2018杭电多校第四场 1002 Harvest of Apples 莫队大法好~

最新推荐文章于 2018-10-08 00:54:34 发布

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本文介绍了一种计算从树上最多摘取m个苹果的方法，使用了莫队算法进行优化，实现了快速的状态转移，并提供了完整的AC代码实现。

Problem B. Harvest of Apples

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 1169 Accepted Submission(s): 443

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≤T≤105) denoting the number of test cases.
Each test case consists of one line with two integers n,m (1≤m≤n≤105).

Output

For each test case, print an integer representing the number of ways modulo 109+7.

Sample Input

5 2

1000 500

Sample Output

924129523

莫队大法好~

先找到状态转移

用s(n,m)表示C(n,0)+C(n,1)+...+C(n,m) 那么

s(n,m)=2*s(n-1,m)-c(n-1,m);

s(n,m)=s(n,m-1)+c(n,m);

也就是说知道一个s(n,m)就能O（1）求出s(n+1,m),s(n-1,m),s(n,m+1),s(n,m-1)

然后瞎文明搞

注意逆元要先预处理出来

然后，就过了(当然是赛后，赛时还不会莫队，弱小可怜无助还特别能wa)

AC代码：

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<vector>
#include<map>
#include<cmath>
using namespace std;
const int maxn = 1e5+10;
const long long ha = 1e9+7;
struct node{
    int pos, n, m;
    bool friend operator < (node na, node nb){
        return (na.m<nb.m||(na.m==nb.m&&na.n<nb.n));
    }
} ex;
long long ans[maxn], c[maxn], inv[maxn];
int T, block;
vector<node> md[1000];
long long quickm(long long a, long long b){
    long long ans=1;
    while(b){
        if (b&1){
            ans=ans*a%ha;
        }
        b>>=1;
        a*=a;   a%=ha;
    }
    return ans;
}
void init()
{
    c[1]=1;
    for(int i=2;i<=maxn;i++)
        c[i]=c[i-1]*i%ha;
    inv[100000]=quickm(c[100000],ha-2);
    for (int i=100000-1;i>=1;--i){
        inv[i]=inv[i+1]*((long long)(i+1))%ha;
    }
    return ;
}
long long C(long long n, long long m){
    if (n==m)
        return 1;
    if (m==0)
        return 1;
    return (((c[n]*inv[m])%ha)*inv[n-m])%ha;
}
void read(){
    scanf("%d",&T);
    block=sqrt(maxn);
    int j,k;
    for (int i=0;i<T;++i){
        scanf("%d%d",&ex.n,&ex.m);
        ex.pos=i;
        j=ex.n/block;
        md[j].push_back(ex);
    }
    return ;
}
void solve(){
    int nn, nm;
    long long nans;
    for (int i=0;i<=block;++i){
        if (md[i].size()==0){
            continue;
        }
        sort(md[i].begin(),md[i].end());
        nans=1; nn=i*block;  nm=0;
        for (int j=0;j<md[i].size();++j){
            while( nn < md[i][j].n ){
                nans= (2*nans+ha-C(nn,nm))%ha;
                nn++;
            }
            while (nn > md[i][j].n){
                nans = (nans+C(nn-1,nm))*inv[2]%ha;
                nn--;
            }
            while (nm < md[i][j].m){
                nm++;
                nans = ( nans + C(nn,nm) )% ha;
            }
            ans[md[i][j].pos]=nans;
        }
    }
    for (int i=0;i<T;++i){
        printf("%lld\n",ans[i]);
    }
    return ;
}
int main(){
    //freopen("b.in","r",stdin);
    //freopen("out.txt","w",stdout);
    init();
    read();
    solve();
    return 0;
}