Problem B. Harvest of Apples

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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chendu   |   We have carefully selected several similar problems for you:   6343  6342  6341  6340  6339 

 

莫隊，O(1)轉移爲：f(n,m+1)=f(n,m)+C(n,m+1)

f(n+1,m)=2f(n,m)-C(n,m).

#include <bits/stdc++.h>
#define maxn 100505
using namespace std;
typedef long long ll;
const ll mod =1e9+7;
struct node
{
    ll l,r,pos;
}query[maxn];
ll block;
ll fact[maxn],inv[maxn];
ll Pow(ll x,ll n){
    ll ans=1,base=x;
    while(n){
        if(n&1) ans=ans*base%mod;
        base=base*base%mod;
        n>>=1;
    }
    return ans;
}
void init(){
    fact[0]=1;
    for (int i = 1; i < maxn; ++i)
    {
        fact[i]=fact[i-1]*i%mod;
    }
    inv[maxn-1]=Pow(fact[maxn-1],mod-2);
    for (int i = maxn-2; i >= 0; --i)
    {
        inv[i]=inv[i+1]*(i+1)%mod;
    }
}
ll C(ll n, ll m)
{
    if(n==m||m==0)
        return 1;
    if(m>n) return 0;
    return ((long long)fact[n]*inv[m]%mod)*inv[n-m]%mod;
}
bool cmp(node a,node b)
{
    if((a.l/block)==(b.l/block))
    {
        return a.r<b.r;
    }
    return ((a.l/block)<(b.l/block));
}
ll ans[maxn];
int main()
{
    ll t,i;
    init();
    //cout<<C(6,4)<<endl;
    scanf("%lld",&t);
    ll maxim=0;
    for(i=1;i<=t;i++)
    {
        scanf("%lld%lld",&query[i].l,&query[i].r);
        query[i].pos=i;
        maxim=max(maxim,query[i].l);
    }
    block=sqrt(maxim);
    sort(query+1,query+1+t,cmp);
    ll le=1,ri=1;
    ll now=2;//le=n,ri=m;
    for(i=1;i<=t;i++)
    {
        while(le<query[i].l)
        {
            now=(2*now-C(le,ri)+mod)%mod;
            le++;
        }
        while(le>query[i].l)
        {   le--;
            now=((now+C(le,ri))%mod*inv[2]%mod)%mod;
        }
        while(ri<query[i].r)
        {   ri++;
            now=(now+C(le,ri))%mod;
        }
        while(ri>query[i].r)
        {
            now=(now-C(le,ri)+5*mod)%mod;
            ri--;
        }
        ans[query[i].pos]=now;
    }
    for(i=1;i<=t;i++)
    {
        printf("%lld\n",ans[i]);
    }
    return 0;
}