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

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

2018 Multi-University Training Contest 4

 

 解析：

　　 

　　

#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 + 100, INF = 0x7fffffff, LL_INF = 0x7fffffffffffffff;
const int MOD = 1e9+7;
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 = 1;
    while(b)
    {
        if(b & 1) res  = res * a % MOD;
        a = a * a % MOD;
        b >>= 1;
    }
    return res;
}

void init()
{
    up[0] = 1;
    down[0] = 1;
    for(int i=1; i<maxn; i++)
    {
        up[i] = up[i-1] * i % MOD;
        down[i] = qp(up[i], MOD - 2) % MOD;
    }
}

LL C(LL n, LL m)
{
    if(n < m) return 0;
    return up[n] * down[n-m] % MOD * down[m] % MOD;
}

int main()
{
    init();
    int block = sqrt(100000);
    for(int i=1; i<=100000; i++)
        pos[i] = (i-1)/block + 1;
    int T;
    rd(T);
    for(int i=1; i<=T; i++)
    {
        rlld(Node[i].r), rlld(Node[i].l);
        Node[i].id = i;
    }
    sort(Node + 1, Node + T + 1, cmp);
    ans = 2;
    int tmp = qp(2, MOD - 2);
    for(int i=1, l=1, r=1; i<=T; i++)
    {
        for(; r < Node[i].r; r++)
            ans = (2 * ans - C(r, l) + MOD) % MOD;
        for(; r > Node[i].r; r--)
            ans = (ans + C(r-1, l)) * tmp % MOD;
        for(; l < Node[i].l; l++)
            ans = (ans + C(r, l+1)) % 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] = 1;
        }

        inc[Node[i].id] = ans;
    }
    for(int i=1; i<=T; i++)
        printf("%lld\n", inc[i]);


    return 0;
}