hdu 6333 Problem B. Harvest of Apples(莫队算法)

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Problem B. Harvest of Apples

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 3573    Accepted Submission(s): 1380

 

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

 

Recommend

chendu

题意：

求的和

思路：

因为可以O（1）的从S(n,m)转移到S(n,m+1),S(n,m-1),S(n+1,m),S(n-1,m)。所以可以考虑莫队。

由S(n,m)到S(n,m+1)有，S(n,m+1)=S(n,m)+C(n,m+1)

而由S(n,m)到S(n+1,m)，根据

S(n+1,m)=2*S(n,m)-C(n,m)

所以我们就可以用莫队算法来求解了。

代码：

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
using namespace std;
const int maxn=1e5+5;
const int mod=1e9+7;
typedef long long ll;
ll ans=1;
ll res[maxn];
ll fac[maxn]={1,1},inv[maxn]={1,1},f[maxn]={1,1};
struct node
{
    int l,r;
    int id;
}q[maxn];
int pos[maxn];
bool cmp(node a,node b)
{
    if(pos[a.l]==pos[b.l])
        return a.r<b.r;
    return pos[a.l]<pos[b.l];
}
ll cal(ll a,ll b)
{
    if(b>a)
        return 0;
    return fac[a]*inv[b]%mod*inv[a-b]%mod;
}
void init()
{
    for(int i=2;i<maxn;i++){
        fac[i]=fac[i-1]*i%mod;
        f[i]=(mod-mod/i)*f[mod%i]%mod;
        inv[i]=inv[i-1]*f[i]%mod;
    }
}
int main()
{
    int t;
    scanf("%d",&t);
    int sz=sqrt(maxn);
    init();
    for(int i=0;i<maxn;i++){
        pos[i]=i/sz;
    }
    for(int i=1;i<=t;i++){
        scanf("%d%d",&q[i].l,&q[i].r);
        q[i].id=i;
    }
    sort(q+1,q+1+t,cmp);
    int L=1,R=0;
    for(int i=1;i<=t;i++){
        while(L<q[i].l) ans=(2ll*ans%mod-cal(L++,R)+mod)%mod;
        while(L>q[i].l) ans=(ans+cal(--L,R))%mod*f[2]%mod;
        while(R<q[i].r) ans=(ans+cal(L,++R))%mod;
        while(R>q[i].r) ans=(ans-cal(L,R--)+mod)%mod;
        res[q[i].id]=ans;
    }
    for(int i=1;i<=t;i++){
        printf("%lld\n",res[i]);
    }
    return 0;
}