这题很经典啊,对于除怎么解决取模的问题.
这种题确实是非常经典,就是这个取模怎么解决有除的问题呢??
一开始,出于某种奇怪的心态(不就是日常装杯吗....),压根没考虑到.以此警醒自己,认真对待,仔细审题,仔细理解.
回归正文:
处理有除法的取模,可用线段树,就是凑区间吗,能想到这点就可以了,线段树还是挺好写的,就是凑区间嘛
// Problem: C. LR-remainders
// Contest: Codeforces - Codeforces Round 927 (Div. 3)
// URL: https://codeforces.com/contest/1932/problem/C
// Memory Limit: 256 MB
// Time Limit: 2000 ms
// Date: 2024-03-11 19:38:23
//
// Powered by CP Editor (https://cpeditor.org)#include<bits/stdc++.h>
#define endl '\n'
#define ls u << 1
#define rs u << 1 | 1
using namespace std;
const int N = 200005;
struct Tree {int l, r;int sum;
}tr[N * 4];
int n, m, a[200005];
void build(int u, int l, int r) {tr[u] = { l,r };if (l == r) {tr[u].sum = a[l] % m;}else {int mid = l + r >> 1;build(ls, l, mid);build(rs, mid + 1, r);tr[u].sum = tr[ls].sum * tr[rs].sum % m;}
}
int query(int u, int x, int y) {int l = tr[u].l, r = tr[u].r, mid = l + r >> 1;if (x <= l && r <= y)return tr[u].sum;int ans = 1;if (x <= mid)ans = ans * query(ls, x, y) % m;if (y >= mid + 1)ans = ans * query(rs, x, y) % m;return ans;
}
void solve() {cin >> n >> m;for (int i = 1; i <= n; ++i) {cin >> a[i];}build(1, 1, n);string s; cin >> s;for (int i = 0, l = 1, r = n; i < s.size(); ++i) {cout << query(1, l, r) % m << " ";if (s[i] == 'L')++l;else --r;}cout << endl;
}
signed main() {ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);int t = 1;cin >> t;while (t--) {solve();}return 0;
}
)
