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  • HDU 6333 莫队+组合数

    Problem B. Harvest of Apples

    Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
    Total Submission(s): 2397    Accepted Submission(s): 934


    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

    解析  不难发现S(n,m)也满足左上角加右上角(杨辉三角)  所以根据公式可以O(1)得到S(n-1,m),S(n+1,m),S(n,m-1),S(n,m+1) 可以看做区间的转移 从而套用莫队实现求解

    AC代码  

    #include <bits/stdc++.h>
    #define pb push_back
    #define mp make_pair
    #define fi first
    #define se second
    #define all(a) (a).begin(), (a).end()
    #define fillchar(a, x) memset(a, x, sizeof(a))
    #define huan prllf("
    ");
    #define debug(a,b) cout<<a<<" "<<b<<" ";
    using namespace std;
    typedef long long ll;
    const ll maxn=1e5+10,inf=0x3f3f3f3f;
    const ll mod=1e9+7;
    ll gcd(ll a,ll b){ return b?gcd(b,a%b):a;}
    ll fac[maxn],inv[maxn],ans[maxn];
    ll chunk;
    struct node
    {
        ll l,r,id,chunk;
    }q[maxn];
    bool cmp(node a,node b)
    {
        if(a.chunk!=b.chunk)
            return a.l<b.l;
        return a.r<b.r;
    }
    void init()
    {
        fac[0]=fac[1]=1;
        inv[0]=inv[1]=1;
        for(ll i=2;i<maxn;i++)
        {
            fac[i]=fac[i-1]*i%mod;
            inv[i]=1ll*(mod-mod/i)*inv[mod%i]%mod;
        }
        for(ll i=2;i<maxn;i++)         //不可以写成一个for inv还会用到
            inv[i]=inv[i-1]*inv[i]%mod;  //可以再开一个数组 写成一个for
    }
    ll C(ll x,ll y)
    {
        if(y>x) return 0;
        return fac[x]*inv[y]%mod*inv[x-y]%mod;
    }
    int main()
    {
        init();//预处理组合数逆元 从而O(1)获得组合数 实现转移
        ll t;
        chunk=sqrt(maxn);
        scanf("%lld",&t);
        for(ll i=1;i<=t;i++)
        {
            ll n,m;
            scanf("%lld%lld",&n,&m);
            q[i]=node{n,m,i,n/chunk+1};
        }
        sort(q+1,q+1+t,cmp);
        ll l=1,r=0,res=1;
        for(ll i=1;i<=t;i++)
        {
            while(l<q[i].l)
            {
                res=(res*2%mod-C(l,r)+mod)%mod;
                l++;
            }
            while(l>q[i].l)
            {
                l--;
                res=(res+C(l,r))%mod*inv[2]%mod;
            }
            while(r>q[i].r)
            {
                res=(res-C(l,r)+mod)%mod;
                r--;
            }
            while(r<q[i].r)
            {
                r++;
                res=(res+C(l,r))%mod;
            }
            ans[q[i].id]=res;
        }
        for(ll i=1;i<=t;i++)
            printf("%lld
    ",ans[i]);
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/stranger-/p/9414187.html
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