hdu 6333

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

 

 

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#include <iostream>
#include <cstring>
#include <cstdio>
#include <vector>
#include <cmath>
#include <algorithm>
using namespace std;
#define N 100005//一开始是10005,超时
#define mod 1000000007
#define gep(i,a,b) for(int i=a;i<=b;i++)
#define mem(a,b) memset(a,b,sizeof(a))
#define ll  long long 
ll fac[N]={1,1},inv[N]={1,1},f[N]={1,1};
ll c(ll a,ll b)
{
   if(b>a)  return 0;
   return fac[a]*inv[b]%mod*inv[a-b]%mod;
}
void init()
{
    gep(i,2,N-1){
        fac[i]=fac[i-1]*i%mod;
        f[i]=(mod-mod/i)*f[mod%i]%mod;
        inv[i]=inv[i-1]*f[i]%mod;
    }
}
struct Ma{
    ll n,m;
    int id;
    bool operator <(const Ma&a)const{
        return n<a.n;//排序
    }
}ma[N];
vector<Ma>ve[N];
ll ans[N];
int t;
int main()
{
  init();  
  scanf("%d",&t);
  ll mx=sqrt(100000);
  gep(i,1,t){
     scanf("%lld%lld",&ma[i].n,&ma[i].m);
     ll x=ma[i].m/mx;
      ma[i].id=i;
     ve[x].push_back(ma[i]);//分块
  } 
  gep(i,0,mx){
      if(!ve[i].size())  continue;
      sort(ve[i].begin(),ve[i].end());//排序
      int k=ve[i].size();
      ll val=0,ik=-1,mn=ve[i][0].n;//一次处理每组数据
      /*
      s(n,m):c(n,0)+c(n,1)+……c(n,m)
      s(n,m)=2*s(n-1,m)-c(n-1,m)
      */
      gep(j,0,k-1){
          while(mn<ve[i][j].n) val=(2ll*val+mod-c(mn++,ik))%mod;//+mod
          while(ik<ve[i][j].m) val=(val+c(mn,++ik))%mod;//,++ik,不是ik++,可以不加mod
          while(ik>ve[i][j].m) val=(val+mod-c(mn,ik--))%mod;//+mod
          ans[ve[i][j].id]=val;
      }  
  }
  gep(i,1,t){
       printf("%lld\n",ans[i]);
      }      
  return 0;
}