Problem B. Harvest of Apples

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 2397    Accepted Submission(s): 934

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  不难发现S(n,m)也满足左上角加右上角(杨辉三角)  所以根据公式可以O(1)得到S(n-1,m),S(n+1,m),S(n,m-1),S(n,m+1) 可以看做区间的转移 从而套用莫队实现求解

AC代码

#include <bits/stdc++.h>
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define all(a) (a).begin(), (a).end()
#define fillchar(a, x) memset(a, x, sizeof(a))
#define huan prllf("\n");
#define debug(a,b) cout<<a<<" "<<b<<" ";
using namespace std;
typedef long long ll;
const ll maxn=1e5+10,inf=0x3f3f3f3f;
const ll mod=1e9+7;
ll gcd(ll a,ll b){ return b?gcd(b,a%b):a;}
ll fac[maxn],inv[maxn],ans[maxn];
ll chunk;
struct node
{
    ll l,r,id,chunk;
}q[maxn];
bool cmp(node a,node b)
{
    if(a.chunk!=b.chunk)
        return a.l<b.l;
    return a.r<b.r;
}
void init()
{
    fac[0]=fac[1]=1;
    inv[0]=inv[1]=1;
    for(ll i=2;i<maxn;i++)
    {
        fac[i]=fac[i-1]*i%mod;
        inv[i]=1ll*(mod-mod/i)*inv[mod%i]%mod;
    }
    for(ll i=2;i<maxn;i++)         //不可以写成一个for inv还会用到
        inv[i]=inv[i-1]*inv[i]%mod;  //可以再开一个数组 写成一个for
}
ll C(ll x,ll y)
{
    if(y>x) return 0;
    return fac[x]*inv[y]%mod*inv[x-y]%mod;
}
int main()
{
    init();//预处理组合数逆元 从而O(1)获得组合数 实现转移
    ll t;
    chunk=sqrt(maxn);
    scanf("%lld",&t);
    for(ll i=1;i<=t;i++)
    {
        ll n,m;
        scanf("%lld%lld",&n,&m);
        q[i]=node{n,m,i,n/chunk+1};
    }
    sort(q+1,q+1+t,cmp);
    ll l=1,r=0,res=1;
    for(ll i=1;i<=t;i++)
    {
        while(l<q[i].l)
        {
            res=(res*2%mod-C(l,r)+mod)%mod;
            l++;
        }
        while(l>q[i].l)
        {
            l--;
            res=(res+C(l,r))%mod*inv[2]%mod;
        }
        while(r>q[i].r)
        {
            res=(res-C(l,r)+mod)%mod;
            r--;
        }
        while(r<q[i].r)
        {
            r++;
            res=(res+C(l,r))%mod;
        }
        ans[q[i].id]=res;
    }
    for(ll i=1;i<=t;i++)
        printf("%lld\n",ans[i]);
    return 0;
}

原文地址：https://www.cnblogs.com/stranger-/p/9414187.html