2016湖南省赛 [Cloned]

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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:

技术图片
#include<bits/stdc++.h>
using namespace std;
#define ll long long
int vis[2020];
int main()
{
    int n,m;
    while(~scanf("%d%d",&n,&m))
    {
        memset(vis,0,sizeof(vis));
        ll ans=0;
        for(int i=2016;i>=1;i--)
        {
            vis[i]+=n/i;
            for(int j=i-1;j>=1;j--)
            {
                if(i%j==0)
                    vis[j]-=vis[i];
            }
            ans+=1LL*vis[i]*(m/(2016/__gcd(2016,i)));
        }
        printf("%lld\\n",ans);
    }
}
View Code

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
 
技术图片
 
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,vi≤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

4
4
250000014

代码:

技术图片
#include<bits/stdc++.h>
using namespace std;
#define ll long long
const int mod=1e9+7;
const int maxn=1e5+10;
ll a[maxn],b[maxn],s[maxn];
vector<int>G[maxn];
int in[maxn],n,m;
void work()
{
    queue<int>P;
    for(int i=1;i<=n;i++)
    {
        if(in[i]==0)
            P.push(i);
    }
    while(!P.empty())
    {
        int v=P.front();P.pop();
        for(int i=G[v].size()-1;i>=0;i--)
        {
            int u=G[v][i];
            in[u]--;
            if(in[u]==0)
                P.push(u);
            s[u]=(s[u]+s[v])%mod;
        }
    }
    ll ans=0;
    for(int i=1;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=1;i<=n;i++)
        {
            scanf("%lld%lld",&a[i],&b[i]);
            s[i]=a[i];
        }
        for(int i=1;i<=m;i++)
        {
            int x,y;scanf("%d%d",&x,&y);
            G[x].push_back(y);
            in[y]++;
        }
        work();
        for(int i=1;i<=n;i++)
        {
            G[i].clear();
            in[i]=0;
        }
    }
    return 0;
}
View Code

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

4 2
(())
1 3
2 3
2 1
()
1 2

Sample Output

No
Yes
No

代码(括号匹配终于会啦)

技术图片
#include <bits/stdc++.h>
using namespace std;

const int maxn = 1e5 + 10;
int N, Q;
string s;
int sum[maxn], vis[maxn], a[maxn];

int main() {
    while(~scanf("%d%d", &N, &Q)) {
            
        memset(sum, 0, sizeof(sum));
        cin >> s;
        for(int i = 0; s[i]; i ++) {
            if(s[i] == () a[i] = 1;
            else a[i] = -1;
        }

        memset(vis, 0, sizeof(vis));
        for(int i = 0; s[i]; i ++) {
            if(i == 0) sum[i] = a[i];
            else sum[i] = sum[i - 1] + a[i];
        }

        for(int q = 0; q < Q; q ++) {
            int l, r;
            scanf("%d%d", &l, &r);
            l -= 1, r -= 1;
            if(s[l] == s[r]) printf("Yes\\n");
            else {
                if(s[r] == )) if(sum[l] - 2 < 0 || sum[r] - 2 < 0) printf("No\\n");
                else printf("Yes\\n");
            }
        }
    }
    
    return 0;
}
View Code

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
技术图片
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

2
12
3
012
10
0123456789

Sample Output

45
369
733424314

代码:

技术图片
#include<bits/stdc++.h>
using namespace std;
#define ll long long
const int mod=1e9+7;
ll pow1(ll a,ll b)
{
    ll r=1;
    while(b)
    {
        if(b&1)
            r=r*a%mod;
        a=a*a%mod;
        b/=2;
    }
    return r;
}
ll inv_9=pow1(9,mod-2);
ll p[100005];
char t[100005];
int main()
{
    p[0]=1;
    for(int i=1;i<=100000;i++)
        p[i]=p[i-1]*10%mod;
    int n;
    while(~scanf("%d",&n))
    {
        scanf("%s",t+1);
        reverse(t+1,t+n+1);
        ll ans=0;
        for(int i=1;i<=n;i++)
        {
            ll v=t[i]-0;
            ans+=(1LL*(i-1)+1LL*(i-1)*(i-2)/2)%mod*v%mod*p[i-1]%mod;
            ans+=(1LL*(n-i)+1LL*(n-i)*(n-i-1)/2)%mod*v%mod*p[i-1]%mod;
            ans%=mod;
            ll tmp=v*(p[i]-1+mod)%mod*inv_9%mod;
            tmp=tmp*(p[n-i+1]-1+mod)%mod*inv_9%mod;
            ans+=tmp;
            ans%=mod;
        }
        printf("%lld\\n",ans);
    }
    return 0;
}
View Code

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(bi,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

4
1 2 2 1
1 2
2 3
3 4
5
1 1 2 1 2
1 3
2 3
3 5
4 5

Sample Output

1
2
1
1
1
2
1

代码:

技术图片
#include<bits/stdc++.h>
using namespace std;
const int maxn=1e5+10;
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]=1;
    son[v]=0;
    for(int i=G[v].size()-1;i>=0;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]]==1&&s[a[v]]>1)
        cnt++;
    else if(now[a[v]]==s[a[v]]&&now[a[v]]>1)
        cnt--;
    for(int i=G[v].size()-1;i>=0;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]]==0&&s[a[v]]>1)
        cnt--;
    else if(now[a[v]]==s[a[v]]-1&&now[a[v]]>0)
        cnt++;
    for(int i=G[v].size()-1;i>=0;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()-1;i>=0;i--)
    {
        int u=G[v][i];
        if(u==son[v])
            I=id[v][i];
        if(u==fa||u==son[v])
            continue;
        dfs(u,v,0,id[v][i]);
    }
    if(son[v])
        dfs(son[v],v,1,I);
    for(int i=G[v].size()-1;i>=0;i--)
    {
        int u=G[v][i];
        if(u==fa||u==son[v])
            continue;
        work1(u,v);
    }
    now[a[v]]++;
    if(now[a[v]]==1&&s[a[v]]>1)
        cnt++;
    else if(now[a[v]]==s[a[v]]&&now[a[v]]>1)
        cnt--;
    ans[q]=cnt;
    if(flag==0)
        work2(v,fa);
}
int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        memset(s,0,sizeof(s));
        memset(now,0,sizeof(now));
        for(int i=1;i<=n;i++)
            scanf("%d",&a[i]),s[a[i]]++;
        for(int i=1,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(1,0);
        cnt=0;
        dfs(1,0,0,0);
        for(int i=1;i<=n;i++)
        {
            G[i].clear();
            id[i].clear();
            if(i<n)
                printf("%d\\n",ans[i]);
            ans[i]=0;
        }
    }
}
View Code

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:

技术图片
#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 0;
    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 0;
    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 0;
    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 0;
    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);
        }
    }
}
View Code

 

明天省赛 加油啦 希望有好结果

FH && 今日份的瘦宅茶 

技术图片

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

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