P4588 [TJOI2018] 数学计算

合集 - 数据结构(28)

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27.P4588 [TJOI2018] 数学计算2024-07-20

28.P3522 [POI2011] TEM-Temperature2024-07-21

收起

原题链接

题解

由于模拟会爆 longlong，所以用线段树维护每次操作的值，初始每次操作的值均为1

操作一令对应节点变为 m

操作二令对应节点变为 1

返回整棵树的值(相乘)

code

#include<bits/stdc++.h>
#define ll long long
using namespace std;

ll tree[400005];
ll q,mod;

void update(int node,int l,int r,int id,int v)
{
    if(l>id||r<id) return;

    if(l==id&&r==id)
    {
        tree[node]=v%mod;
        return;
    }

    int mid=(l+r)/2;
    update(2*node,l,mid,id,v);
    update(2*node+1,mid+1,r,id,v);

    tree[node]=tree[node*2]*tree[node*2+1]%mod;
}


void solve()
{
    cin>>q>>mod;

    fill(tree,tree+4*q+1,1);

    ll x=1;
    for(int i=1;i<=q;i++)
    {
        int op;
        cin>>op;
        if(op==1)
        {
            int m;
            cin>>m;

            update(1,1,q,i,m);
            cout<<tree[1]<<'\n';
        }
        else
        {
            int pos;
            cin>>pos;
            update(1,1,q,pos,1);
            cout<<tree[1]<<'\n';
        }
    }
}
int main()
{
    ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
    int t=1;
    cin>>t;
    while(t--) solve();
    return 0;
}