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.
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).
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
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 ",ans[i]); 57 return 0; 58 }