bzoj 5334 [Tjoi2018]数学计算

Posted 2021-01-29 wawawa8

bzoj 5334 [Tjoi2018]数学计算

Link

Solution

直接上线段树

相当于单点修改，只查询区间 ((1,n)) 的积

Code

// Copyright lzt
#include<stdio.h>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<vector>
#include<map>
#include<set>
#include<cmath>
#include<iostream>
#include<queue>
#include<string>
#include<ctime>
using namespace std;
typedef long long ll;
typedef std::pair<int, int> pii;
typedef long double ld;
typedef unsigned long long ull;
typedef std::pair<long long, long long> pll;
#define fi first
#define se second
#define pb push_back
#define mp make_pair
#define rep(i, j, k)  for (register int i = (int)(j); i <= (int)(k); i++)
#define rrep(i, j, k) for (register int i = (int)(j); i >= (int)(k); i--)
#define Debug(...) fprintf(stderr, __VA_ARGS__)
 
inline ll read() {
  ll x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch <= '9' && ch >= '0') {
    x = 10 * x + ch - '0';
    ch = getchar();
  }
  return x * f;
}
 
#define lc (i << 1)
#define rc (i << 1 | 1)
const int maxn = 100100;
int tc, q, m;
struct Node {
  int l, r, val;
} tr[maxn << 2];
 
void build(int i, int l, int r) {
  tr[i].l = l; tr[i].r = r; tr[i].val = 1;
  if (l == r) return;
  int md = (l + r) >> 1;
  build(lc, l, md); build(rc, md + 1, r);
}
 
void upd(int i, int p, int v) {
  if (tr[i].l == tr[i].r) {
    tr[i].val = v;
    return;
  }
  int md = (tr[i].l + tr[i].r) >> 1;
  if (p <= md) upd(lc, p, v); else upd(rc, p, v);
  tr[i].val = tr[lc].val * 1ll * tr[rc].val % m;
}
 
void work() {
  tc = read();
  build(1, 1, 100000);
  while (tc--) {
    q = read(), m = read();
    rep(i, 1, q) {
      int op = read(), x = read();
      if (op == 1) {
        upd(1, i, x);
      }
      else if (op == 2) {
        upd(1, x, 1);
      }
      printf("%d
", tr[1].val);
    }
    rep(i, 1, 100000) upd(1, i, 1);
  }
}
 
int main() {
  #ifdef LZT
    freopen("in", "r", stdin);
  #endif
 
  work();
 
  #ifdef LZT
    Debug("My Time: %.3lfms
", (double)clock() / CLOCKS_PER_SEC);
  #endif
}