Testing Round #12

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本文深入探讨了三个核心问题：寻找特定区间内被整除的数字数量，餐厅订单的最大接受数量，以及序列中递增子序列的计数。通过优化算法和数据结构的应用，实现高效求解。对于每个问题，我们详细阐述了解决策略，包括状态转移方程的构建、树状数组与线段树的使用等关键步骤。

Find the number of k-divisible numbers on the segment [a, b]. In other words you need to find the number of such integer values x that a ≤ x ≤ b and x is divisible by k.

Input

The only line contains three space-separated integers k, a and b (1 ≤ k ≤ 10¹⁸; - 10¹⁸ ≤ a ≤ b ≤ 10¹⁸).

Output

Print the required number.

Sample test(s)

Input

1 1 10

Output

Input

2 -4 4

Output

5


题解: 简单水题

AC代码:
/* ***********************************************
Author        :xdlove
Created Time  :2015年11月12日 星期四 10时28分42秒
File Name     :xdlove/codeforces/Testing_Round_#12/A/A.cpp
************************************************ */

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <memory.h>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <time.h>

using namespace std;


#define REP_ab(i,a,b) for(int i = a; i <= b; i++)
#define REP(i, n) for(int i = 0; i < n; i++)
#define REP_1(i,n) for(int i = 1; i <= n; i++)
#define DEP(i,n) for(int i = n - 1; i >= 0; i--)
#define DEP_N(i,n) for(int i = n; i >= 1; i--)
#define CPY(A,B) memcpy(A,B,sizeof(B))
#define MEM(A) memset(A,0,sizeof(A))
#define MEM_1(A) memset(A,-1,sizeof(A))
#define MEM_INF(A) memset(A,0x3f,sizeof(A))
#define MEM_INFLL(A) memset(A,0x3f3f,sizeof(A))
#define mid (((l + r) >> 1))
#define lson l, mid, u << 1
#define rson mid + 1, r, u << 1 | 1
#define ls (u << 1)
#define rs (u << 1 | 1)


typedef long long ll;
typedef unsigned long long ull;
const int INF = 0x3f3f3f3f;
const ll INFLL = 0x3f3f3f3f3f3f3f3f;
const int MAXN = 1e5 + 5;
const int MAXM = MAXN;
const int mod = 1e9 + 7;


int main()
{
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    ll k,a,b,ans = 0;
    cin>>k>>a>>b;
    if(a == b && a == 0) ans = 1;
    else 
    {
        if(b >= 0) 
        {
            ans += b / k;
            if(a > 0) ans -= (a - 1) / k;
            else ans += -a / k + 1;
        }
        else 
            ans += -a / k - (-b - 1) / k;
    }
    cout<<ans<<endl;
    return 0;
}


                   B. Restaurant

A restaurant received n orders for the rental. Each rental order reserve the restaurant for a continuous period of time, the i-th order is characterized by two time values — the start time l_i and the finish time r_i (l_i ≤ r_i).
Restaurant management can accept and reject orders. What is the maximal number of orders the restaurant can accept?
No two accepted orders can intersect, i.e. they can't share even a moment of time. If one order ends in the moment other starts, they can't be accepted both.
Input
The first line contains integer number n (1 ≤ n ≤ 5·10⁵) — number of orders. The following n lines contain integer values l_i and r_i each (1 ≤ l_i ≤ r_i ≤ 10⁹).
Output
Print the maximal number of orders that can be accepted.
Sample test(s)
Input
2
7 11
4 7
Output
1
Input
5
1 2
2 3
3 4
4 5
5 6
Output
3
Input
6
4 8
1 5
4 7
2 5
1 3
6 8
Output
2

题解: 简单水题

AC代码:
/* ***********************************************
Author        :xdlove
Created Time  :2015年11月12日 星期四 10时39分50秒
File Name     :xdlove/codeforces/Testing_Round_#12/B/B.cpp
************************************************ */

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <memory.h>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <time.h>

using namespace std;


#define REP_ab(i,a,b) for(int i = a; i <= b; i++)
#define REP(i, n) for(int i = 0; i < n; i++)
#define REP_1(i,n) for(int i = 1; i <= n; i++)
#define DEP(i,n) for(int i = n - 1; i >= 0; i--)
#define DEP_N(i,n) for(int i = n; i >= 1; i--)
#define CPY(A,B) memcpy(A,B,sizeof(B))
#define MEM(A) memset(A,0,sizeof(A))
#define MEM_1(A) memset(A,-1,sizeof(A))
#define MEM_INF(A) memset(A,0x3f,sizeof(A))
#define MEM_INFLL(A) memset(A,0x3f3f,sizeof(A))
#define mid (((l + r) >> 1))
#define lson l, mid, u << 1
#define rson mid + 1, r, u << 1 | 1
#define ls (u << 1)
#define rs (u << 1 | 1)


typedef long long ll;
typedef unsigned long long ull;
const int INF = 0x3f3f3f3f;
const ll INFLL = 0x3f3f3f3f3f3f3f3f;
const int MAXN = 5e5 + 5;
const int MAXM = MAXN;
const int mod = 1e9 + 7;

struct RE
{
    int l,r;
    void in()
    {
        scanf("%d %d",&l,&r);
    }
    bool operator < (const RE &a) const
    {
        return r < a.r;
    }
}p[MAXN];

int main()
{
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    int n;
    cin>>n;
    REP(i,n) p[i].in();
    sort(p,p + n);
    int num = 1,pos = p[0].r;
    REP_1(i,n - 1)
    {
        if(p[i].l > pos)
        {
            num++;
            pos = p[i].r;
        }
    }
    cout<<num<<endl;
    return 0;
}

C. Subsequences

For the given sequence with n different elements find the number of increasing subsequences with k + 1 elements. It is guaranteed that the answer is not greater than 8·10¹⁸.
Input
First line contain two integer values n and k (1 ≤ n ≤ 10⁵, 0 ≤ k ≤ 10) — the length of sequence and the number of elements in increasing subsequences.
Next n lines contains one integer a_i (1 ≤ a_i ≤ n) each — elements of sequence. All values a_i are different.
Output
Print one integer — the answer to the problem.
Sample test(s)
Input
5 2
1
2
3
5
4
Output
7

题解:
       很好的一道用数据结构来优化DP的一道题,首先很容易想到DP的状态转移方程,以dp[i][j]表示前i个数以a[i]结尾的递增序列长度为j的方案数,则状态的转移方程就可以写为
 dp[i][j] += dp[x][j - 1],其中满足a[x]<a[i];这部分的处理如果不用数据结构优化,时间复杂度是n^2的,因为注意到我们是统计所有小于a[i]的DP值的和,所以可以利用树状数组或者
线段树优化;

AC代码:
/* ***********************************************
Author        :xdlove
Created Time  :2015年11月12日 星期四 10时49分32秒
File Name     :xdlove/codeforces/Testing_Round_#12/C/c.cpp
************************************************ */

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <memory.h>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <time.h>

using namespace std;


#define REP_ab(i,a,b) for(int i = a; i <= b; i++)
#define REP(i, n) for(int i = 0; i < n; i++)
#define REP_1(i,n) for(int i = 1; i <= n; i++)
#define DEP(i,n) for(int i = n - 1; i >= 0; i--)
#define DEP_N(i,n) for(int i = n; i >= 1; i--)
#define CPY(A,B) memcpy(A,B,sizeof(B))
#define MEM(A) memset(A,0,sizeof(A))
#define MEM_1(A) memset(A,-1,sizeof(A))
#define MEM_INF(A) memset(A,0x3f,sizeof(A))
#define MEM_INFLL(A) memset(A,0x3f3f,sizeof(A))
#define mid (((l + r) >> 1))
#define lson l, mid, u << 1
#define rson mid + 1, r, u << 1 | 1
#define ls (u << 1)
#define rs (u << 1 | 1)


typedef long long ll;
typedef unsigned long long ull;
const int INF = 0x3f3f3f3f;
const ll INFLL = 0x3f3f3f3f3f3f3f3f;
const int MAXN = 1e5 + 5;
const int MAXM = MAXN;
const int mod = 1e9 + 7;
int n,k;
ll dp[MAXN][20],val[20][MAXN];

int lowbit(int x)
{
    return x & -x;
}

void add(ll c,ll b[],int v)
{
    while(v <= n)
    {
        b[v] += c;
        v += lowbit(v);
    }
}

ll sum(ll b[],int v)
{
    ll s = 0;
    while(v > 0)
    {
        s += b[v];
        v -= lowbit(v);
    }
    return s;
}

int main()
{
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    cin>>n>>k;
    k++;
    REP_1(i,n)
    {
        int x;
        scanf("%d",&x);
        dp[i][1] = 1;
        REP_1(j,k)
        {
            if(j > i) break;
            ll tp = sum(val[j - 1],x - 1);
            dp[i][j] += tp;
            add(dp[i][j],val[j],x);
        }
    }
    ll ans = 0;
    REP_1(i,n) ans += dp[i][k];
    cout<<ans<<endl;
    return 0;
}