HDU 6333:Harvest of Apples _Apple

五陵年少金市东，银鞍白马渡春风。这篇文章主要讲述HDU 6333:Harvest of Apples相关的知识，希望能为你提供帮助。
Problem B. Harvest of Apples Time Limit: 4000/2000 MS (java/Others) Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 347 Accepted Submission(s): 121

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 $T$ $(1 \\le T \\le 10 ^ 5)$ denoting the number of test cases.
Each test case consists of one line with two integers $n, m$ $(1 \\le m \\le n \\le 10 ^ 5)$. 【HDU 6333:Harvest of Apples】
Output For each test case, print an integer representing the number of ways modulo $10 ^ 9 + 7$.
Sample Input2 5 2 1000 500
Sample Output16 924129523
Source 2018 Multi-University Training Contest 4
Recommend chendu 分析：这题显然暴力O(n^2)是要TLE的，比赛的时候考虑简单数学优化一下变成2.5e9，显然肯定也是要TLE的。后来考虑深度数学优化得出：

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其中F1表示超几何函数。。。然而找了半个多小时没有找到C++的模板。。就很绝望补题的时候发现可以用莫队和分块去搞，学了半天嘤嘤嘤，太菜了。

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#include < iostream> #include < string> #include < cstdio> #include < cmath> #include < cstring> #include < algorithm> #include < vector> #include < queue> #include < deque> #include < stack> #include < map> #define LL long long #define elif else if #define range(i,a,b) for(auto i=a; i< =b; ++i) #define rerange(i,a,b) for(auto i=a; i> =b; --i) #define itrange(i,a,b) for(auto i=a; i!=b; ++i) #define fill(arr,tmp) memset(arr,tmp,sizeof(arr)) using namespace std; int t,n,m,block[int(2e5)],interval,id=1; struct query{ int n,k,i; }Q[int(1e5+5)]; vector< query> ll[int(1e5+5)]; bool cmp(query a,query b){ return a.n< b.n; } const int mod=int(1e9+7); LL fac[int(1e5+5)],inv[int(1e5+5)],ans[int(1e5+5)]; LL q_pow(LL a,LL b){ if(b< 0)return 0; LL ret=1; a%=mod; while(b){ ret=b& 1?ret*a%mod:ret; b> > =1; a=a*a%mod; } return ret; } LL C(LL n,LL m){ if(n< m)return 0; return fac[n]*inv[m]%mod*inv[n-m]%mod; } void init(){ fac[0]=1; range(i,1,int(1e5))fac[i]=fac[i-1]*i%mod; inv[int(1e5)]=q_pow(fac[int(1e5)],mod-2); rerange(i,int(1e5-1),0)inv[i]=inv[i+1]*(i+1)%mod; interval=int(sqrt(1e5)); for(int i=1; i< =int(1e5); i+=interval,++id) for(int j=i; j< i+interval and j< =int(1e5); ++j)block[j]=id; --id; scanf("%d",& t); range(i,1,t){ scanf("%d%d",& (Q+i)-> n,& (Q+i)-> k); (Q+i)-> i=i; ll[block[Q[i].k]].push_back(Q[i]); } } void solve(){ range(i,1,id) if(ll[i].size()){ sort(ll[i].begin(),ll[i].end(),cmp); LL val=0,in=ll[i][0].n,ik=-1; range(j,0,ll[i].size()-1){ while(in< ll[i][j].n)val=((val< < 1)+mod-C(in++,ik))%mod; while(ik< ll[i][j].k)val=(val+C(in,++ik))%mod; while(ik> ll[i][j].k)val=(val+mod-C(in,ik--))%mod; ans[ll[i][j].i]=val; } } range(i,1,t)printf("%lld\\n",ans[i]); } int main() { init(); solve(); return 0; }

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