【洛谷】P4588 [TJOI2018] 数学计算 的题解 + 线段树板子代码

【洛谷】P4588 [TJOI2018] 数学计算 的题解 + 线段树板子代码

洛谷传送门

题解

论我的快读快写的神奇故事

用快读快写直接整出了这种逆天不用快读,神奇!!!

划水一个线段树 + + + 离散化。

一如既往的打了一个暴力,神奇的一分不得。果然线段树的威力在这里qaq

总体思路就是将数据按时间排序,建线段树,维护区间乘。这样的话根节点就是到现在为止的所有数的乘积。

代码

#include <bits/stdc++.h>
#define lowbit(x) x & (-x)
#define endl "\n"
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
namespace fastIO {
	inline int read() {
		register int x = 0, f = 1;
		register char c = getchar();
		while (c < '0' || c > '9') {
			if(c == '-') f = -1;
			c = getchar();
		}
		while (c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
		return x * f;
	}
	inline void write(int x) {
		if(x < 0) putchar('-'), x = -x;
		if(x > 9) write(x / 10);
		putchar(x % 10 + '0');
		return;
	}
}
using namespace fastIO;
int T, q, op, m;
ll mod, sum[1000005];
void update(int now) {
	sum[now] = (sum[now << 1] * sum[now << 1|1]) % mod;
}
void build(int now, int l, int r) {
	if(l == r) {
		sum[now] = 1;
		return;
	}
	int mid = (l + r) >> 1;
	build(now << 1, l, mid);
	build(now << 1|1, mid + 1, r);
	update(now);
}
void change(int c, int l, int r, int L, int R, int p) {
	if(l >= L && r <= R) {
		sum[c] = p;
		return;
	}
	int mid = (l + r) >> 1;
	if(L <= mid) 
		change(c << 1, l, mid, L, R, p);
	if(R > mid) 
		change(c << 1 | 1, mid + 1, r, L, R, p);
	update(c);
}
int main() {
	//freopen(".in","r",stdin);
	//freopen(".out","w",stdout);
	ios::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
	cin >> T;
	for(int i = 1; i <= T; i ++) {
		cin >> q >> mod;
		build(1, 1, q); 
		for(int i = 1; i <= q; i ++) {
			cin >> op >> m;
			if(op == 1) {
				change(1, 1, q, i, i, m);
				sum[1] %= mod;
			}
			else {
				change(1, 1, q, m, m, 1);
			}
			cout << sum[1] % mod << endl;
		}
	}
	return 0;
}

附加

其实这就是一个有一点点像线段树板子,总体来说没有什么难度,线段树的话,其实只要写熟练了以后,基本就很简单了。

很多大佬在 NOIP 的比赛前都会打一遍线段树,可见这个的重要性了吧。

这里也给出两个线段树的板子题的代码。

【洛谷】P3372 【模板】线段树 1

题目传送门

代码

#include <bits/stdc++.h>
#define endl "\n"
using namespace std;
typedef long long ll;
ll n, a[100005], d[270005], b[270005];
void build(ll l, ll r, ll p) {  
	if (l == r) {
    	d[p] = a[l];  
    	return;
	}
	ll m = l + ((r - l) >> 1);
	build(l, m, p << 1), build(m + 1, r, (p << 1) | 1); 
	d[p] = d[p << 1] + d[(p << 1) | 1];
}
void update(ll l, ll r, ll c, ll s, ll t, ll p) {
	if (l <= s && t <= r) {
    	d[p] += (t - s + 1) * c, b[p] += c;
    	return;
	}
	ll m = s + ((t - s) >> 1);
	if (b[p])
    	d[p << 1] += b[p] * (m - s + 1), d[(p << 1) | 1] += b[p] * (t - m), b[p << 1] += b[p], b[(p << 1) | 1] += b[p];
	b[p] = 0;
	if (l <= m)
    	update(l, r, c, s, m, p << 1);
	if (r > m) update(l, r, c, m + 1, t, (p << 1) | 1);
	d[p] = d[p << 1] + d[(p << 1) | 1]; 
}
ll getsum(ll l, ll r, ll s, ll t, ll p) {
	if (l <= s && t <= r) return d[p];
	ll m = s + ((t - s) >> 1);
	if (b[p])
    	d[p << 1] += b[p] * (m - s + 1), d[(p << 1) | 1] += b[p] * (t - m), b[p << 1] += b[p], b[(p << 1) | 1] += b[p];
	b[p] = 0;
	ll sum = 0;
	if (l <= m)
    	sum =getsum(l, r, s, m, p << 1);
	if (r > m) sum += getsum(l, r, m + 1, t, (p << 1) | 1);
	return sum;
}
int main() {
	//freopen(".in","r",stdin);
    //freopen(".out","w",stdout);
	ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
	ll q, i1, i2, i3, i4;
	cin >> n >> q;
	for (ll i = 1; i <= n; i++) cin >> a[i];
	build(1, n, 1);
	while (q--) {
        cin >> i1 >> i2 >> i3;
        if (i1 == 2)
    		cout << getsum(i2, i3, 1, n, 1) << endl;
        else
          cin >> i4, update(i2, i3, i4, 1, n, 1);
    }
	return 0;
}

【洛谷】P3373 【模板】线段树 2

题目传送门

代码

#include <bits/stdc++.h>
#define endl "\n"
using namespace std;
typedef long long ll;
inline ll read() {
    register ll x = 0, f = 1;
    register char c = getchar();
    while (c < '0' || c > '9') {
        if(c == '-') f = -1;
        c = getchar();
    }
    while (c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
    return x * f;
}
int n, m;
ll mod;
ll a[100005], sum[400005], mul[400005], laz[400005];
void up(int i) { sum[i] = (sum[(i << 1)] + sum[(i << 1) | 1]) % mod; }
void pd(int i, int s, int t) {
	int l = (i << 1), r = (i << 1) | 1, mid = (s + t) >> 1;
	if (mul[i] != 1) {
	    mul[l] *= mul[i];
	    mul[l] %= mod;
	    mul[r] *= mul[i];
	    mul[r] %= mod;
	    laz[l] *= mul[i];
	    laz[l] %= mod;
	    laz[r] *= mul[i];
	    laz[r] %= mod;
	    sum[l] *= mul[i];
	    sum[l] %= mod;
	    sum[r] *= mul[i];
	    sum[r] %= mod;
	    mul[i] = 1;
	}
	if (laz[i]) { 
	    sum[l] += laz[i] * (mid - s + 1);
	    sum[l] %= mod;
	    sum[r] += laz[i] * (t - mid);
	    sum[r] %= mod;
	    laz[l] += laz[i];
	    laz[l] %= mod;
	    laz[r] += laz[i];
	    laz[r] %= mod;
	    laz[i] = 0;
	}
	return;
}
void build(int s, int t, int i) {
	mul[i] = 1;
	if (s == t) {
    	sum[i] = a[s];
    	return;
	}
	int mid = s + ((t - s) >> 1);
	build(s, mid, i << 1);
	build(mid + 1, t, (i << 1) | 1);
	up(i);
}
void chen(int l, int r, int s, int t, int i, ll z) {
	int mid = s + ((t - s) >> 1);
	if (l <= s && t <= r) {
	    mul[i] *= z;
	    mul[i] %= mod;
	    laz[i] *= z;
	    laz[i] %= mod;
	    sum[i] *= z;
	    sum[i] %= mod; 
	    return;
	}
	pd(i, s, t);
	if (mid >= l) chen(l, r, s, mid, (i << 1), z);
	if (mid + 1 <= r) chen(l, r, mid + 1, t, (i << 1) | 1, z);
	up(i);
}
void add(int l, int r, int s, int t, int i, ll z) {
	int mid = s + ((t - s) >> 1);
	if (l <= s && t <= r) {
	    sum[i] += z * (t - s + 1);
	    sum[i] %= mod;
	    laz[i] += z;
	    laz[i] %= mod;
	    return;
	}
	pd(i, s, t);
	if (mid >= l) add(l, r, s, mid, (i << 1), z);
	if (mid + 1 <= r) add(l, r, mid + 1, t, (i << 1) | 1, z);
	up(i);
}
ll getans(int l, int r, int s, int t,int i) {  
	int mid = s + ((t - s) >> 1);
	ll tot = 0;
	if (l <= s && t <= r) return sum[i];
	pd(i, s, t);
	if (mid >= l) tot += getans(l, r, s, mid, (i << 1));
	tot %= mod;
	if (mid + 1 <= r) tot += getans(l, r, mid + 1, t, (i << 1) | 1);
	return tot % mod;
}
int main() {
	//freopen(".in","r",stdin);
    //freopen(".out","w",stdout);
	ios::sync_with_stdio(false);
    cout.tie(0);
	int x, y, bh;
	ll z;
	n = read(), m = read(), mod = read();
	for(int i = 1; i <= n; i ++) a[i] = read();
	build(1, n, 1);
	for(int i = 1; i <= m; i ++) {
    	bh = read();
    	if(bh == 1) {
    		x = read(), y = read(), z = read();
    		chen(x, y, 1, n, 1, z);
    	} 
		else if(bh == 2) {
			x = read(), y = read(), z = read();
    		add(x, y, 1, n, 1, z);
    	} 
		else if(bh == 3) {
    		x = read(), y = read();
    		cout << getans(x, y, 1, n, 1) << endl;
    	}
	}
	return 0;
}
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