ZOJ 3940 Modulo Query(区间取模)

最新推荐文章于 2020-08-16 14:41:00 发布

还是太年轻

最新推荐文章于 2020-08-16 14:41:00 发布

阅读量400

点赞数

CC 4.0 BY-SA版权

分类专栏：竞赛

本文链接：https://blog.youkuaiyun.com/albertluf/article/details/80085565

竞赛专栏收录该内容

87 篇文章

订阅专栏

本文介绍了一种解决特定模运算问题的高效算法，通过构造优先队列和区间更新的方法，实现了对一系列模运算的快速求解。适用于处理大规模数据集上的模运算问题。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Modulo Query

Time Limit: 2 Seconds Memory Limit: 65536 KB

One day, Peter came across a function which looks like:

F(1, X) = X mod A₁.
F(i, X) = F(i - 1, X) mod A_i, 2 ≤ i ≤ N.

Where A is an integer array of length N, X is a non-negative integer no greater than M.

Peter wants to know the number of solutions for equation F(N, X) = Y, where Y is a given number.

Input

There are multiple test cases. The first line of input contains an integer T, indicating the number of test cases. For each test case:

The first line contains two integers N and M (2 ≤ N ≤ 10⁵, 0 ≤ M ≤ 10⁹).

The second line contains N integers: A₁, A₂, ..., A_N (1 ≤ A_i ≤ 10⁹).

The third line contains an integer Q (1 ≤ Q ≤ 10⁵) - the number of queries. Each of the following Q lines contains an integer Y_i (0 ≤ Y_i ≤ 10⁹), which means Peter wants to know the number of solutions for equation F(N, X) = Y_i.

Output

For each test cases, output an integer S = (1 ⋅ Z₁ + 2 ⋅ Z₂ + ... + Q ⋅ Z_Q) mod (10⁹ + 7), where Z_i is the answer for the i-th query.

Sample Input

Sample Output

Hint

The answer for each query is: 4, 2, 0, 0, 0.

题解：
整个区间n个数取模，因为每个数取模后至少变为原来的二分之一，因此复杂度n*logn
node right表示区间的最右端点，cnt表示此区间的个数
代码：

#include<bits/stdc++.h>
using namespace std;
#define ll long long
const int maxn=1e5+7;
const int mod=1e9+7;
int n,m,t[maxn],rr[maxn];
struct node
{
    int right,cnt;
}c[maxn];
bool operator<(const node &a,const node &b)
{
    return a.right<b.right;
}
bool cmp(const node &a,const node &b)
{
    return a.right<b.right;
}
int main()
{
    int T;scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d",&n,&m);
        priority_queue<node>P;
        node r;r.right=m,r.cnt=1;
        P.push(r);
        for(int i=0;i<n;i++)
        {
            int x;scanf("%d",&x);
            while(P.top().right>=x)
            {
                node e=P.top();P.pop();
                int cnt=e.cnt;
                while(!P.empty()&&P.top().right==e.right)
                    cnt+=P.top().cnt,P.pop();
                r.right=x-1,r.cnt=(e.right+1)/x*cnt;
                P.push(r);
                if((e.right+1)%x)
                {
                    r.right=e.right%x;r.cnt=cnt;
                    P.push(r);
                }
            }
        }
        int tol=0;
        while(!P.empty())
        {
            c[tol].cnt=P.top().cnt;
            c[tol++].right=P.top().right;
            P.pop();
        }
        ll ans=0,Q;scanf("%lld",&Q);
        sort(c,c+tol,cmp);
        memset(t,0,sizeof(t));
        t[tol-1]=c[tol-1].cnt;rr[tol-1]=c[tol-1].right;
        for(int i=tol-2;i>=0;i--)t[i]=t[i+1]+c[i].cnt,rr[i]=c[i].right;
        for(ll i=1;i<=Q;i++)
        {
            int x;scanf("%d",&x);
            int id=lower_bound(rr,rr+tol,x)-rr;
            ans=(ans+1LL*t[id]*i)%mod;
        }
        printf("%lld\n",ans);
    }
    return 0;
}