Day 26 - 模拟赛

合集 - 2024 暑假(51)

1.2024 暑假记2024-08-162.Day 1 - 二分2024-07-083.Day 1 - 题目集2024-07-084.Day 2 - 分治、倍增、LCA 与树链剖分2024-07-095.Day 2 - 题目集2024-07-096.Day 3 - 单调栈、单调队列、凸包与斜率优化2024-07-107.Day 3 - 题目集2024-07-108.Day 4 - 搜索进阶与模拟2024-07-119.Day 4 - 题目集2024-07-1110.Day 5 - 双指针与折半搜索2024-07-1211.Day 5 - 题目集2024-07-1212.Day 7 - 哈希与 KMP2024-07-1413.Day 7 - 题目集2024-07-1414.Day 8 - 并查集、堆、set 与 map2024-07-1515.Day 8 - 题目集2024-07-1516.Day 9 - 线段树2024-07-1617.Day 9 - 题目集2024-07-1718.Day 10 - 动态规划与树状数组2024-07-1719.Day 10 - 题目集2024-07-1720.Day 11 - 模拟赛2024-07-1821.Day 11 - 题目集2024-07-1822.Day 13 - 树形 DP 与换根 DP2024-07-2023.Day 13 - 题目集2024-07-2024.Day 14 - 模拟赛2024-07-2125.Day 14 - 题目集2024-07-2126.Day 15 - 数位 DP 与状压 DP2024-07-2227.Day 15 - 题目集2024-07-2228.Day 16 - 模拟赛2024-07-2329.Day 16 - 题目集2024-07-2330.Day 17 - 贪心2024-07-2431.Day 17 - 题目集2024-07-2432.Day 19 - 构造、转换与模拟2024-07-2633.Day 19 - 题目集2024-07-2634.Day 20 - 最短路与差分约束2024-07-2735.Day 20 - 题目集2024-07-2736.Day 21 - 最小生成树2024-07-2837.Day 21 - 题目集2024-07-2838.Day 22 - 图论联通性2024-07-2939.Day 22 - 题目集2024-07-2940.Day 23 - 模拟赛2024-07-3041.Day 23 - 题目集2024-07-3042.Day 25 - 同余与乘法逆元2024-08-0143.Day 25 - 题目集2024-08-01

44.Day 26 - 模拟赛2024-08-02

45.Day 26 - 题目集2024-08-0346.Day 27 - 欧几里得算法与中国剩余定理2024-08-0347.Day 27 - 题目集2024-08-0348.Day 28 - 组合数学与矩阵初步2024-08-0449.Day 28 - 题目集2024-08-0450.Day 29 - 结营测试2024-08-0551.Day 29 - 题目集2024-08-05

收起

1|0热门景点（hot）

题目描述

输入格式

输出格式

$\text{input1}$

10 5
1 7 4 0 9 4 8 8 2 4
6 6
2 8
2 2
3 6
7 8

$\text{output1}$

Y
N
Y
N
Y

数据范围

#include<iostream>
#include<cstdio>
#include<ctime>
using namespace std;
#define MAXN 5000005

int read() {
	int x = 0; bool f = 0;
	char ch = getchar();
	while(ch < 48 || ch > 57) {
		if(ch == 45) f = 1;
		ch = getchar();
	}
	while(ch >= 48 && ch <= 57) {
		x = (x << 3) + (x << 1) + (ch - 48);
		ch = getchar();
	}
	return f ? -x : x;
}

int n, q, a[MAXN], pl[MAXN], pr[MAXN], cnt = 0, l, r;
struct node { int lm, rm; } b[MAXN];

int main() {
	freopen("hot.in", "r", stdin);
	freopen("hot.out", "w", stdout);
	
	n = read(); q = read();
	for(int i = 1; i <= n; i ++) a[i] = read();
	pl[1] = 1;
	for(int i = 2; i <= n; i ++) 
		if(a[i] <= a[i - 1]) pl[i] = pl[i - 1];
		else pl[i] = i;
	pr[n] = n;
	for(int i = n - 1; i >= 1; i --) 
		if(a[i] <= a[i + 1]) pr[i] = pr[i + 1];
		else pr[i] = i;
//	for(int i = 1; i <= n; i ++) cout << pl[i] << " ";
//	cout << "\n";
//	for(int i = 1; i <= n; i ++) cout << pr[i] << " ";
//	cout << "\n";
	while(q --) {
		l = read(); r = read();
		if(pr[l] >= pl[r]) printf("Y\n");
		else printf("N\n");
	}
//	for(int i = 1; i <= n; i ++) 
//		if(a[i] >= a[i - 1]) pl[i] = pl[i - 1] + 1;
//		else pl[i] = 1;
//	for(int i = n; i >= 1; i --) 
//		if(a[i] >= a[i + 1]) pr[i] = pr[i + 1] + 1;
//		else pr[i] = 1;
////	for(int i = 1; i <= n; i ++) cout << pl[i] << " ";
////	cout << "\n";
////	for(int i = 1; i <= n; i ++) cout << pr[i] << " ";
////	cout << "\n";
//	int tmp = 1; while(pl[tmp] > pl[tmp - 1]) tmp ++;
////	cout << tmp - 1 << "\n"; 
//	pl[n + 1] = 1;
//	for(int i = tmp - 1; i <= n; ) {
//		b[++ cnt].lm = i - pl[i] + 1, b[cnt].rm = i + pr[i] - 1;
//		i += pr[i]; while(pl[i + 1] != 1) i ++; 
//	}
////	for(int i = 1; i <= cnt; i ++) cout << b[i].lm << " " << b[i].rm << "\n"; 
//	int l = 0, r = 0;
//	while(q --) {
//		l = read(); r = read();
//		if(r - l + 1 == 1 || r - l + 1 == 2) { printf("Y\n"); continue; }
//		int ll = 1, lr = cnt, ansl = -1;
//		while(ll <= lr) {
//			int mid = (ll + lr) >> 1;
//			if(b[mid].lm <= l) ansl = mid, ll = mid + 1;
//			else lr = mid - 1;
//		}
//		int rl = 1, rr = cnt, ansr = -1;
//		while(rl <= rr) {
//			int mid = (rl + rr) >> 1;
//			if(b[mid].rm >= r) ansr = mid, rr = mid - 1;
//			else rl = mid + 1;
//		}
////		cout << ansl << " " << ansr << "\n";
//		if(ansr + 1 == ansl && b[ansr].lm <= l && b[ansr].rm >= r) printf("Y\n");
//		else if(ansl == ansr && ansl != -1) printf("Y\n");
//		else printf("N\n");
//	}
	return 0;
}

2|0管道（tunnel）

题目描述

输入格式

输出格式

$\text{input1}$

1 7
15 -5 100 -40 10 10 10

$\text{output1}$

115

$\text{input2}$

3 3
-10 4 -10
3 1 -10
6 2 8

$\text{output2}$

28

样例解释

数据范围

tunnel.zip

#include<iostream>
using namespace std;
#define MAXN 1005

long long read() {
	long long x = 0; bool f = 0;
	char ch = getchar();
	while(ch < 48 || ch > 57) {
		if(ch == 45) f = 1;
		ch = getchar();
	}
	while(ch >= 48 && ch <= 57) {
		x = (x << 3) + (x << 1) + (ch - 48);
		ch = getchar();
	}
	return f ? -x : x;
}

long long n, m, a[MAXN][MAXN], b[MAXN][MAXN], ans = 0;

void update() {
	long long res = 0;
	for(int i = 1; i <= n; i ++) for(int j = 1; j <= m; j ++) 
		if(b[i][j] == 1) {
			if(b[i][j + 1] == 2 || b[i][j + 1] == 4) res += max(0ll, a[i][j] + a[i][j + 1]);
			if(b[i + 1][j] == 3 || b[i + 1][j] == 4) res += max(0ll, a[i][j] + a[i + 1][j]);
		}
		else if(b[i][j] == 2) {
			if(b[i][j - 1] == 1 || b[i][j - 1] == 3) res += max(0ll, a[i][j] + a[i][j - 1]);
			if(b[i + 1][j] == 3 || b[i + 1][j] == 4) res += max(0ll, a[i][j] + a[i + 1][j]);
		}
		else if(b[i][j] == 3) {
			if(b[i][j + 1] == 2 || b[i][j + 1] == 4) res += max(0ll, a[i][j] + a[i][j + 1]);
			if(b[i - 1][j] == 1 || b[i - 1][j] == 2) res += max(0ll, a[i][j] + a[i - 1][j]);
		}
		else if(b[i][j] == 4) {
			if(b[i][j - 1] == 1 || b[i][j - 1] == 3) res += max(0ll, a[i][j] + a[i][j - 1]);
			if(b[i - 1][j] == 1 || b[i - 1][j] == 2) res += max(0ll, a[i][j] + a[i - 1][j]);
		}
//	if(res / 2 > ans) {
//		for(int i = 1; i <= n; i ++) {
//			for(int j = 1; j <= m; j ++) cout << b[i][j] << " ";
//			cout << "\n";
//		} 
//		cout << res / 2 << "\n";
//	}
	ans = max(ans, res / 2); return;
}

void dfs(long long x, long long y) {
//	cout << x << " " << y << "\n"; 
	for(int i = 1; i <= 4; i ++) {
		b[x][y] = i;
		if(x == n && y == m) update();
		else if(y == m) dfs(x + 1, 1);
		else dfs(x, y + 1);
	}
	return;
}

int main() {
	freopen("tunnel.in", "r", stdin);
	freopen("tunnel.out", "w", stdout);
	
	n = read(); m = read();
	for(int i = 1; i <= n; i ++) for(int j = 1; j <= m; j ++) a[i][j] = read();
	dfs(1, 1);
	cout << ans << "\n"; 
	return 0;
}

#include<iostream>
#include<cmath>
using namespace std;
#define MAXN 1005

long long read() {
	long long x = 0; bool f = 0;
	char ch = getchar();
	while(ch < 48 || ch > 57) {
		if(ch == 45) f = 1;
		ch = getchar();
	}
	while(ch >= 48 && ch <= 57) {
		x = (x << 3) + (x << 1) + (ch - 48);
		ch = getchar();
	}
	return f ? -x : x;
}

long long n, m, a[MAXN][MAXN], f[MAXN][MAXN][2], g[MAXN][MAXN][2], ans = 0;

int main() {
	freopen("tunnel.in", "r", stdin);
	freopen("tunnel.out", "w", stdout);
	
	n = read(); m = read();
	for(int i = 1; i <= n; i ++) for(int j = 1; j <= m; j ++) a[i][j] = read();
	for(int i = 2; i <= n; i ++) for(int j = 1; j <= m; j ++) 
        f[i][j][0] = max(f[i - 1][j][0], f[i - 1][j][1] + max(0ll, a[i - 1][j] + a[i][j])), 
        f[i][j][1] = max(f[i - 1][j][0], f[i - 1][j][1]);
    for(int i = 1; i <= n; i ++) for(int j = 2; j <= m; j ++)
        g[i][j][0] = max(g[i][j - 1][0], g[i][j - 1][1] + max(0ll, a[i][j - 1] + a[i][j])), 
        g[i][j][1] = max(g[i][j - 1][0], g[i][j - 1][1]);
    for(int i = 1; i <= m; i ++) ans += f[n][i][0];
    for(int i = 1; i <= n; i ++) ans += g[i][m][0];
    cout << ans << "\n";
	return 0;
}

3|0序列上的树（tree）

题目描述

输入格式

输出格式

$\text{input1}$

5
1 2 1 2 1

$\text{output1}$

8

$\text{input2}$

9
1 2 1 3 4 3 5 3 1

$\text{output2}$

15

样例解释

数据范围

tree.zip

#include<iostream>
using namespace std;
#define MAXN 200005

long long read() {
	long long x = 0; bool f = 0;
	char ch = getchar();
	while(ch < 48 || ch > 57) {
		if(ch == 45) f = 1;
		ch = getchar();
	}
	while(ch >= 48 && ch <= 57) {
		x = (x << 3) + (x << 1) + (ch - 48);
		ch = getchar();
	}
	return f ? -x : x;
}

long long n, a[MAXN], ans = 0, cnt = 0, b[MAXN], p[MAXN];
bool c[MAXN];

int main() {
	freopen("tree.in", "r", stdin);
	freopen("tree.out", "w", stdout);
	
	n = read();
	for(int i = 1; i <= n; i ++) a[i] = read();
	for(int i = 1; i <= n; i ++) {
		b[++ cnt] = a[i];
		if(b[cnt] == b[cnt - 2] && !c[b[cnt - 1]]) 
			c[b[cnt - 1]] = true, cnt -= 2;
	}
	if(cnt == 1) {
		for(int i = 1; i <= n; i ++) p[a[i]] ++;
		for(int i = 1; i <= n; i ++) ans += (p[i] - 1) * p[i] / 2;
		cout << ans + n << "\n";
		return 0;
	}
	for(int l = 1; l < n - 1; l ++) for(int r = l + 2; r <= n; r ++) {
		if(a[l] != a[r]) continue; cnt = 0;
		for(int i = l; i <= r; i ++) c[a[i]] = false;
		for(int i = l; i <= r; i ++) {
			b[++ cnt] = a[i];
			if(b[cnt] == b[cnt - 2] && !c[b[cnt - 1]]) 
				c[b[cnt - 1]] = true, cnt -= 2;
		}
		if(cnt == 1) ans ++;
	}
	cout << ans + n << "\n";
	return 0;
}

4|0生存（survive）

题目描述

输入格式

输出格式

$\text{input1}$

3 2
1 2 3
1 2
2 3
1 2
2 1 2

$\text{output1}$

5

样例解释

数据范围

survive.zip

#include<iostream>
#include<vector>
using namespace std;
#define MAXN 1000005

long long read() {
	long long x = 0; bool f = 0;
	char ch = getchar();
	while(ch < 48 || ch > 57) {
		if(ch == 45) f = 1;
		ch = getchar();
	}
	while(ch >= 48 && ch <= 57) {
		x = (x << 3) + (x << 1) + (ch - 48);
		ch = getchar();
	}
	return f ? -x : x;
}

long long n, m, a[MAXN], ans = 0;

int main() {
	freopen("survive.in", "r", stdin);
	freopen("survive.out", "w", stdout);
	
	n = read(); m = read();
	for(int i = 1; i <= n; i ++) a[i] = read();
	for(int i = 1; i <= n; i ++) ans += a[i];
	cout << ans << "\n";
	return 0;
}

$$100+35+50+0=185$$

__EOF__

本文作者：So_noSlack
本文链接：https://www.cnblogs.com/So-noSlack/p/18338616.html
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