A

水题

/*Huyyt*/

#include<bits/stdc++.h>

#define mem(a,b) memset(a,b,sizeof(a))

using namespace std;

typedef long long ll;

typedef unsigned long long ull;

const int dir[][] = {{, }, {, }, {, -}, { -, }, {, }, {, -}, { -, -}, { -, }};

const int mod = 1e9 + , gakki =  +  +  +  + 1e9;

const int MAXN = 1e5 + , MAXM = 1e5 + , N = 1e5 + ;

const int MAXQ = ;

int to[MAXM << ], nxt[MAXM << ], Head[MAXN], tot = ;

inline void addedge(int u, int v)

{

        to[++tot] = v;

        nxt[tot] = Head[u];

        Head[u] = tot;

}

int main()

{

        ios_base::sync_with_stdio();

        cin.tie();

        ll n, m, a, b;

        cin >> n >> m >> a >> b;

        if (n % m == )

        {

                cout <<  << endl;

                return ;

        }

        ll now = n / m;

        ll ans1 = (n - now * m) * b;

        ll ans2 = ((now + ) * m - n) * a;

        cout << min(ans1, ans2) << endl;

        return ;

}

B

题目说了 在某个数之前K范围内的都能被消除掉 所以直接作一个前缀和即可

/*Huyyt*/

#include<bits/stdc++.h>

#define mem(a,b) memset(a,b,sizeof(a))

using namespace std;

typedef long long ll;

typedef unsigned long long ull;

const int dir[][] = {{, }, {, }, {, -}, { -, }, {, }, {, -}, { -, -}, { -, }};

const int mod = 1e9 + , gakki =  +  +  +  + 1e9;

const int MAXN = 1e5 + , MAXM = 1e5 + , N = 2e5 + ;

const int MAXQ = ;

int to[MAXM << ], nxt[MAXM << ], Head[MAXN], tot = ;

inline void addedge(int u, int v)

{

        to[++tot] = v;

        nxt[tot] = Head[u];

        Head[u] = tot;

}

int num[N];

int visit[];

priority_queue<int, vector<int>, greater<int> >que;

int main()

{

        ios_base::sync_with_stdio();

        cin.tie();

        int n, k;

        cin >> n >> k;

        int anser = ;

        for (int i = ; i <= n; i++)

        {

                cin >> num[i];

                int l = max(num[i] - k, );

                visit[l]++;

                visit[num[i]]--;

        }

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

        {

                visit[i] += visit[i - ];

        }

        for (int i = ; i <= n; i++)

        {

                if (!visit[num[i]])

                {

                        anser++;

                }

        }

        cout << anser << endl;

        return ;

}

C

卡题意题..最后弱智答案没遍历完全FST了QAQ

/*Huyyt*/

#include<bits/stdc++.h>

#define mem(a,b) memset(a,b,sizeof(a))

using namespace std;

typedef long long ll;

typedef unsigned long long ull;

const int dir[][] = {{, }, {, }, {, -}, { -, }, {, }, {, -}, { -, -}, { -, }};

const int mod = 1e9 + , gakki =  +  +  +  + 1e9;

const int MAXN = 1e5 + , MAXM = 1e5 + , N = 2e5 + ;

const int MAXQ = ;

int to[MAXM << ], nxt[MAXM << ], Head[MAXN], tot = ;

inline void addedge(int u, int v)

{

        to[++tot] = v;

        nxt[tot] = Head[u];

        Head[u] = tot;

}

ll n, flag, cnt, pop;

ll anser = ;

string ch[];

ll number[];

ll ansl[];

ll ansr[];

void init()

{

        pop = , cnt = , flag = ;

}

int main()

{

        int n;

        cin >> n;

        for (int i = ; i <= n; i++)

        {

                init();

                cin >> ch[i];

                for (int j = ; j < ch[i].size(); j++)

                {

                        if (ch[i][j] == '(')

                        {

                                number[++pop] = ;

                        }

                        else

                        {

                                number[++pop] = -;

                        }

                }

                for (int i = ; i <= pop; i++)

                {

                        cnt += number[i];

                        if (cnt < )

                        {

                                flag = ;

                                break;

                        }

                }

                if (!flag)

                {

                        ansl[cnt]++;

                }

                flag = cnt = ;

                for (int i = pop; i >= ; i--)

                {

                        cnt -= number[i];

                        if (cnt < )

                        {

                                flag = ;

                                break;

                        }

                }

                if (!flag)

                {

                        ansr[cnt]++;

                }

        }

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

        {

                anser += ansl[i] * ansr[i];

        }

        cout << anser << endl;

}

D

构造题

/*Huyyt*/

#include<bits/stdc++.h>

#define mem(a,b) memset(a,b,sizeof(a))

using namespace std;

typedef long long ll;

typedef unsigned long long ull;

const int dir[][] = {{, }, {, }, {, -}, { -, }, {, }, {, -}, { -, -}, { -, }};

const int mod = 1e9 + , gakki =  +  +  +  + 1e9;

const int MAXN = 1e5 + , MAXM = 1e5 + , N = 2e5 + ;

const int MAXQ = ;

int to[MAXM << ], nxt[MAXM << ], Head[MAXN], tot = ;

inline void addedge(int u, int v)

{

        to[++tot] = v;

        nxt[tot] = Head[u];

        Head[u] = tot;

}

int ans[][];

int main()

{

        int n;

        int a, b;

        cin >> n >> a >> b;

        if (min(a, b) != )

        {

                cout << "NO" << endl;

                return ;

        }

        if ((n ==  && a ==  && b == ) || (n ==  && a ==  && b == ))

        {

                cout << "NO" << endl;

                return ;

        }

        if (a >  && b == )

        {

                for (int i = ; i <= n; i++)

                {

                        for (int j = ; j <= n; j++)

                        {

                                if (i != j)

                                {

                                        ans[i][j] = ;

                                }

                        }

                }

                for (int i = ; i <= a - ; i++)

                {

                        for (int j = ; j <= n; j++)

                        {

                                ans[i][j] = ;

                                ans[j][i] = ;

                        }

                }

                cout << "YES" << endl;

                for (int i = ; i <= n; i++)

                {

                        for (int j = ; j <= n; j++)

                        {

                                cout << ans[i][j];

                        }

                        cout << endl;

                }

                return ;

        }

        if (a ==  && b == )

        {

                for (int i = ; i <= n; i++)

                {

                        ans[i][i + ] = ans[i + ][i] = ;

                }

                cout << "YES" << endl;

                for (int i = ; i <= n; i++)

                {

                        for (int j = ; j <= n; j++)

                        {

                                cout << ans[i][j];

                        }

                        cout << endl;

                }

                return ;

        }

        if (a ==  && b > )

        {

                for (int i = ; i <= b - ; i++)

                {

                        for (int j = ; j <= n; j++)

                        {

                                if (i != j)

                                {

                                        ans[i][j] = ans[j][i] = ;

                                }

                        }

                }

                cout << "YES" << endl;

                for (int i = ; i <= n; i++)

                {

                        for (int j = ; j <= n; j++)

                        {

                                cout << ans[i][j];

                        }

                        cout << endl;

                }

                return ;

        }

}

E

直接把灯尽量往右放 贪心暴力即可 复杂度和调和级数有关 并不会炸

有些人暴力的姿势不对会FST

/*Huyyt*/

#include<bits/stdc++.h>

#define mem(a,b) memset(a,b,sizeof(a))

using namespace std;

typedef long long ll;

typedef unsigned long long ull;

const int dir[][] = {{, }, {, }, {, -}, { -, }, {, }, {, -}, { -, -}, { -, }};

const int mod = 1e9 + , gakki =  +  +  +  + 1e9;

const int MAXN = 1e5 + , MAXM = 1e5 + , N = 2e6 + ;

const int MAXQ = ;

int to[MAXM << ], nxt[MAXM << ], Head[MAXN], tot = ;

inline void addedge(int u, int v)

{

        to[++tot] = v;

        nxt[tot] = Head[u];

        Head[u] = tot;

}

int n, m, k;

int visit[], has_block[], lighttt[];

ll anser;

int pop, summm, a1, a2;

void init()

{

        summm = pop = ;

}

int main()

{

        ios_base::sync_with_stdio();

        cin.tie();

        anser = LLONG_MAX;

        int n, m, k, now;

        cin >> n >> m >> k;

        n++;

        for (int i = ; i <= m; i++)

        {

                cin >> now;

                visit[now + ] = ;

        }

        if (visit[])

        {

                cout << - << endl;

                return ;

        }

        for (int i = ; i <= k; i++)

        {

                cin >> lighttt[i];

        }

        for (int i = ; i <= n; i++)

        {

                if (visit[i])

                {

                        has_block[i] = has_block[i - ];

                }

                else

                {

                        has_block[i] = i;

                }

        }

        for (int i = ; i <= k; i++)

        {

                init();

                while ()

                {

                        if (pop + i >= n)

                        {

                                ll cnt = 1LL*summm * lighttt[i];

                                anser = min(anser, cnt);

                                break;

                        }

                        if (pop == has_block[pop + i])

                        {

                                break;

                        }

                        summm++;

                        pop = has_block[pop + i];

                }

        }

        if (anser == LLONG_MAX)

        {

                cout << - << endl;

        }

        else

        {

                cout << anser << endl;

        }

}

F

首先很显然 因为传递不改变总值而要求是每个点的值都为0 所以当点的总值不为0时 是Impossible

然后我们要做的就是 DFS出一个生成树 然后从叶子到根传递出以一个顶点为根的子树其值的总值

因为每条边连接有两个顶点 这条边可以把这颗树分为两个子树 但题目要求是全部节点的值为0

所以最终要求是这两个子树的值也为0 则这条边传递的值 一定是 |其中一个子树的总值-0|  再判一下方向即可

其他没用到的边 答案是0 直接输出就行了

/*Huyyt*/

#include<bits/stdc++.h>

#define mem(a,b) memset(a,b,sizeof(a))

using namespace std;

typedef long long ll;

typedef unsigned long long ull;

const int dir[][] = {{, }, {, }, {, -}, { -, }, {, }, {, -}, { -, -}, { -, }};

const int mod = 1e9 + , gakki =  +  +  +  + 1e9;

const int MAXN = 2e5 + , MAXM = 2e5 + , N = 2e5 + ;

const int MAXQ = ;

int to[MAXM << ], nxt[MAXM << ], Head[MAXN], tot = ;

inline void addedge(int u, int v)

{

        to[++tot] = v;

        nxt[tot] = Head[u];

        Head[u] = tot;

}

inline void read(int &v)

{

        v = ;

        char c = ;

        int p = ;

        while (c < '' || c > '')

        {

                if (c == '-')

                {

                        p = -;

                }

                c = getchar();

        }

        while (c >= '' && c <= '')

        {

                v = (v << ) + (v << ) + c - '';

                c = getchar();

        }

        v *= p;

}

int value[MAXN], visit[MAXN], ans[MAXM << ];

int dfs(int x)

{

        int now = value[x];

        visit[x] = ;

        for (int i = Head[x]; i; i = nxt[i])

        {

                int v = to[i];

                if (!visit[v])

                {

                        int nownxt = dfs(v);

                        now += nownxt;

                        if (i & )

                        {

                                ans[i ^ ] = -nownxt;

                        }

                        else

                        {

                                ans[i] = nownxt;

                        }

                }

        }

        return now;

}

int main()

{

        ios_base::sync_with_stdio();

        cin.tie();

        int sum = ;

        int n, m, u, v;

        read(n);

        for (int i = ; i <= n; i++)

        {

                read(value[i]);

                sum += value[i];

        }

        read(m);

        for (int i = ; i <= m; i++)

        {

                read(u), read(v);

                addedge(u, v), addedge(v, u);

        }

        if (sum != )

        {

                printf("Impossible\n");

                return ;

        }

        printf("Possible\n");

        dfs();

        for (int i = ; i <= m; i++)

        {

                printf("%d\n", ans[i << ]);

        }

        return ;

}

G

也是个暴力题..把GCD搞出来 然后看连通块就行了

#include<iostream>

#include<cstdio>

#include<cstdlib>

#include<cstring>

#include<cmath>

#include<algorithm>

#include<set>

#include<map>

#include<vector>

#include<queue>

using namespace std;

#define ll long long

#define RG register

#define MAX 222222

inline int read()

{

    RG int x=,t=;RG char ch=getchar();

    while((ch<''||ch>'')&&ch!='-')ch=getchar();

    if(ch=='-')t=-,ch=getchar();

    while(ch<=''&&ch>='')x=x*+ch-,ch=getchar();

    return x*t;

}

struct Line{int v,next;}e[MAX<<];

int h[MAX],cnt=,mx;

inline void Add(int u,int v){e[cnt]=(Line){v,h[u]};h[u]=cnt++;}

vector<int> ys[MAX];

int n,a[MAX],tot;

bool book[MAX],vis[MAX];

ll ans[MAX];

void Div(int x,int id)

{

    for(int i=,m=sqrt(x);i<=m;++i)

        if(x%i==)

        {

            ys[i].push_back(id);

            if(i*i!=x)ys[x/i].push_back(id);

        }

}

void dfs(int u,int ff)

{

    if(!book[u])return;

    vis[u]=true;++tot;

    for(int i=h[u];i;i=e[i].next)

        if(e[i].v!=ff)dfs(e[i].v,u);

}

int main()

{

    n=read();

    for(int i=;i<=n;++i)mx=max(mx,a[i]=read());

    for(int i=;i<=n;++i)Div(a[i],i);

    for(int i=;i<n;++i)

    {

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

        Add(u,v);Add(v,u);

    }

    for(int i=;i<=mx;++i)

    {

        for(int j=,l=ys[i].size();j<l;++j)

        {

            int v=ys[i][j];

            vis[v]=false;book[v]=true;

        }

        for(int j=,l=ys[i].size();j<l;++j)

        {

            tot=;int v=ys[i][j];

            if(!vis[v])dfs(v,);book[v]=false;

            ans[i]+=1ll*tot*(tot-)/+tot;

        }

    }

    for(int i=mx;i;--i)

        if(ans[i])for(int j=i+i;j<=mx;j+=i)ans[i]-=ans[j];

    for(int i=;i<=mx;++i)

        if(ans[i])printf("%d %I64d\n",i,ans[i]);

    return ;

}

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