hdu 6333

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
 

 

Recommend
chendu   |   We have carefully selected several similar problems for you:  6343 6342 6341 6340 6339 
 
 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\n",ans[i]);
66       }      
67   return 0;
68 }

 

posted on 2018-08-02 23:38  cltt  阅读(355)  评论(0)  编辑  收藏  举报

导航