Problem B. Harvest of Apples HDU - 6333（莫队）

时间：2021-10-03 20:17:00
Problem Description
There are n apples on a tree, numbered from 1 to n.
Count the number of ways to pick at most m apples.
 
Input
The first line of the input contains an integer T (1≤T≤105) denoting the number of test cases.
Each test case consists of one line with two integers n,m (1≤m≤n≤105).
 
Output
For each test case, print an integer representing the number of ways modulo 109+7.
 
Sample Input
2
5 2
1000 500
 
Sample Output
16
924129523
 
Source
 

解析：

　　 Problem B. Harvest of Apples HDU - 6333（莫队）

　　

#include <iostream>

#include <cstdio>

#include <sstream>

#include <cstring>

#include <map>

#include <cctype>

#include <set>

#include <vector>

#include <stack>

#include <queue>

#include <algorithm>

#include <cmath>

#include <bitset>

#define rap(i, a, n) for(int i=a; i<=n; i++)

#define rep(i, a, n) for(int i=a; i<n; i++)

#define lap(i, a, n) for(int i=n; i>=a; i--)

#define lep(i, a, n) for(int i=n; i>a; i--)

#define rd(a) scanf("%d", &a)

#define rlld(a) scanf("%lld", &a)

#define rc(a) scanf("%c", &a)

#define rs(a) scanf("%s", a)

#define pd(a) printf("%d\n", a);

#define plld(a) printf("%lld\n", a);

#define pc(a) printf("%c\n", a);

#define ps(a) printf("%s\n", a);

#define LL long long

#define ULL unsigned long long

#define Pair pair<int, int>

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

#define _  ios_base::sync_with_stdio(0),cin.tie(0)

//freopen("1.txt", "r", stdin);

using namespace std;

const int maxn = 1e5 + , INF = 0x7fffffff, LL_INF = 0x7fffffffffffffff;

const int MOD = 1e9+;

LL n, m, ans;

LL up[maxn], down[maxn], pos[maxn], inc[maxn], inv[maxn];

struct node

{

    LL l, r;

    int id;

}Node[maxn];

bool cmp(node a, node b)

{

    return pos[a.l] == pos[b.l] ? (a.r < b.r) : (a.l < b.l);

}

LL qp(LL a, LL b)

{

    LL res = ;

    while(b)

    {

        if(b & ) res  = res * a % MOD;

        a = a * a % MOD;

        b >>= ;

    }

    return res;

}

void init()

{

    up[] = ;

    down[] = ;

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

    {

        up[i] = up[i-] * i % MOD;

        down[i] = qp(up[i], MOD - ) % MOD;

    }

}

LL C(LL n, LL m)

{

    if(n < m) return ;

    return up[n] * down[n-m] % MOD * down[m] % MOD;

}

int main()

{

    init();

    int block = sqrt();

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

        pos[i] = (i-)/block + ;

    int T;

    rd(T);

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

    {

        rlld(Node[i].r), rlld(Node[i].l);

        Node[i].id = i;

    }

    sort(Node + , Node + T + , cmp);

    ans = ;

    int tmp = qp(, MOD - );

    for(int i=, l=, r=; i<=T; i++)

    {

        for(; r < Node[i].r; r++)

            ans = ( * ans - C(r, l) + MOD) % MOD;

        for(; r > Node[i].r; r--)

            ans = (ans + C(r-, l)) * tmp % MOD;

        for(; l < Node[i].l; l++)

            ans = (ans + C(r, l+)) % MOD;

        for(; l > Node[i].l; l--)

            ans = (ans - C(r, l) + MOD) % MOD;

        if(Node[i].l == Node[i].r)

        {

            inc[Node[i].id] = ;

        }

        inc[Node[i].id] = ans;

    }

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

        printf("%lld\n", inc[i]);

    return ;

}

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