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
Recommend
1 #include <iostream> 2 #include <cstring> 3 #include <cstdio> 4 #include <vector> 5 #include <cmath> 6 #include <algorithm> 7 using namespace std; 8 #define N 100005//一开始是10005,超时 9 #define mod 1000000007 10 #define gep(i,a,b) for(int i=a;i<=b;i++) 11 #define mem(a,b) memset(a,b,sizeof(a)) 12 #define ll long long 13 ll fac[N]={1,1},inv[N]={1,1},f[N]={1,1}; 14 ll c(ll a,ll b) 15 { 16 if(b>a) return 0; 17 return fac[a]*inv[b]%mod*inv[a-b]%mod; 18 } 19 void init() 20 { 21 gep(i,2,N-1){ 22 fac[i]=fac[i-1]*i%mod; 23 f[i]=(mod-mod/i)*f[mod%i]%mod; 24 inv[i]=inv[i-1]*f[i]%mod; 25 } 26 } 27 struct Ma{ 28 ll n,m; 29 int id; 30 bool operator <(const Ma&a)const{ 31 return n<a.n;//排序 32 } 33 }ma[N]; 34 vector<Ma>ve[N]; 35 ll ans[N]; 36 int t; 37 int main() 38 { 39 init(); 40 scanf("%d",&t); 41 ll mx=sqrt(100000); 42 gep(i,1,t){ 43 scanf("%lld%lld",&ma[i].n,&ma[i].m); 44 ll x=ma[i].m/mx; 45 ma[i].id=i; 46 ve[x].push_back(ma[i]);//分块 47 } 48 gep(i,0,mx){ 49 if(!ve[i].size()) continue; 50 sort(ve[i].begin(),ve[i].end());//排序 51 int k=ve[i].size(); 52 ll val=0,ik=-1,mn=ve[i][0].n;//一次处理每组数据 53 /* 54 s(n,m):c(n,0)+c(n,1)+……c(n,m) 55 s(n,m)=2*s(n-1,m)-c(n-1,m) 56 */ 57 gep(j,0,k-1){ 58 while(mn<ve[i][j].n) val=(2ll*val+mod-c(mn++,ik))%mod;//+mod 59 while(ik<ve[i][j].m) val=(val+c(mn,++ik))%mod;//,++ik,不是ik++,可以不加mod 60 while(ik>ve[i][j].m) val=(val+mod-c(mn,ik--))%mod;//+mod 61 ans[ve[i][j].id]=val; 62 } 63 } 64 gep(i,1,t){ 65 printf("%lld ",ans[i]); 66 } 67 return 0; 68 }