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  • 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
     
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    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~
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  • 原文地址:https://www.cnblogs.com/MengX/p/9681077.html
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