赛后经验：div3过于简单，以后不做了

A.存在以下情况即为NO

1.存在相同字母 2.最大字母-最小字母 != 字符串长度

#include <map>

#include <set>

#include <ctime>

#include <cmath>

#include <queue>

#include <stack>

#include <vector>

#include <string>

#include <cstdio>

#include <cstdlib>

#include <cstring>

#include <sstream>

#include <iostream>

#include <algorithm>

#include <functional>

using namespace std;

#define For(i, x, y) for(int i=x;i<=y;i++)

#define _For(i, x, y) for(int i=x;i>=y;i--)

#define Mem(f, x) memset(f,x,sizeof(f))

#define Sca(x) scanf("%d", &x)

#define Sca2(x,y) scanf("%d%d",&x,&y)

#define Sca3(x,y,z) scanf("%d%d%d",&x,&y,&z)

#define Scl(x) scanf("%lld",&x);

#define Pri(x) printf("%d\n", x)

#define Prl(x) printf("%lld\n",x);

#define CLR(u) for(int i=0;i<=N;i++)u[i].clear();

#define LL long long

#define ULL unsigned long long

#define mp make_pair

#define PII pair<int,int>

#define PIL pair<int,long long>

#define PLL pair<long long,long long>

#define pb push_back

#define fi first

#define se second

typedef vector<int> VI;

int read(){int x = ,f = ;char c = getchar();while (c<'' || c>''){if (c == '-') f = -;c = getchar();}

while (c >= ''&&c <= ''){x = x *  + c - '';c = getchar();}return x*f;}

const double eps = 1e-;

const int maxn = 1e5 + ;

const int INF = 0x3f3f3f3f;

const int mod = 1e9 + ;

int N,M,K;

bool vis[maxn];

char str[maxn];

int main(){

    int T = read();

    while(T--){

        scanf("%s",str);

        int l = strlen(str);

        int MIN = INF,MAX = ,flag = ;

        for(int i =  ; i < ; i ++) vis[i] = ;

        for(int i =  ; i < l ; i ++){

            if(vis[str[i] - 'a']){

                flag = ;

                continue;

            }

            vis[str[i] - 'a'] = ;

            MIN = min(MIN,str[i] - 'a');

            MAX = max(MAX,str[i] - 'a');

        }

        if(MAX - MIN +  == l && flag){

            puts("Yes");

        }else{

            puts("No");

        }

    }

    return ;

}

A

B.设奇数字母cnt1个,偶数字母cnt2个

假设cnt1 >= cnt2 最多留下(cnt1 - cnt2 - 1)个奇数字母，偶数同理

取最小的保留就可以了

#include <map>

#include <set>

#include <ctime>

#include <cmath>

#include <queue>

#include <stack>

#include <vector>

#include <string>

#include <cstdio>

#include <cstdlib>

#include <cstring>

#include <sstream>

#include <iostream>

#include <algorithm>

#include <functional>

using namespace std;

#define For(i, x, y) for(int i=x;i<=y;i++)

#define _For(i, x, y) for(int i=x;i>=y;i--)

#define Mem(f, x) memset(f,x,sizeof(f))

#define Sca(x) scanf("%d", &x)

#define Sca2(x,y) scanf("%d%d",&x,&y)

#define Sca3(x,y,z) scanf("%d%d%d",&x,&y,&z)

#define Scl(x) scanf("%lld",&x);

#define Pri(x) printf("%d\n", x)

#define Prl(x) printf("%lld\n",x);

#define CLR(u) for(int i=0;i<=N;i++)u[i].clear();

#define LL long long

#define ULL unsigned long long

#define mp make_pair

#define PII pair<int,int>

#define PIL pair<int,long long>

#define PLL pair<long long,long long>

#define pb push_back

#define fi first

#define se second

typedef vector<int> VI;

int read(){int x = ,f = ;char c = getchar();while (c<'' || c>''){if (c == '-') f = -;c = getchar();}

while (c >= ''&&c <= ''){x = x *  + c - '';c = getchar();}return x*f;}

const double eps = 1e-;

const int maxn = ;

const int INF = 0x3f3f3f3f;

const int mod = 1e9 + ;

int N,M,K;

int even[maxn],odd[maxn];

int main(){

    Sca(N);

    int cnt1 = ,cnt2 = ;

    for(int i = ; i <= N ; i ++){

        int x = read();

        if(x & ) odd[++cnt1] = x;

        else even[++cnt2] = x;

    }

    sort(odd + ,odd + cnt1 + );

    sort(even + ,even +  + cnt2);

    if(abs(cnt1 - cnt2) <= ){

        puts("");

    }else if(cnt1 > cnt2){

        LL sum = ;

        for(int i = ; i <= cnt1 - cnt2 - ;i ++) sum += odd[i];

        Prl(sum);

    }else{

        LL sum = ;

        for(int i = ; i <= cnt2 - cnt1 - ;i ++) sum += even[i];

        Prl(sum);

    }

    return ;

}

B

C.很显然一个数出现次数超过3次就NO

第一个数列输出所有次数 >= 1的数

第二个数列输出所有次数 >= 2的数即可

#include <map>

#include <set>

#include <ctime>

#include <cmath>

#include <queue>

#include <stack>

#include <vector>

#include <string>

#include <cstdio>

#include <cstdlib>

#include <cstring>

#include <sstream>

#include <iostream>

#include <algorithm>

#include <functional>

using namespace std;

#define For(i, x, y) for(int i=x;i<=y;i++)

#define _For(i, x, y) for(int i=x;i>=y;i--)

#define Mem(f, x) memset(f,x,sizeof(f))

#define Sca(x) scanf("%d", &x)

#define Sca2(x,y) scanf("%d%d",&x,&y)

#define Sca3(x,y,z) scanf("%d%d%d",&x,&y,&z)

#define Scl(x) scanf("%lld",&x);

#define Pri(x) printf("%d\n", x)

#define Prl(x) printf("%lld\n",x);

#define CLR(u) for(int i=0;i<=N;i++)u[i].clear();

#define LL long long

#define ULL unsigned long long

#define mp make_pair

#define PII pair<int,int>

#define PIL pair<int,long long>

#define PLL pair<long long,long long>

#define pb push_back

#define fi first

#define se second

typedef vector<int> VI;

int read(){int x = ,f = ;char c = getchar();while (c<'' || c>''){if (c == '-') f = -;c = getchar();}

while (c >= ''&&c <= ''){x = x *  + c - '';c = getchar();}return x*f;}

const double eps = 1e-;

const int maxn = 2e5 + ;

const int INF = 0x3f3f3f3f;

const int mod = 1e9 + ;

int N,M,K;

int a[maxn];

int ans[maxn];

int main(){

    Sca(N);

    int MAX = -,MIN = INF;

    bool flag = ;

    for(int i = ; i <= N; i ++){

        int x = read();

        a[x]++;

        if(a[x] > ) flag = ;

        MAX = max(MAX,x); MIN = min(MIN,x);

    }

    if(!flag){

        puts("NO");

        return ;

    }

    puts("YES");

    int top = ;

    for(int i = MIN; i <= MAX; i ++){

        if(a[i]){

            a[i]--;

            ans[++top] = i;

        }

    }

    Pri(top);

    for(int i = ; i <= top; i ++) printf("%d ",ans[i]);

    puts("");

    top = ;

    for(int i = MAX; i >= MIN; i --){

        if(a[i]) ans[++top] = i;

    }

    Pri(top);

    for(int i = ; i <= top; i ++) printf("%d ",ans[i]);

    puts("");

    return ;

}

C

D.仔细一看就会发现操作就是把一个数变为相邻的一个数。

那么出现最多次数的数不变，其余的都变成他就可以了。

#include <map>

#include <set>

#include <ctime>

#include <cmath>

#include <queue>

#include <stack>

#include <vector>

#include <string>

#include <cstdio>

#include <cstdlib>

#include <cstring>

#include <sstream>

#include <iostream>

#include <algorithm>

#include <functional>

using namespace std;

#define For(i, x, y) for(int i=x;i<=y;i++)

#define _For(i, x, y) for(int i=x;i>=y;i--)

#define Mem(f, x) memset(f,x,sizeof(f))

#define Sca(x) scanf("%d", &x)

#define Sca2(x,y) scanf("%d%d",&x,&y)

#define Sca3(x,y,z) scanf("%d%d%d",&x,&y,&z)

#define Scl(x) scanf("%lld",&x);

#define Pri(x) printf("%d\n", x)

#define Prl(x) printf("%lld\n",x);

#define CLR(u) for(int i=0;i<=N;i++)u[i].clear();

#define LL long long

#define ULL unsigned long long

#define mp make_pair

#define PII pair<int,int>

#define PIL pair<int,long long>

#define PLL pair<long long,long long>

#define pb push_back

#define fi first

#define se second

typedef vector<int> VI;

int read(){int x = ,f = ;char c = getchar();while (c<'' || c>''){if (c == '-') f = -;c = getchar();}

while (c >= ''&&c <= ''){x = x *  + c - '';c = getchar();}return x*f;}

const double eps = 1e-;

const int maxn = 2e5 + ;

const int INF = 0x3f3f3f3f;

const int mod = 1e9 + ;

int N,M,K;

int a[maxn];

int num[maxn];

int main(){

    Sca(N);

    int MAX = ;

    for(int i = ; i <= N ; i ++){

        Sca(a[i]); num[a[i]]++;

        if(num[MAX] < num[a[i]]) MAX = a[i];

    }

    int root = ;

    int ans = ;

    Pri(N - num[MAX]);

    for(int i = ; i <= N ; i ++){

        if(a[i] == MAX){

            root = i;

            break;

        }

    }

    for(int i = root + ; i <= N; i ++){

        if(a[i] > a[i - ]) printf("2 %d %d\n",i,i - );

        else if(a[i] < a[i - ]) printf("1 %d %d\n",i,i - );

        a[i] = a[i - ];

    }

    for(int i = root - ; i >= ; i --){

        if(a[i] > a[i + ]) printf("2 %d %d\n",i,i + );

        else if(a[i] < a[i + ]) printf("1 %d %d\n",i,i + );

        a[i] = a[i + ];

    }

    return ;

}

D

E.模拟26进制加减法即可。具体操作就是(a + b) / 2

#include <map>

#include <set>

#include <ctime>

#include <cmath>

#include <queue>

#include <stack>

#include <vector>

#include <string>

#include <cstdio>

#include <cstdlib>

#include <cstring>

#include <sstream>

#include <iostream>

#include <algorithm>

#include <functional>

using namespace std;

#define For(i, x, y) for(int i=x;i<=y;i++)

#define _For(i, x, y) for(int i=x;i>=y;i--)

#define Mem(f, x) memset(f,x,sizeof(f))

#define Sca(x) scanf("%d", &x)

#define Sca2(x,y) scanf("%d%d",&x,&y)

#define Sca3(x,y,z) scanf("%d%d%d",&x,&y,&z)

#define Scl(x) scanf("%lld",&x);

#define Pri(x) printf("%d\n", x)

#define Prl(x) printf("%lld\n",x);

#define CLR(u) for(int i=0;i<=N;i++)u[i].clear();

#define LL long long

#define ULL unsigned long long

#define mp make_pair

#define PII pair<int,int>

#define PIL pair<int,long long>

#define PLL pair<long long,long long>

#define pb push_back

#define fi first

#define se second

typedef vector<int> VI;

int read(){int x = ,f = ;char c = getchar();while (c<'' || c>''){if (c == '-') f = -;c = getchar();}

while (c >= ''&&c <= ''){x = x *  + c - '';c = getchar();}return x*f;}

const double eps = 1e-;

const int maxn = 2e5 + ;

const int INF = 0x3f3f3f3f;

const int mod = 1e9 + ;

int N,M,K;

char a[maxn],b[maxn];

int c[maxn];

int main(){

    N = read();

    scanf("%s%s",a + ,b + );

    for(int i = N ; i >= ; i--){

        c[i] += b[i] + a[i] - 'a' - 'a' + ;

        while(c[i] > ){

            c[i] -= ;

            c[i - ]++;

        }

    }

    int zero = ;

    for(int i =  ; i <= N ; i ++){

        if(c[i] & ){

            c[i]--;

            c[i + ] += ;

        }

        c[i] /= ;

        if(c[i]) zero = ;

        if(!zero) printf("%c",c[i] + 'a' - );

    }

    return ;

}

E

F.意识到一个点不会即有入度又有出度，那么所有点就分为了出度点和入度点。

直接图上染色即可。

#include <map>

#include <set>

#include <ctime>

#include <cmath>

#include <queue>

#include <stack>

#include <vector>

#include <string>

#include <cstdio>

#include <cstdlib>

#include <cstring>

#include <sstream>

#include <iostream>

#include <algorithm>

#include <functional>

using namespace std;

#define For(i, x, y) for(int i=x;i<=y;i++)

#define _For(i, x, y) for(int i=x;i>=y;i--)

#define Mem(f, x) memset(f,x,sizeof(f))

#define Sca(x) scanf("%d", &x)

#define Sca2(x,y) scanf("%d%d",&x,&y)

#define Sca3(x,y,z) scanf("%d%d%d",&x,&y,&z)

#define Scl(x) scanf("%lld",&x);

#define Pri(x) printf("%d\n", x)

#define Prl(x) printf("%lld\n",x);

#define CLR(u) for(int i=0;i<=N;i++)u[i].clear();

#define LL long long

#define ULL unsigned long long

#define mp make_pair

#define PII pair<int,int>

#define PIL pair<int,long long>

#define PLL pair<long long,long long>

#define pb push_back

#define fi first

#define se second

typedef vector<int> VI;

int read(){int x = ,f = ;char c = getchar();while (c<'' || c>''){if (c == '-') f = -;c = getchar();}

while (c >= ''&&c <= ''){x = x *  + c - '';c = getchar();}return x*f;}

const double eps = 1e-;

const int maxn = 2e5 + ;

const int INF = 0x3f3f3f3f;

const int mod = 1e9 + ;

int N,M,K;

struct Edge{

    int to,next;

}edge[maxn * ];

int head[maxn],tot;

void init(){

    for(int i =  ; i <= N; i ++) head[i] = -;

    tot = ;

}

void add(int u,int v){

    edge[tot].to = v;

    edge[tot].next = head[u];

    head[u] = tot++;

}

int color[maxn];

bool flag = ;

void dfs(int t){

    if(!flag) return;

    for(int i = head[t]; ~i ; i = edge[i].next){

        int v = edge[i].to;

        if(color[v] == color[t]){

            flag = ;

            return;

        }

        if(color[v] ==  - color[t]) continue;

        color[v] =  - color[t];

        dfs(v);

    }

}

PII E[maxn];

int main(){

    Sca2(N,M); init();

    for(int i = ; i <= M; i ++){

        int u = read(),v = read();

        add(u,v); add(v,u);

        E[i] = mp(u,v);

    }

    for(int i = ; i <= N && flag; i ++){

        if(!color[i]){

            color[i] = ;

            dfs(i);

        }

    }

    if(!flag){

        puts("NO"); return ;

    }else{

        puts("YES");

    }

    for(int i = ; i <= M ; i ++){

        if(color[E[i].fi] == ) printf("");

        else printf("");

    }

    return ;

}

F

G.题目的一个子问题是序列中的一个区间能否分为递增和递减

意识到一个区间最小的点，只能作为递增的起点或者递减的终点，在知道了这个条件下，另一边就已知必须递增或者递减。

预处理出每个点为起点的最大递增终点和最大递减终点。

用solve(l,r,lMIN,lMAX,rMIN,rMAX)表示在l到r这段区间内，递减段的数字被限制在lMIN到lMAX内，递增段的数字被限制在rMIN，rMAX内，在此基础上求是否可行。

#include <map>

#include <set>

#include <ctime>

#include <cmath>

#include <queue>

#include <stack>

#include <vector>

#include <string>

#include <cstdio>

#include <cstdlib>

#include <cstring>

#include <sstream>

#include <iostream>

#include <algorithm>

#include <functional>

using namespace std;

#define For(i, x, y) for(int i=x;i<=y;i++)

#define _For(i, x, y) for(int i=x;i>=y;i--)

#define Mem(f, x) memset(f,x,sizeof(f))

#define Sca(x) scanf("%d", &x)

#define Sca2(x,y) scanf("%d%d",&x,&y)

#define Sca3(x,y,z) scanf("%d%d%d",&x,&y,&z)

#define Scl(x) scanf("%lld",&x);

#define Pri(x) printf("%d\n", x)

#define Prl(x) printf("%lld\n",x);

#define CLR(u) for(int i=0;i<=N;i++)u[i].clear();

#define LL long long

#define ULL unsigned long long

#define mp make_pair

#define PII pair<int,int>

#define PIL pair<int,long long>

#define PLL pair<long long,long long>

#define pb push_back

#define fi first

#define se second

typedef vector<int> VI;

int read(){int x = ,f = ;char c = getchar();while (c<'' || c>''){if (c == '-') f = -;c = getchar();}

while (c >= ''&&c <= ''){x = x *  + c - '';c = getchar();}return x*f;}

const double eps = 1e-;

const int maxn = 8e5 + ;

const int INF = 0x3f3f3f3f;

const int mod = 1e9 + ;

int N,M,K;

int a[maxn],Hash[maxn];

struct tree{

    int l,r,MIN;

}tree[maxn << ];

void Pushup(int t){

    if(a[tree[t << ].MIN] < a[tree[t <<  | ].MIN]) tree[t].MIN = tree[t << ].MIN;

    else tree[t].MIN = tree[t <<  | ].MIN;

}

void Build(int t,int l,int r){

    tree[t].l = l; tree[t].r = r;

    if(l == r){

        tree[t].MIN = l;

        return;

    }

    int m = l + r >> ;

    Build(t << ,l,m); Build(t <<  | ,m + ,r);

    Pushup(t);

}

int query(int t,int l,int r){

    if(l <= tree[t].l && tree[t].r <= r) return tree[t].MIN;

    int m = tree[t].l + tree[t].r >> ;

    if(r <= m) return query(t << ,l,r);

    else if(l > m) return query(t <<  | ,l,r);

    else{

        int p1 = query(t << ,l,m),p2 = query(t <<  | ,m + ,r);

        if(a[p1] > a[p2]) return p2;

        return p1;

    }

}

int ans[maxn];

int radd[maxn],rdel[maxn];

bool solve(int L,int R,int lMIN,int lMAX,int rMIN,int rMAX){

    if(L > R) return true;

    if(L == R){

        if(rMIN < a[L] && a[L] < rMAX){

            ans[L] = ;

            return true;

        }

        if(lMIN < a[L] && a[L] < lMAX){

            ans[L] = ;

            return true;

        }

        return false;

    }

    int p = query(,L,R);

    if(lMIN < a[p] && a[p] < lMAX && (p == R || (radd[p + ] >= R && a[p + ] > rMIN && a[p + ] < rMAX && a[R] < rMAX && a[R] > rMIN))){

        if(solve(L,p - ,a[p],lMAX,rMIN,p == R?rMAX:a[p + ])){

            ans[p] = ;

            for(int j = p + ; j <= R; j ++) ans[j] = ;

            return true;

        }

    }

    if(rMIN < a[p] && a[p] < rMAX && (p == L || (rdel[L] >= p -  && a[L] > lMIN && a[L] < lMAX && a[p - ] < lMAX && a[p - ] > lMIN))){

        if(solve(p + ,R,lMIN,p == L?lMAX:a[p - ],a[p],rMAX)){

            ans[p] = ;

            for(int j = L; j <= p - ; j ++) ans[j] = ;

            return true;

        }

    }

    return false;

}

int main(){

    Sca(N);

    for(int i = ; i <= N ; i ++) a[i] = read();

    radd[N] = rdel[N] = N;

    for(int i = N - ; i >= ; i --){

        if(a[i] < a[i + ]) radd[i] = radd[i + ];

        else radd[i] = i;

        if(a[i] > a[i + ]) rdel[i] = rdel[i + ];

        else rdel[i] = i;

    }

    Build(,,N);

    if(solve(,N,-INF,INF,-INF,INF)){

        puts("YES");

        for(int i = ; i <= N ; i ++) printf("%d ",ans[i]);

    }else{

        puts("NO");

    }

    return ;

}

G

 

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