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=+;
 LL len,p[maxn]={,},inv[maxn]={,},p_inv[maxn]={,},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 ;
     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=;i<=;++i){
     p[i]=p[i-]*i%mod;
     inv[i]=(mod-mod/i)*inv[mod%i]%mod;
     p_inv[i]=p_inv[i-]*inv[i]%mod;
     }
     scanf("%d",&t);
     for(i=;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+,q++t);
     int L=,R=;
     LL res=;
     for(i=;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*+mod-cal(R++,L))%mod;
         }
         while(R>q[i].n){
             res=(res+cal(--R,L))%mod*inv[]%mod;
         }
         ans[q[i].id]=res;
     }
     for(i=;i<=t;++i)printf("%lld\n",ans[i]);
     return ;
 }