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;
}