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。
　　 
 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 #define LL long long 
 4 #define eps 1e-6
 5 #define mod 1000000007
 6 //LL mod=1e9+7;
 7 const int maxn=100000+5;
 8 LL len,p[maxn]={1,1},inv[maxn]={0,1},p_inv[maxn]={1,1},ans[maxn];
 9 int t;
10 struct Query{
11     LL n,m,blo;
12     int id;
13     bool operator<(const Query&C)const{
14         if(blo==C.blo) return n<C.n;
15         return blo<C.blo;    
16     }
17 }q[maxn];
18 LL cal(LL a,LL b)  
19 {
20     if(b>a)
21         return 0;
22     return p[a]*p_inv[b]%mod*p_inv[a-b]%mod;
23 }
24 int main(){
25     int n,m,i,j,k;
26     len=sqrt(maxn);
27     for(i=2;i<=100000;++i){
28     p[i]=p[i-1]*i%mod;
29     inv[i]=(mod-mod/i)*inv[mod%i]%mod;
30     p_inv[i]=p_inv[i-1]*inv[i]%mod;
31     }
32     scanf("%d",&t);
33     for(i=1;i<=t;++i){
34         scanf("%lld %lld",&q[i].n,&q[i].m);
35         q[i].id=i;
36         q[i].blo=q[i].m/len;
37     }
38     sort(q+1,q+1+t);
39     int L=0,R=1;
40     LL res=1;
41     for(i=1;i<=t;++i){
42         while(L<q[i].m){
43             res=(res+cal(R,++L))%mod;
44         }
45         while(L>q[i].m){
46             res=(res+mod-cal(R,L--))%mod;
47         }
48         while(R<q[i].n){
49             res=(res*2+mod-cal(R++,L))%mod;
50         }
51         while(R>q[i].n){
52             res=(res+cal(--R,L))%mod*inv[2]%mod;
53         }
54         ans[q[i].id]=res;
55     }
56     for(i=1;i<=t;++i)printf("%lld\n",ans[i]);
57     return 0;
58 }

 

 
posted @ 2018-08-06 18:59  *zzq  阅读(324)  评论(0)    收藏  举报