zoukankan      html  css  js  c++  java
  • Problem B. Harvest of Apples

    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 (1T105) denoting the number of test cases.
    Each test case consists of one line with two integers n,m (1mn105).
     
    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
    chendu   |   We have carefully selected several similar problems for you:  6343 6342 6341 6340 6339 
     
    莫队,O(1)转移为:f(n,m+1)=f(n,m)+C(n,m+1)
    f(n+1,m)=2f(n,m)-C(n,m).
    #include <bits/stdc++.h>
    #define maxn 100505
    using namespace std;
    typedef long long ll;
    const ll mod =1e9+7;
    struct node
    {
        ll l,r,pos;
    }query[maxn];
    ll block;
    ll fact[maxn],inv[maxn];
    ll Pow(ll x,ll n){
        ll ans=1,base=x;
        while(n){
            if(n&1) ans=ans*base%mod;
            base=base*base%mod;
            n>>=1;
        }
        return ans;
    }
    void init(){
        fact[0]=1;
        for (int i = 1; i < maxn; ++i)
        {
            fact[i]=fact[i-1]*i%mod;
        }
        inv[maxn-1]=Pow(fact[maxn-1],mod-2);
        for (int i = maxn-2; i >= 0; --i)
        {
            inv[i]=inv[i+1]*(i+1)%mod;
        }
    }
    ll C(ll n, ll m)
    {
        if(n==m||m==0)
            return 1;
        if(m>n) return 0;
        return ((long long)fact[n]*inv[m]%mod)*inv[n-m]%mod;
    }
    bool cmp(node a,node b)
    {
        if((a.l/block)==(b.l/block))
        {
            return a.r<b.r;
        }
        return ((a.l/block)<(b.l/block));
    }
    ll ans[maxn];
    int main()
    {
        ll t,i;
        init();
        //cout<<C(6,4)<<endl;
        scanf("%lld",&t);
        ll maxim=0;
        for(i=1;i<=t;i++)
        {
            scanf("%lld%lld",&query[i].l,&query[i].r);
            query[i].pos=i;
            maxim=max(maxim,query[i].l);
        }
        block=sqrt(maxim);
        sort(query+1,query+1+t,cmp);
        ll le=1,ri=1;
        ll now=2;//le=n,ri=m;
        for(i=1;i<=t;i++)
        {
            while(le<query[i].l)
            {
                now=(2*now-C(le,ri)+mod)%mod;
                le++;
            }
            while(le>query[i].l)
            {   le--;
                now=((now+C(le,ri))%mod*inv[2]%mod)%mod;
            }
            while(ri<query[i].r)
            {   ri++;
                now=(now+C(le,ri))%mod;
            }
            while(ri>query[i].r)
            {
                now=(now-C(le,ri)+5*mod)%mod;
                ri--;
            }
            ans[query[i].pos]=now;
        }
        for(i=1;i<=t;i++)
        {
            printf("%lld
    ",ans[i]);
        }
        return 0;
    }
    

      

     
  • 相关阅读:
    SQL Server2016 AlwaysOn无域高可用
    Windows Server 2016 无域故障转移群集
    SQL Server高可用实现方案
    oracle11g RMAN catalog的基本使用
    Oracle_Windows server ORA-01031: insufficient privileges
    MySQL MGR 单主模式下master角色切换规则
    SQL Server AlwaysOn原理简介
    DB2创建视图并授权给其他用户
    Oracle数据库用户的密码过期问题处理
    访问GitLab的PostgreSQL数据库
  • 原文地址:https://www.cnblogs.com/zyf3855923/p/9406286.html
Copyright © 2011-2022 走看看