zoukankan      html  css  js  c++  java
  • HDU

    Problem B. Harvest of Apples

    Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
    Total Submission(s): 3752    Accepted Submission(s): 1442


    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:  6447 6446 6445 6444 6443 
    题意:给出n,m n个苹果最多取m个 求方案数
     

    题解:

    贴个官方题解,需要用到哦Lucas计算组合数
    #include<bits/stdc++.h>
    #define ll long long
    #define mod 1000000007
    using namespace std;
    const int maxn = 100000+5;
    int unit;
    ll ans,res[maxn],inv[maxn],fac[maxn],inv_fac[maxn];
    struct node
    {
        ll l,r,id;
        bool friend operator < (const node u,const node v)
        {
            if(u.l/unit==v.l/unit)
                return u.r<v.r;
            else return u.l<v.l;
        }
    } p[maxn];
    void init()
    {
        inv[0]=inv[1]=inv_fac[0]=fac[0]=1;
        for(int i=2; i<maxn; i++) inv[i]=inv[mod%i]*(mod-mod/i)%mod;
        for(int i=1; i<maxn; i++) fac[i]=fac[i-1]*i%mod;
        for(int i=1; i<maxn; i++) inv_fac[i]=inv_fac[i-1]*inv[i]%mod;
    }
    ll C(ll n,ll m)
    {
        return fac[n]*inv_fac[m]%mod*inv_fac[n-m]%mod;
    }
    int main()
    {
        int n,t;
        init();
        scanf("%d",&t);
        unit = sqrt(t)+0.5;
        for(int i=1; i<=t; i++)
        {
            scanf("%lld %lld",&p[i].r,&p[i].l);
            p[i].id=i*1ll;
        }
        sort(p+1,p+1+t);
        ll l=0,r=1;
        ans=1;
        for(int i=1; i<=t; i++)
        {
            while(r<p[i].r)
            {
                ans=(ans*2-C(r,l)+mod)%mod;
                r++;
            }
            while(r>p[i].r)
            {
                ans=1;
                l=0;
                r=p[i].r;
            }
            while(l<p[i].l)
            {
                ans=(ans+C(r,l+1)+mod)%mod;
                l++;
            }
            while(l>p[i].l)
            {
                ans=(ans-C(r,l)+mod)%mod;
                l--;
            }
            res[p[i].id]=ans;
    
        }
        for(int i=1; i<=t; i++)
            printf("%lld
    ",res[i]);
    }
    View Code

    PS:摸鱼怪的博客分享,欢迎感谢各路大牛的指点~
  • 相关阅读:
    腾讯2016春招安全岗笔试题解析
    AlgorithmVisualizer
    agentzh --春哥--调试专家
    大话Java性能优化 BOOK
    《Linux内核分析》-----张超
    ROS中Mangle解析
    shell中trap捕获信号
    虚拟化技术性能总结:Zones, KVM, Xen
    Dtrace on Mac OS X
    linux内核学习-建议路线
  • 原文地址:https://www.cnblogs.com/MengX/p/9681077.html
Copyright © 2011-2022 走看看