BZOJ 5334: [Tjoi2018]数学计算

题目链接：​​传送门​​

一眼看过去很简单的样子

根节点维护线段树乘积
每个叶节点对应每一个操作
如果是2操作则该叶节点为1
否则就是就是要乘的m

/**
 * @Date:   2019-03-27T19:21:25+08:00
 * @Last modified time: 2019-03-27T19:21:26+08:00
 */
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <complex>
#include <algorithm>
#include <climits>
#include <queue>
#include <map>
#include <set>
#include <vector>
#include <iomanip>
#define
#define

using namespace std;
typedef long long ll;
#define
struct node {
    int l, r, w;
}tree[A];
int T, q, mod, a, b;
void build(int k, int l, int r) {
    tree[k].l = l; tree[k].r = r; tree[k].w = 1;
    if (l == r) return;
    int m = (l + r) >> 1;
    build(k << 1, l, m);
    build(k << 1 | 1, m + 1, r);
}
void change(int k, int val, int pos) {
    if (tree[k].l == tree[k].r) {
        tree[k].w = val;
        return;
    }
    int m = (tree[k].l + tree[k].r) >> 1;
    if (pos <= m) change(k << 1, val, pos);
    else change(k << 1 | 1, val, pos);
    tree[k].w = tree[k << 1].w * tree[k << 1 | 1].w % mod;
}

signed main() {
    cin >> T;
    while (T--) {
        memset(tree, 0, sizeof tree);
        scanf("%lld%lld", &q, &mod);
        build(1, 1, q);
        for (int i = 1; i <= q; i++) {
            scanf("%lld%lld", &a, &b);
            if (a == 1) change(1, b, i);
            else change(1, 1, b);
            printf("%lld\n", tree[1].w);
        }
    }
    return 0;
}