理到第i位用了j根火柴式子的最大值

    代码:

    #pragma GCC optimize(2)
    #pragma GCC optimize(3)
    #pragma GCC optimize(4)
    #include<bits/stdc++.h>
    using namespace std;
    #define y1 y11
    #define fi first
    #define se second
    #define pi acos(-1.0)
    #define LL long long
    #define ls rt<<1, l, m
    #define rs rt<<1|1, m+1, r
    //#define mp make_pair
    #define pb push_back
    #define ULL unsigned LL
    #define pll pair<LL, LL>
    #define pli pair<LL, int>
    #define pii pair<int, int>
    #define piii pair<pii, int>
    #define pdi pair<double, int>
    #define pdd pair<double, double>
    #define mem(a, b) memset(a, b, sizeof(a))
    #define debug(x) cerr << #x << " = " << x << "
    ";
    #define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
    //head
    
    const int N = 105, M = 70;
    char s[N];
    int a[N];
    int n, T;
    int mx[M][10], mn[M][10];
    map<char, int>mp;
    int t[10] = {-1, -1, 1, 7, 4, 5, 9, 8};
    LL dp[N][705];
    void init() {
        for (int i = 1; i < M; ++i) {
            for (int j = 0; j <= 9; ++j) mx[i][j] = INT_MIN, mn[i][j] = INT_MAX;
        }
        mx[0][0] = mn[0][0] = 0;
        for (int i = 1; i <= M; ++i) {
            for (int j = 1; j <= 9; ++j) {
                for (int k = 2; k <= 7 && k <= i; ++k) {
                    if(mx[i-k][j-1] != INT_MIN) mx[i][j] = max(mx[i-k][j-1]*10+t[k], mx[i][j]);
                    if(mn[i-k][j-1] != INT_MAX) mn[i][j] = min(mn[i-k][j-1]*10+t[k], mn[i][j]);
                }
            }
        }
    }
    int main() {
        mp['0']=6; mp['1']=2; mp['2']=5; mp['3']=5;
        mp['4']=4; mp['5']=5; mp['6']=6; mp['7']=3;
        mp['8']=7; mp['9']=6; mp['+']=2; mp['-']=1;
        init();
        scanf("%d", &T);
        while(T--) {
            scanf("%d", &n);
            scanf("%s", s);
            int c1 = 1, cnt = 0;
            for (int i = 0; i < n; ++i) {
                if(isdigit(s[i])) cnt += mp[s[i]], a[c1]++;
                else cnt += mp[s[i]], ++c1;
            }
            n = c1;
            for (int i = 0; i <= n; ++i) for (int j = 0; j < 705; ++j) dp[i][j] = LONG_MIN;
            dp[0][0] = 0;
            for (int i = 1; i <= n; ++i) {
                if(i == 1) {
                    for (int j = 1; j <= cnt; ++j) {
                        for (int k = 1; k < M && k <= j; ++k) {
                            if(dp[i-1][j-k] != LONG_MIN && mx[k][a[i]] != INT_MIN) dp[i][j] = max(dp[i-1][j-k] + mx[k][a[i]], dp[i][j]);
                        }
                    }
                }
                else {
                    for (int j = 1; j <= cnt; ++j) {
                        for (int k = 1; k < M; ++k) {
                            if(k+2 <= j && dp[i-1][j-k-2] != LONG_MIN && mx[k][a[i]] != INT_MIN) dp[i][j] = max(dp[i-1][j-k-2] + mx[k][a[i]], dp[i][j]);
                            if(k+1 <= j && dp[i-1][j-k-1] != LONG_MIN && mn[k][a[i]] != INT_MAX) dp[i][j] = max(dp[i-1][j-k-1] - mn[k][a[i]], dp[i][j]);
                        }
                    }
                }
            }
            printf("%lld
    ", dp[n][cnt]);
            for (int i = 1; i <= n; ++i) a[i] = 0;
        }
        return 0;
    }
    View Code

    E. Card Game 题库链接

    F. Information Transmitting 题库链接

    G. tsy's number 题库链接 

    H. Coloring Game 题库链接

    思路:4*3^(n-2)

    代码:

    #include<bits/stdc++.h>
    #define ll long long
    using namespace std;
    const int mod=1e9+7;
    ll quick(ll x,ll n)
    {
        ll ans=1;
        while(n)
        {
            if(n%2==0) { x=x*x%mod; n=n/2; }
            else       {ans=ans*x%mod; n--; }
        }
        return ans;
    }
    int main()
    {
       ll n; cin>>n;
       if(n==1) { cout<<1<<endl; }
       else
       {
           cout<<(quick(2,2)%mod*quick(3,n-2)%mod+mod)%mod<<endl;
       }
       return 0;
    }
    View Code

     I. Max answer 题库链接

    思路:单调栈+区间RMQ

    代码:

    #include <bits/stdc++.h>
    
    using namespace std;
    #define LL long long
    #define fi first
    #define se second
    #define pb push_back
    #define ls rt<<1, l, m
    #define rs rt<<1|1, m+1, r
    const int N = 5e5 + 5;
    int a[N], pre[N], suf[N];
    LL sum[N], mx[N<<2], mn[N<<2];
    stack<int> st;
    int n;
    void push_up(int rt) {
        mn[rt] = min(mn[rt<<1], mn[rt<<1|1]);
        mx[rt] = max(mx[rt<<1], mx[rt<<1|1]);
    }
    void build(int rt, int l, int r) {
        if(l == r) {
            mx[rt] = mn[rt] = sum[l];
            return ;
        }
        int m = l+r >> 1;
        build(ls);
        build(rs);
        push_up(rt);
    }
    LL querymx(int L, int R, int rt, int l, int r) {
        if(L <= l && r <= R) return mx[rt];
        int m = l+r >> 1;
        LL ans = LONG_MIN;
        if(L <= m) ans = max(ans, querymx(L, R, ls));
        if(R > m) ans = max(ans, querymx(L, R, rs));
        return ans;
    }
    LL querymn(int L, int R, int rt, int l, int r) {
        if(L <= l && r <= R) return mn[rt];
        int m = l+r >> 1;
        LL ans = LONG_MAX;
        if(L <= m) ans = min(ans, querymn(L, R, ls));
        if(R > m) ans = min(ans, querymn(L, R, rs));
        return ans;
    }
    int main() {
        scanf("%d", &n);
        for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
        for (int i = 1; i <= n; ++i) sum[i] = sum[i-1] + a[i];
        build(1, 0, n);
        a[0] = a[n+1] = INT_MIN;
        st.push(0);
        for (int i = 1; i <= n; ++i) {
            while(!st.empty() && a[st.top()] >= a[i]) st.pop();
            pre[i] = st.top();
            st.push(i);
        }
        while(!st.empty()) st.pop();
        st.push(n+1);
        for (int i = n; i >= 1; --i) {
            while(!st.empty() && a[st.top()] >= a[i]) st.pop();
            suf[i] = st.top();
            st.push(i);
        }
        LL ans = 0;
        for (int i = 1; i <= n; ++i) {
            int l = pre[i], r = i-1;
            int ll = i, rr = suf[i]-1;
            if(a[i] < 0) {
                ans = max(ans, a[i]*(querymn(ll, rr, 1, 0, n)-querymx(l, r, 1, 0, n)));
            }
            else if(a[i] > 0) {
                ans = max(ans, a[i]*(querymx(ll, rr, 1, 0, n)-querymn(l, r, 1, 0, n)));
            }
        }
        printf("%lld
    ", ans);
        return 0;
    }
    View Code

    J. Distance on the tree 题库链接

    思路:离线+树剖

    代码:

    #pragma GCC optimize(2)
    #pragma GCC optimize(3)
    #pragma GCC optimize(4)
    #include<bits/stdc++.h>
    using namespace std;
    #define y1 y11
    #define fi first
    #define se second
    #define pi acos(-1.0)
    #define LL long long
    #define ls rt<<1, l, m
    #define rs rt<<1|1, m+1, r
    //#define mp make_pair
    #define pb push_back
    #define ULL unsigned LL
    #define pll pair<LL, LL>
    #define pli pair<LL, int>
    #define pii pair<int, int>
    #define piii pair<pii, int>
    #define pdi pair<double, int>
    #define pdd pair<double, double>
    #define mem(a, b) memset(a, b, sizeof(a))
    #define debug(x) cerr << #x << " = " << x << "
    ";
    #define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
    //head
    
    const int N = 1e5 + 5;
    vector<int> g[N];
    int fa[N], dp[N], sz[N], son[N], top[N], dfn[N], to[N], cnt = 0, n, m;
    int tree[N<<2], ans[N];
    struct edge {
        int u, v, w;
        int id;
        bool operator < (const edge & rhs) const {
            return w < rhs.w;
        }
    }e[N], q[N];
    void push_up(int rt) {
        tree[rt] = tree[rt<<1] + tree[rt<<1|1];
    }
    void update(int p, int x, int rt, int l, int r) {
        if(l == r) {
            tree[rt] += x;
            return ;
        }
        int m = l+r >> 1;
        if(p <= m) update(p, x, ls);
        else update(p, x, rs);
        push_up(rt);
    }
    int query(int L, int R, int rt, int l, int r) {
        if(L <= l && r <= R) return tree[rt];
        int m = l+r >> 1;
        int ans = 0;
        if(L <= m) ans += query(L, R, ls);
        if(R > m) ans += query(L, R, rs);
        return ans;
    }
    void dfs1(int u, int o) {
        fa[u] = o;
        sz[u] = 1;
        dp[u] = dp[o] + 1;
        for (int i = 0; i < g[u].size(); ++i) {
            int v = g[u][i];
            if(v != o) {
                dfs1(v, u);
                sz[u] += sz[v];
                if(sz[v] > sz[son[u]]) son[u] = v;
            }
        }
    }
    void dfs2(int u, int t) {
        top[u] = t;
        dfn[u] = ++cnt;
        to[cnt] = u;
        if(!son[u]) return ;
        dfs2(son[u], t);
        for (int i = 0; i < g[u].size(); ++i) {
            int v = g[u][i];
            if(v != fa[u] && v != son[u]) dfs2(v, v);
        }
    }
    int sum(int u, int v) {
        int ans = 0, fu = top[u], fv = top[v];
        while(fu != fv) {
            if(dp[fu] >= dp[fv]) ans += query(dfn[fu], dfn[u], 1, 1, n), u = fa[fu], fu = top[u];
            else ans += query(dfn[fv], dfn[v], 1, 1, n), v = fa[fv], fv = top[v];
        }
        if(dfn[u] < dfn[v]) ans += query(dfn[u]+1, dfn[v], 1, 1, n);
        else if(dfn[v] < dfn[u]) ans += query(dfn[v]+1, dfn[u], 1, 1, n);
        return ans;
    }
    int main() {
        scanf("%d %d", &n, &m);
        for (int i = 1; i < n; ++i) {
            scanf("%d %d %d", &e[i].u, &e[i].v, &e[i].w);
            g[e[i].u].pb(e[i].v);
            g[e[i].v].pb(e[i].u);
        }
        dfs1(1, 1);
        dfs2(1, 1);
        for (int i = 1; i <= m; ++i) scanf("%d %d %d", &q[i].u, &q[i].v, &q[i].w), q[i].id = i;
        sort(e+1, e+n);
        sort(q+1, q+1+m);
        int now = 1;
        for (int i = 1; i <= m; ++i) {
            while(now < n && e[now].w <= q[i].w) {
                int u = e[now].u, v = e[now].v;
                if(dp[u] < dp[v]) swap(u, v);
                update(dfn[u], 1, 1, 1, n);
                ++now;
            }
            ans[q[i].id] = sum(q[i].u, q[i].v);
        }
        for (int i = 1; i <= m; ++i) printf("%d
    ", ans[i]);
        return 0;
    }
    View Code

     K. MORE XOR 题库链接

    思路:打表找规律+前缀亦或和

    代码:

    #include<bits/stdc++.h>
    #define ll long long
    using namespace std;
    const int mod=1e9+7;
    const int maxn=1e5+10;
    int a[maxn];
    int n;
    int a1000[maxn];
    int a0100[maxn];
    int a0010[maxn];
    int a0001[maxn];
    void make1()
    {
        for(int i=1;i<=n;i++)
        {
            if(i%4==1) a1000[i]=a1000[i-1]^a[i];
            else       a1000[i]=a1000[i-1];
        }
        for(int i=1;i<=n;i++)
        {
            if(i%4==2) a0100[i]=a0100[i-1]^a[i];
            else       a0100[i]=a0100[i-1];
        }
        for(int i=1;i<=n;i++)
        {
            if(i%4==3) a0010[i]=a0010[i-1]^a[i];
            else       a0010[i]=a0010[i-1];
        }
        for(int i=1;i<=n;i++)
        {
            if(i%4==0) a0001[i]=a0001[i-1]^a[i];
            else       a0001[i]=a0001[i-1];
        }
    }
    int work(int l,int r)
    {
        int d=r-l+1;
        int ans1=0;
        int ans2=0;
    
            if(l%4==1)      ans1^=a1000[r]^a1000[l-1];
            else if(l%4==2) ans1^=a0100[r]^a0100[l-1];
            else if(l%4==3) ans1^=a0010[r]^a0010[l-1];
            else            ans1^=a0001[r]^a0001[l-1];
    
            if(l%4==1)      ans2^=a0100[r]^a0100[l-1];
            else if(l%4==2) ans2^=a0010[r]^a0010[l-1];
            else if(l%4==3) ans2^=a0001[r]^a0001[l-1];
            else            ans2^=a1000[r]^a1000[l-1];
       if(d%4==0)      return 0;
       else if(d%4==1) return ans1;
       else if(d%4==2) return ans1^ans2;
       else            return ans2;
    
    }
    int main()
    {
    int T; scanf("%d",&T);
    while(T--)
    {
       scanf("%d",&n);
       for(int i=1;i<=n;i++) scanf("%d",&a[i]);  make1();
       int q; scanf("%d",&q);
       while(q--)
       {
          int l,r; scanf("%d %d",&l,&r);
          printf("%d
    ",work(l,r));
       }
       for(int i=1;i<=n;i++)
       {
           a0001[i]=0;  a1000[i]=0;a0010[i]=0;a0100[i]=0;
       }
    }
    return 0;
    }
    View Code

     L. qiqi'tree 题库链接

    思路:暴力计算几何

    代码:

    #include<bits/stdc++.h>
    const double PI=acos(-1.0);
    const double TT=PI/3;
    using namespace std;
    struct Point  { double x,y; Point(){} Point(double x, double y) :x(x), y(y){}  };;
    struct Segment{ Point a,b;  Segment(Point x,Point y )     { a=x;b=y; };  Segment(){}; };;
    struct Line   { Point a,b;  Line(Point x,Point y )        { a=x;b=y; };  Line(){}; };;
    typedef Point Point;
    Point operator + (Point A, Point B){ return Point(A.x+B.x, A.y+B.y); } // 向量相加
    Point operator - (Point  A, Point  B){ return Point(B.x-A.x, B.y-A.y); } // 向量生成   A-B;
    double operator * (Point A, Point B){ return A.x*B.x-A.y*B.y;          } // 点积
    double operator ^ (Point A, Point B){ return A.x*B.y-A.y*B.x;          } // 叉积
    double Dot(Point A, Point B)   { return A.x*B.x + A.y*B.y; }  // 点积
    double Cross(Point A, Point B) { return A.x*B.y-A.y*B.x;   }  // 叉积
    double Length(Point A)          { return sqrt(Dot(A, A));   } // 向量长度
    double dis(Point a,Point b)      { return sqrt( (a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y) ); }
    Line qq;
    bool check(Segment A,Line B)
    {
        Point a=A.a; Point b=A.b;
        Point c=B.a; Point d=B.b;
        if(Cross(a-c,a-d)*Cross(b-c,b-d)>0 ) return 1;
        else return 0;
        return 0;
    }
    Point make(Segment A,Line B)
    {
        Point a=A.a; Point b=A.b;Point c=B.a; Point d=B.b;
        double A1=b.y-a.y,B1=-(b.x-a.x),C1=b.y*a.x-b.x*a.y;
        double A2=d.y-c.y,B2=-(d.x-c.x),C2=d.y*c.x-d.x*c.y;
        double k=A1*B2-A2*B1;
        double x=-(B1*C2-C1*B2)*1.000000000/k;
        double y=(A1*C2-C1*A2)*1.00000000/k;
        Point kk; kk.x=x; kk.y=y;  return kk;
    }
    double dfs(double l,int k,int d,Point x,double ff)
    {
        if(k==d) return 0;
        double ans=0;
        Segment x1;
        x1.a=x;
        x1.b.x=x.x+l*sin(ff); x1.b.y=x.y+l*cos(ff);
        Point W;
        W.x=x.x+l*sin(ff); W.y=x.y+l*cos(ff);
        if(check(x1,qq)==1)
        {
            ans+=l;
            ans+=dfs(l/4,k+1,d,W,ff-TT);
            ans+=dfs(l/4,k+1,d,W,ff);
            ans+=dfs(l/4,k+1,d,W,ff+TT);
        }
        else
        {
            Point P=make(x1,qq); ans+=dis(x,P);
        }
        return ans;
    }
    void work(double l,int d)
    {
        Segment x1;
        x1.a.x=0; x1.a.y=0;
        x1.b.x=0; x1.b.y=l;
        double ans=0;
        Point W;  W.x=0;  W.y=l;
        if(check(x1,qq)==1)
        {
            ans+=l;
            ans+=dfs(l/4,1,d,W,-TT);
            ans+=dfs(l/4,1,d,W,0);
            ans+=dfs(l/4,1,d,W,TT);
        }
        else
        {
            Point x;
            x.x=0;
            x.y=0;
            Point P=make(x1,qq); ans+=dis(x,P);
        }
        cout<<ans<<endl;
    }
    int main()
    {
        cout<<fixed<<setprecision(6);
        int T; cin>>T;
        while(T--)
        {
            int d;
            double l,x1,y1,k; cin>>l>>d>>x1>>y1>>k;
            qq.a.x=x1; qq.a.y=y1;
            qq.b.x=x1+1; qq.b.y=y1+k*1;
            work(l,d);
        }
    }
    View Code

     M. Subsequence 题库链接

    思路:预处理

    代码:

    #include <bits/stdc++.h>
    
    using namespace std;
    
    #define fi first
    #define se second
    
    #define pb push_back
    
    const int inf = 0x3f3f3f3f;
                char str[100009],s1[100009];
                int cnt1[300],cc[300];
                int nxt[100009][30];
                int pt[30];
    
                int getid(char c) {
                    return c - 'a' + 1;
                }
    int main(){
                scanf("%s", str);
                int len = strlen(str);
                for(int i=0; i<30; i++) pt[i] = len + 1;
                for(int i=0; i<30; i++) nxt[len+1][i] = len+1;
                for(int i=len-1; i>=0; i--) {
                    for(int j=0; j<30; j++)
                        nxt[i][j] = pt[j];
                    pt[getid(str[i])] = i;
                }
                int T;  scanf("%d", &T);
                while(T--) {
                    scanf("%s", s1);
                    int l = strlen(s1);
                    int flag = 1,id = pt[getid(s1[0])];
    
                    for(int i=1; i<l; i++) {
                        int d = nxt[id][getid(s1[i])];
                        if(d > id) {
                            id = d;
                        }
                        else {
                            flag = 0;
                            break;
                        }
                    }
                    if(flag && id <= len) puts("YES");
                    else puts("NO");
                }
                return 0;
    }
    View Code
  • 相关阅读:
    4个小时实现一个HTML5音乐播放器
    一款好看+极简到不行的HTML5音乐播放器-skPlayer
    操纵浏览器的历史记录
    基于jQuery查找dom的多种方式性能问题
    你真的了解console吗?
    关于overflow:hidden和bfc
    jQuery插件开发
    深入浅出jsonp
    jQuery.extend 函数详解
    [转] Hibernate一级缓存、二级缓存
  • 原文地址:https://www.cnblogs.com/widsom/p/10742707.html
Copyright © 2020-2023  润新知