2016湖南省赛 [Cloned]

A.2016

给出正整数 n 和 m，统计满足以下条件的正整数对 (a,b) 的数量：

1. 1≤a≤n,1≤b≤m;

2. a×b 是 2016 的倍数。

Input

输入包含不超过 30 组数据。

每组数据包含两个整数 n,m (1≤n,m≤10 ⁹).

Output对于每组数据，输出一个整数表示满足条件的数量。Sample Input

32 63

2016 2016

1000000000 1000000000

Sample Output

1

30576

7523146895502644

代码：

#include<bits/stdc++.h>

using namespace std;

#define ll long long

int vis[];

int main()

{

    int n,m;

    while(~scanf("%d%d",&n,&m))

    {

        memset(vis,,sizeof(vis));

        ll ans=;

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

        {

            vis[i]+=n/i;

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

            {

                if(i%j==)

                    vis[j]-=vis[i];

            }

            ans+=1LL*vis[i]*(m/(/__gcd(,i)));

        }

        printf("%lld\n",ans);

    }

}

B. 有向无环图

Bobo 有一个 n 个点，m 条边的有向无环图（即对于任意点 v，不存在从点 v 开始、点 v 结束的路径）。

为了方便，点用 1,2,…,n 编号。设 count(x,y) 表示点 x 到点 y 不同的路径数量（规定 count(x,x)=0），Bobo 想知道

除以 (10 ⁹+7) 的余数。

其中，a _i,b _j 是给定的数列。

Input

输入包含不超过 15 组数据。

每组数据的第一行包含两个整数 n,m (1≤n,m≤10 ⁵).

接下来 n 行的第 i 行包含两个整数 a _i,b _i (0≤a _i,b _i≤10 ⁹).

最后 m 行的第 i 行包含两个整数 u _i,v _i，代表一条从点 u _i 到 v _i 的边 (1≤u _i,v_i≤n)。

Output对于每组数据，输出一个整数表示要求的值。Sample Input

3 3

1 1

1 1

1 1

1 2

1 3

2 3

2 2

1 0

0 2

1 2

1 2

2 1

500000000 0

0 500000000

1 2

Sample Output

代码：

#include<bits/stdc++.h>

using namespace std;

#define ll long long

const int mod=1e9+;

const int maxn=1e5+;

ll a[maxn],b[maxn],s[maxn];

vector<int>G[maxn];

int in[maxn],n,m;

void work()

{

    queue<int>P;

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

    {

        if(in[i]==)

            P.push(i);

    }

    while(!P.empty())

    {

        int v=P.front();P.pop();

        for(int i=G[v].size()-;i>=;i--)

        {

            int u=G[v][i];

            in[u]--;

            if(in[u]==)

                P.push(u);

            s[u]=(s[u]+s[v])%mod;

        }

    }

    ll ans=;

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

        ans=(ans+(s[i]-a[i]+mod)*b[i])%mod;

    printf("%lld\n",ans);

}

int main()

{

    while(~scanf("%d%d",&n,&m))

    {

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

        {

            scanf("%lld%lld",&a[i],&b[i]);

            s[i]=a[i];

        }

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

        {

            int x,y;scanf("%d%d",&x,&y);

            G[x].push_back(y);

            in[y]++;

        }

        work();

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

        {

            G[i].clear();

            in[i]=;

        }

    }

    return ;

}

G.Parentthesis

Bobo has a balanced parenthesis sequence P=p ₁ p ₂…p _n of length n and q questions.

The i-th question is whether P remains balanced after p _ai and p _bi swapped. Note that questions are individual so that they have no affect on others.

Parenthesis sequence S is balanced if and only if:

1. S is empty;

2. or there exists balanced parenthesis sequence A,B such that S=AB;

3. or there exists balanced parenthesis sequence S' such that S=(S').

Input

The input contains at most 30 sets. For each set:

The first line contains two integers n,q (2≤n≤10 ⁵,1≤q≤10 ⁵).

The second line contains n characters p ₁ p ₂…p _n.

The i-th of the last q lines contains 2 integers a _i,b _i (1≤a _i,b _i≤n,a _i≠b _i).

OutputFor each question, output " Yes" if P remains balanced, or " No" otherwise.Sample Input

Sample Output

No

Yes

No

代码（括号匹配终于会啦）：

#include <bits/stdc++.h>

using namespace std;

const int maxn = 1e5 + ;

int N, Q;

string s;

int sum[maxn], vis[maxn], a[maxn];

int main() {

    while(~scanf("%d%d", &N, &Q)) {

        memset(sum, , sizeof(sum));

        cin >> s;

        for(int i = ; s[i]; i ++) {

            if(s[i] == '(') a[i] = ;

            else a[i] = -;

        }

        memset(vis, , sizeof(vis));

        for(int i = ; s[i]; i ++) {

            if(i == ) sum[i] = a[i];

            else sum[i] = sum[i - ] + a[i];

        }

        for(int q = ; q < Q; q ++) {

            int l, r;

            scanf("%d%d", &l, &r);

            l -= , r -= ;

            if(s[l] == s[r]) printf("Yes\n");

            else {

                if(s[r] == ')') if(sum[l] -  <  || sum[r] -  < ) printf("No\n");

                else printf("Yes\n");

            }

        }

    }

    return ;

}

H. Reverse

Bobo has a n digits decimal number D=d ₁ d ₂…d _n (It may have leading zeros).

Let R(i,j) denotes number D with digits between the i-th position and j-th position reversed. That is, R(i,j)=d ₁…d _i-1 d _j d _j-1…d _i d _j+1 d _j+2…d _n.

Bobo would like to find

modulo (10 ⁹+7).

Input

The input contains at most 30 sets. For each set:

The first line contains an integer n (1≤n≤10 ⁵).

The second line contains n digits d ₁ d ₂…d _n (0≤d _i≤9).

OutputFor each set, an integer denotes the result.Sample Input

Sample Output

代码：

#include<bits/stdc++.h>

using namespace std;

#define ll long long

const int mod=1e9+;

ll pow1(ll a,ll b)

{

    ll r=;

    while(b)

    {

        if(b&)

            r=r*a%mod;

        a=a*a%mod;

        b/=;

    }

    return r;

}

ll inv_9=pow1(,mod-);

ll p[];

char t[];

int main()

{

    p[]=;

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

        p[i]=p[i-]*%mod;

    int n;

    while(~scanf("%d",&n))

    {

        scanf("%s",t+);

        reverse(t+,t+n+);

        ll ans=;

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

        {

            ll v=t[i]-'';

            ans+=(1LL*(i-)+1LL*(i-)*(i-)/)%mod*v%mod*p[i-]%mod;

            ans+=(1LL*(n-i)+1LL*(n-i)*(n-i-)/)%mod*v%mod*p[i-]%mod;

            ans%=mod;

            ll tmp=v*(p[i]-+mod)%mod*inv_9%mod;

            tmp=tmp*(p[n-i+]-+mod)%mod*inv_9%mod;

            ans+=tmp;

            ans%=mod;

        }

        printf("%lld\n",ans);

    }

    return ;

}

I.Tree Intersection

Bobo has a tree with n vertices numbered by 1,2,…,n and (n-1) edges. The i-th vertex has color c _i, and the i-th edge connects vertices a _i and b _i.

Let C(x,y) denotes the set of colors in subtree rooted at vertex x deleting edge (x,y).

Bobo would like to know R_i which is the size of intersection of C(a _i,b _i) and C(b_i,a _i) for all 1≤i≤(n-1). (i.e. |C(a _i,b _i)∩C(b _i,a _i)|)

Input

The input contains at most 15 sets. For each set:

The first line contains an integer n (2≤n≤10 ⁵).

The second line contains n integers c ₁,c ₂,…,c _n (1≤c_i≤n).

The i-th of the last (n-1) lines contains 2 integers a _i,b _i (1≤a _i,b _i≤n).

OutputFor each set, (n-1) integers R ₁,R ₂,…,R _n-1.Sample Input

Sample Output

代码：

#include<bits/stdc++.h>

using namespace std;

const int maxn=1e5+;

vector<int>G[maxn],id[maxn];

int a[maxn],ans[maxn],s[maxn],now[maxn],son[maxn],sz[maxn],cnt;

void get_son(int v,int fa)

{

    sz[v]=;

    son[v]=;

    for(int i=G[v].size()-;i>=;i--)

    {

        int u=G[v][i];

        if(u==fa)

            continue;

        get_son(u,v);

        sz[v]+=sz[u];

        if(sz[u]>sz[son[v]])

            son[v]=u;

    }

}

void work1(int v,int fa)

{

    now[a[v]]++;

    if(now[a[v]]==&&s[a[v]]>)

        cnt++;

    else if(now[a[v]]==s[a[v]]&&now[a[v]]>)

        cnt--;

    for(int i=G[v].size()-;i>=;i--)

    {

        int u=G[v][i];

        if(u==fa)

            continue;

        work1(u,v);

    }

}

void work2(int v,int fa)

{

    now[a[v]]--;

    if(now[a[v]]==&&s[a[v]]>)

        cnt--;

    else if(now[a[v]]==s[a[v]]-&&now[a[v]]>)

        cnt++;

    for(int i=G[v].size()-;i>=;i--)

    {

        int u=G[v][i];

        if(u==fa)

            continue;

        work2(u,v);

    }

}

void dfs(int v,int fa,int flag,int q)

{

    int I;

    for(int i=G[v].size()-;i>=;i--)

    {

        int u=G[v][i];

        if(u==son[v])

            I=id[v][i];

        if(u==fa||u==son[v])

            continue;

        dfs(u,v,,id[v][i]);

    }

    if(son[v])

        dfs(son[v],v,,I);

    for(int i=G[v].size()-;i>=;i--)

    {

        int u=G[v][i];

        if(u==fa||u==son[v])

            continue;

        work1(u,v);

    }

    now[a[v]]++;

    if(now[a[v]]==&&s[a[v]]>)

        cnt++;

    else if(now[a[v]]==s[a[v]]&&now[a[v]]>)

        cnt--;

    ans[q]=cnt;

    if(flag==)

        work2(v,fa);

}

int main()

{

    int n;

    while(~scanf("%d",&n))

    {

        memset(s,,sizeof(s));

        memset(now,,sizeof(now));

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

            scanf("%d",&a[i]),s[a[i]]++;

        for(int i=,x,y;i<n;i++)

        {

            scanf("%d%d",&x,&y);

            G[x].push_back(y);

            id[x].push_back(i);

            G[y].push_back(x);

            id[y].push_back(i);

        }

        get_son(,);

        cnt=;

        dfs(,,,);

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

        {

            G[i].clear();

            id[i].clear();

            if(i<n)

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

            ans[i]=;

        }

    }

}

J.三角形和矩形

Bobo 有一个三角形和一个矩形，他想求他们交的面积。

具体地，三角形和矩形由 8 个整数 x ₁,y ₁,x ₂,y ₂,x ₃,y ₃,x ₄,y ₄ 描述。表示三角形的顶点坐标是 (x ₁,y ₁),(x ₁,y ₂),(x ₂,y ₁)，矩形的顶点坐标是 (x ₃,y ₃),(x ₃,y ₄),(x ₄,y₄),(x ₄,y ₃).

Input

输入包含不超过 30000 组数据。

每组数据的第一行包含 4 个整数 x ₁,y ₁,x ₂,y ₂ (x ₁≠x ₂,y ₁≠y ₂).

第二行包含 4 个整数 x ₃,y ₃,x ₄,y ₄ (x ₃<x ₄,y ₃<y ₄).

(0≤x _i,y _i≤10 ⁴)

Output对于每组数据，输出一个实数表示交的面积。绝对误差或相对误差小于 10 ^-6 即认为正确。Sample Input

1 1 3 3

0 0 2 2

0 3 3 1

0 0 2 2

4462 1420 2060 2969

4159 257 8787 2970

Sample Output

1.00000000

0.75000000

439744.13967527

代码：

#include<bits/stdc++.h>

using namespace std;

double k,b;

double cal1(double Y1,double Y2,double y,double xx1,double xx2)

{

    if(y>=Y2) return ;

    double h=(y-b)/k;

    if(y>=Y1&&y<Y2)

        return (xx2-h)*(Y2-y)*0.5;

    return ((Y1-y)+(Y2-y))*(xx2-xx1)*0.5;

}

double cal2(double Y1,double Y2,double y,double xx1,double xx2)

{

    if(y<=Y2) return ;

    double h=(y-b)/k;

    if(y>Y2&&y<=Y1)

        return (y-Y2)*(xx2-h)*0.5;

    return ((y-Y1)+(y-Y2))*(xx2-xx1)*0.5;

}

double cal3(double Y1,double Y2,double y,double xx1,double xx2)

{

    if(y>=Y1) return ;

    double h=(y-b)/k;

    if(y>=Y2&&y<Y1)

        return (Y1-y)*(h-xx1)*0.5;

    return ((Y1-y)+(Y2-y))*(xx2-xx1)*0.5;

}

double cal4(double Y1,double Y2,double y,double xx1,double xx2)

{

    if(y<=Y1) return ;

    double h=(y-b)/k;

    if(y>Y1&&y<=Y2)

        return (y-Y1)*(h-xx1)*0.5;

    return ((y-Y1)+(y-Y2))*(xx2-xx1)*0.5;

}

int main()

{

    int x1,y1,x2,y2,x3,y3,x4,y4;

    while(~scanf("%d%d%d%d%d%d%d%d",&x1,&y1,&x2,&y2,&x3,&y3,&x4,&y4))

    {

        int xx1=max(min(x1,x2),x3),xx2=min(max(x1,x2),x4);

        int yy1=max(min(y1,y2),y3),yy2=min(max(y1,y2),y4);

        k=1.0*(y2-y1)/(x1-x2),b=1.0*y2-k*x1;

        if(xx1>=xx2||yy1>=yy2)

            printf("0.0000000000000\n");

        else if(y2>y1&&x1>x2)

        {

            double Y1=k*xx1+b,Y2=k*xx2+b;

            double ans=cal1(Y1,Y2,1.0*yy1,1.0*xx1,1.0*xx2)-cal1(Y1,Y2,1.0*yy2,1.0*xx1,1.0*xx2);

            printf("%.10lf\n",ans);

        }

        else if(y2<y1&&x2<x1)

        {

            double Y1=k*xx1+b,Y2=k*xx2+b;

            double ans=cal2(Y1,Y2,1.0*yy2,1.0*xx1,1.0*xx2)-cal2(Y1,Y2,1.0*yy1,1.0*xx1,1.0*xx2);

            printf("%.10lf\n",ans);

        }

        else if(x2>x1&&y2>y1)

        {

            double Y1=k*xx1+b,Y2=k*xx2+b;

            double ans=cal3(Y1,Y2,1.0*yy1,1.0*xx1,1.0*xx2)-cal3(Y1,Y2,1.0*yy2,1.0*xx1,1.0*xx2);

            printf("%.10lf\n",ans);

        }

        else

        {

            double Y1=k*xx1+b,Y2=k*xx2+b;

            double ans=cal4(Y1,Y2,1.0*yy2,1.0*xx1,1.0*xx2)-cal4(Y1,Y2,1.0*yy1,1.0*xx1,1.0*xx2);

            printf("%.10lf\n",ans);

        }

    }

}

明天省赛加油啦希望有好结果

今日份的瘦宅茶

喜茶最近出的芝芝桃桃和多肉粉荔都没时间去喝！！！不是合格的 HEYTEA Girl 了！！！快放假吧