Problem B. Harvest of Apples(莫队+数学）

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

 

SOLUTION:

神仙题目

 

设s （n,m) 为所要求的值

发现不同的 s 之间可以互相的转移

应该可以联想到莫队离线

CODE:

#include <iostream>
#include <stdio.h>
#include <math.h>
#include <stdlib.h>
#include <algorithm>
#include <queue>
#include <vector>
#include <map>
using namespace std;
#define ll long long
#define mod 1000000007

const int maxn =100000+5;

ll ans[maxn],fac[maxn]={1,1},inv[maxn]={1,1},f[maxn]={1,1},len;

int T;

ll cal(ll a,ll b)  //求组合数
{
    if(b>a)
        return 0;
    return fac[a]*inv[b]%mod*inv[a-b]%mod;
}

void init()
{
    for(int i=2; i<=maxn-1; i++)
    {
        fac[i]=fac[i-1]*i%mod;//阶乘
        f[i]=(mod-mod/i)*f[mod%i]%mod;//逆元
        inv[i]=inv[i-1]*f[i]%mod;//逆元的阶乘
    }
}
struct Query
{
    ll L,R,block;
    int ID;
    Query(){};
    Query(ll l, ll r, int ID) :L(l), R(r), ID(ID){
        block=l/len;
    }
   bool operator<(const Query rhs)const{
        if(block ==rhs.block)
          {
             if(block&1) return R>rhs.R;
             else return R<rhs.R;
          }
        return block <rhs.block;
   }
}queries[maxn];


int main()
{
    len=sqrt(maxn);///块长度
    init();

    scanf("%d",&T);
    for(int i=1; i<=T; i++)
    {
        ll l,r;
        scanf("%lld %lld",&l,&r);
         queries[i]=Query(l,r,i);
    }
    sort(queries+1,queries+T+1);

    int l=1 ,r=1;
    ll val=2;

    for(int i=1; i<=T; i++)
    {


      while(l<queries[i].L)
      {

        val=(2ll*val%mod+mod-cal(l,r))%mod;
        l++;

      }

       while(l>queries[i].L)
      {
          --l;
          val=(val+cal(l,r))%mod *f[2]%mod;

      }

    while(r<queries[i].R)
      {
          val=(val+cal(l,++r))%mod;

      }

      while(r>queries[i].R)
      {
          val=(val-cal(l,r--))%mod;
          val+=mod;
          val%=mod;

      }

        ans[queries[i].ID]=val;
    }

    for(int i=1; i<=T; i++)
        printf("%lld
",ans[i]);

    return 0;
}