洛谷P4588 [TJOI2018]数学计算 【线段树】

题目链接

洛谷P4588

题解

用线段树维护即可

#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<map>
#define Redge(u) for (int k = h[u],to; k; k = ed[k].nxt)
#define REP(i,n) for (int i = 1; i <= (n); i++)
#define mp(a,b) make_pair<int,int>(a,b)
#define cls(s) memset(s,0,sizeof(s))
#define cp pair<int,int>
#define LL long long int
#define ls (u << 1)
#define rs (u << 1 | 1)
using namespace std;
const int maxn = 100005,maxm = 100005,INF = 1000000000;
inline LL read(){
	LL out = 0,flag = 1; char c = getchar();
	while (c < 48 || c > 57){if (c == '-') flag = -1; c = getchar();}
	while (c >= 48 && c <= 57){out = (out << 3) + (out << 1) + c - 48; c = getchar();}
	return out * flag;
}
int Q,P;
LL val[maxn << 2];
void build(int u,int l,int r){
	if (l == r){val[u] = 1; return;}
	int mid = l + r >> 1;
	build(ls,l,mid);
	build(rs,mid + 1,r);
	val[u] = 1ll * val[ls] * val[rs] % P;
}
void modify(int u,int l,int r,int pos,LL v){
	if (l == r){val[u] = v; return;}
	int mid = l + r >> 1;
	if (mid >= pos) modify(ls,l,mid,pos,v);
	else modify(rs,mid + 1,r,pos,v);
	val[u] = 1ll * val[ls] * val[rs] % P;
}
LL query(int u,int l,int r,int L,int R){
	if (l >= L && r <= R) return val[u];
	int mid = l + r >> 1;
	if (mid >= R) return query(ls,l,mid,L,R);
	if (mid < L) return query(rs,mid + 1,r,L,R);
	return 1ll * query(ls,l,mid,L,R) * query(rs,mid + 1,r,L,R) % P;
}
int main(){
	int T = read();
	while (T--){
		Q = read(); P = read(); LL opt,v;
		build(1,1,Q);
		REP(i,Q){
			opt = read(); v = read();
			if (opt & 1) modify(1,1,Q,i,v % P);
			else modify(1,1,Q,v,1);
			printf("%lld\n",query(1,1,Q,1,i));
		}
	}
	return 0;
}