【BZOJ 5334 [Tjoi2018]数学计算】

BZOJ 5334
题意 给你n个操作 1 y 操作是把原值乘上一个 y
2 y 操作是把原值除以第y次操作的值
输出每次操作后的值
因为是乘积
所以你按照修改次数建一颗线段树
那么你 1 操作就是把第 i 次操作单点修改为 y
2 操作就是把第 y 次操作单点修改为1
然后每次输出s[1]即可

/*
    if you can't see the repay
    Why not just work step by step
    rubbish is relaxed
    to ljq
*/
#include <cstdio>
#include <cstring>
#include <iostream>
#include <queue>
#include <cmath>
#include <map>
#include <stack>
#include <set>
#include <sstream>
#include <vector>
#include <stdlib.h>
#include <algorithm>
using namespace std;

#define dbg(x) cout<<#x<<" = "<< (x)<< endl
#define dbg2(x1,x2) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<endl
#define dbg3(x1,x2,x3) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<" "<<#x3<<" = "<<x3<<endl
#define max3(a,b,c) max(a,max(b,c))
#define min3(a,b,c) min(a,min(b,c))
#define lc (rt<<1)
#define rc (rt<<11)
#define mid ((l+r)>>1)

typedef pair<int,int> pll;
typedef long long ll;
const int inf = 0x3f3f3f3f;
const int _inf = 0xc0c0c0c0;
const ll INF = 0x3f3f3f3f3f3f3f3f;
const ll _INF = 0xc0c0c0c0c0c0c0c0;

ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll ksm(ll a,ll b,ll mod){int ans=1;while(b){if(b&1) ans=(ans*a)%mod;a=(a*a)%mod;b>>=1;}return ans;}
ll inv2(ll a,ll mod){return ksm(a,mod-2,mod);}
void exgcd(ll a,ll b,ll &x,ll &y,ll &d){if(!b) {d = a;x = 1;y=0;}else{exgcd(b,a%b,y,x,d);y-=x*(a/b);}}//printf("%lld*a + %lld*b = %lld\n", x, y, d);
const int MAX_N = 100025;
int mod;
ll s[MAX_N<<2];
void up(int rt)
{
    s[rt] = s[rt<<1] * s[rt<<1|1] %mod;
}
void build(int rt,int l,int r)
{
    s[rt] = 1;
    if(l==r)
    {
        return ;
    }
    build(rt<<1,l,mid);
    build(rt<<1|1,mid+1,r);
    up(rt);
}
void update(int rt,int l,int r,int x,int v)
{
    if(l==r)
    {
        s[rt] = v%mod;
        return ;
    }
    if(x<=mid) update(rt<<1,l,mid,x,v);
    else update(rt<<1|1,mid+1,r,x,v);
    up(rt);
}

int main()
{
    //ios::sync_with_stdio(false);
    //freopen("a.txt","r",stdin);
    //freopen("b.txt","w",stdout);
    int t,n,x,y;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&n,&mod);
        build(1,1,n);
        for(int i = 1;i<=n;++i)
        {
            scanf("%d%d",&x,&y);
            if(x==1)
            {
                update(1,1,n,i,y);
                printf("%lld\n",s[1]);
            }
            else
            {
                update(1,1,n,y,1);
                printf("%lld\n",s[1]);
            }
        }
    }
    //fclose(stdin);
    //fclose(stdout);
    //cout << "time: " << (long long)clock() * 1000 / CLOCKS_PER_SEC << " ms" << endl;
    return 0;
}