Harvest of Apples

问题 B: Harvest of Apples

时间限制: 1 Sec  内存限制: 128 MB
提交: 18  解决: 11
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题目描述

There are n apples on a tree, numbered from 1 to n.
Count the number of ways to pick at most m apples.

输入

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).

输出

For each test case, print an integer representing the number of ways modulo 109+7.

样例输入

2
5 2
1000 500

样例输出

16
924129523

莫队（分块），组合数学。

AC代码：

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn=1e5+10;
const int mod=1e9+7;
ll fac[maxn],inv[maxn],ans[maxn];
ll rev2,res;
int pos[maxn];
ll qpow(ll b,int n)
{
    ll res=1;
    while(n)
    {
        if(n&1) res=res*b%mod;
        b=b*b%mod;
        n>>=1;
    }
    return res;
}
ll Comb(int n,int k)
{
    return fac[n]*inv[k]%mod*inv[n-k]%mod;
}
void pre()
{
    rev2=qpow(2,mod-2);
    fac[0]=fac[1]=1;
    for(int i=2; i<maxn; ++i) fac[i]=i*fac[i-1]%mod;
    inv[maxn-1]=qpow(fac[maxn-1],mod-2);
    for(int i=maxn-2;i>=0;i--) inv[i]=inv[i+1]*(i+1)%mod;
}
struct Query
{
    int L,R,id;
    bool operator <(const Query& p) const
    {
        if(pos[L]==pos[p.L]) return R<p.R;
        return L<p.L;
    }
} Q[maxn];
inline void addN(int posL,int posR)
{
    res=(2*res%mod-Comb(posL-1,posR)+mod)%mod;
}
inline void addM(int posL,int posR)
{
    res=(res+Comb(posL,posR))%mod;
}
inline void delN(int posL,int posR)
{
    res=(res+Comb(posL-1,posR))%mod*rev2%mod;
}
inline void delM(int posL,int posR)
{
    res=(res-Comb(posL,posR)+mod)%mod;
}
int main()
{
    int T,curL,curR;
    int block=(int)sqrt(1.0*maxn);
    pre();
    scanf("%d",&T);
    for(int i=1;i<=T;++i)
    {
        scanf("%d %d",&Q[i].L,&Q[i].R);
        pos[i]=i/block;
        Q[i].id=i;
    }
    sort(Q+1,Q+T+1);
    res=2;
    curL=1,curR=1;
    for(int i=1;i<=T;++i)
    {
        while(curL<Q[i].L) addN(++curL,curR);
        while(curR<Q[i].R) addM(curL,++curR);
        while(curL>Q[i].L) delN(curL--,curR);
        while(curR>Q[i].R) delM(curL,curR--);
        ans[Q[i].id] = res;
    }
    for(int i=1;i<=T;++i) printf("%lld\n",ans[i]);
    return 0;
}

原文地址：https://www.cnblogs.com/lglh/p/9406264.html