hdu-6333-莫队

Problem B. Harvest of Apples

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 2970    Accepted Submission(s): 1153

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

 

　　经过观察可以发现，设S(m,n)=C(0,n)+C(1,n)+.....+C(m,n)的话，S(m,n+1)=2*S(m,n)-C(m,n) , S(m,n-1)=(S(m,n)+C(m,n-1))/2。

　　S(m-1,n)=S(m,n)-C(m,n) ,S(m+1,n)=S(m,n)+C(m+1,n),也就是说我们能在O(1)求出来这四个式子，这样就可以用莫队处理了。

分块的时候我错把 q[i].m/M写成了 i/M导致一直T ，是对mi所在的块作为关键字而不是输入的次序。

　　

#include<bits/stdc++.h>
using namespace std;
#define LL long long 
#define eps 1e-6
#define mod 1000000007
//LL mod=1e9+7;
const int maxn=100000+5;
LL len,p[maxn]={1,1},inv[maxn]={0,1},p_inv[maxn]={1,1},ans[maxn];
int t;
struct Query{
    LL n,m,blo;
    int id;
    bool operator<(const Query&C)const{
        if(blo==C.blo) return n<C.n;
        return blo<C.blo;    
    }
}q[maxn];
LL cal(LL a,LL b)  
{
    if(b>a)
        return 0;
    return p[a]*p_inv[b]%mod*p_inv[a-b]%mod;
}
int main(){
    int n,m,i,j,k;
    len=sqrt(maxn);
    for(i=2;i<=100000;++i){
    p[i]=p[i-1]*i%mod;
    inv[i]=(mod-mod/i)*inv[mod%i]%mod;
    p_inv[i]=p_inv[i-1]*inv[i]%mod;
    }
    scanf("%d",&t);
    for(i=1;i<=t;++i){
        scanf("%lld %lld",&q[i].n,&q[i].m);
        q[i].id=i;
        q[i].blo=q[i].m/len;
    }
    sort(q+1,q+1+t);
    int L=0,R=1;
    LL res=1;
    for(i=1;i<=t;++i){
        while(L<q[i].m){
            res=(res+cal(R,++L))%mod;
        }
        while(L>q[i].m){
            res=(res+mod-cal(R,L--))%mod;
        }
        while(R<q[i].n){
            res=(res*2+mod-cal(R++,L))%mod;
        }
        while(R>q[i].n){
            res=(res+cal(--R,L))%mod*inv[2]%mod;
        }
        ans[q[i].id]=res;
    }
    for(i=1;i<=t;++i)printf("%lld
",ans[i]);
    return 0;
}