8.13日测试内容

T1：P1571 眼红的Medusa

题目传送门

一遍过

实现方法

用~~陈老师~~二分函数，找到~~陈老师~~目标数字

再按照科技创新奖的顺序输出

$AC$ $Code:$

#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 7;
int a[maxn], b[maxn];
int n, m;
int chenlaoshi(int x)
{
	int l = 1 - 1;
	int r = m + 1;
	b[l] = INT_MIN;
	b[r] = INT_MAX;
	while (l + 1 < r)
	{
		int mid = (l + r) / 2;
		if (b[mid] < x)
		{
			l = mid;
		}
		if (b[mid] > x)
		{
			r = mid;
		}
		if (b[mid] == x)
		{
			return mid;
		}
	}
	return -1;
}
int main()
{
	ios::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0); 
    cin >> n >> m;
    for (int i = 1; i <= n; ++i)
    {
    	cin >> a[i];
	}
	for (int i = 1; i <= m; ++i)
    {
    	cin >> b[i];
	}
	sort(b + 1, b + 1 + m);
	for (int i = 1; i <= n; ++i)
	{
		if (chenlaoshi(a[i]) != -1)
		{
			cout << a[i] << ' ';
		}
	}
	return 0;
}

T2：P2249 【深基13.例1】查找

题目传送门

$Wrong$ $Code:$

#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e6 + 7;
int a[maxn];
int n;
int chenlaoshi(int x)
{
	int l = 1 - 1;
	int r = n + 1;
	a[0] = -1e9;
	a[n + 1] = 1e9;
	while (l + 1 < r)
	{
		int mid = (l + r) / 2;
		if (a[mid] > x)
		{
			r = mid;
		}
		if (a[mid] < x)
		{
			l = mid;
		}
		if (a[mid] == x)
		{
			r = mid;
		}
	}
	if (a[r] == x)
	{
		return r;
	}
	return -1;
}
int main()
{
	cin >> n;
	int m;
	cin >> m;
	for (int i = 1; i <= n; i++)
	{
		cin >> a[i];
	}
	while (m--)
	{
		int x;
		cin >> x;
		cout << chenlaoshi(x) << ' ';
	}
	return 0;
}

错误原因

数组开小了

错误信息：

Runtime Error.

Received signal 11:

Segmentation fault with invalid memory reference.

翻译过来就是：

运行时错误。

接收到的信号11：

分段故障，内存引用无效。

这不就是数组开小了吗 (^^_^^^)

$AC$ $Code:$

#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e6 + 7;
int a[maxn];
int n;
int chenlaoshi(int x)
{
	int l = 1 - 1;
	int r = n + 1;
	a[0] = -1e9;
	a[n + 1] = 1e9;
	while (l + 1 < r)
	{
		int mid = (l + r) / 2;
		if (a[mid] > x)
		{
			r = mid;
		}
		if (a[mid] < x)
		{
			l = mid;
		}
		if (a[mid] == x)
		{
			r = mid;
		}
	}
	if (a[r] == x)
	{
		return r;
	}
	return -1;
}
int main()
{
	cin >> n;
	int m;
	cin >> m;
	for (int i = 1; i <= n; i++)
	{
		cin >> a[i];
	}
	while (m--)
	{
		int x;
		cin >> x;
		cout << chenlaoshi(x) << ' ';
	}
	return 0;
}

T3：P1678 烦恼的高考志愿

题目传送门

不会做 ~~前世记忆忘了~~

实现方法

用 lower_bound 找最接近的答案

$But$ 答案可能在左边也可能在右边

$So$ 两个做差值

如果是左侧边界，只有一个答案，右侧边界也只有一个答案。

$AC$ $Code:$

#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 7;
int a[maxn];
int b[maxn];
int m, n;
int main()
{
    cin >> m >> n;
    for (int i = 1; i <= m; ++i)
    {
    	cin >> a[i];
	}
	sort(a + 1, a + 1 + m);
	long long sum = 0;
	for (int i = 1; i <= n; ++i)
	{
		cin >> b[i];
		int pos = lower_bound(a + 1, a + 1 + m, b[i]) - a;
		if (pos == m + 1)
		{
			sum += abs(b[i] - a[m]);
			continue;
		}
		if (pos == 1)
		{
			sum += abs(b[i] - a[1]);
			continue;
		}
		if (abs(b[i] - a[pos]) < abs(b[i] - a[pos - 1]))
		{
			sum += abs(b[i] - a[pos]);
		}
		else
		{
			sum += abs(b[i] - a[pos - 1]);
		}
	}
	cout << sum << "\n";
	return 0;
}

T4：P1918 保龄球

题目传送门

一遍过

实现方法

定义一个 Balls 类型的结构体数组

输入每一个球道的球数，每一个球道的 id $=$ i

用 sort 把 a 数组排序

再输入每一个要找的瓶子数

用陈老师函数查找对应的瓶子数量是否存在

$AC$ $Code:$

#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 7;
struct balls
{
	int id;
	int x;
};
bool cmp(balls a, balls b)
{
	return a.x < b.x;
}
balls a[maxn];
int n;
int chenlaoshi(int x)
{
	int l = 1 - 1;
	int r = n + 1;
	a[l].x = INT_MIN;
	a[r].x = INT_MAX;
	while (l + 1 < r)
	{
		int mid = (l + r) / 2;
		if (a[mid].x > x)
		{
			r = mid;
		}
		if (a[mid].x < x)
		{
			l = mid;
		}
		if (a[mid].x == x)
		{
			return a[mid].id;
		}
	}
	return 0;
}
int main()
{
	ios::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0); 
	cin >> n;
	for (int i = 1; i <= n; ++i)
	{
		cin >> a[i].x;
		a[i].id = i;
	}
	sort(a + 1, a + 1 + n, cmp);
	int m;
	cin >> m;
	for (int i = 1; i <= m; ++i)
	{
		int op;
		cin >> op;
		cout << chenlaoshi(op) << "\n";
	}
	return 0;
}

T5：P1102 A-B 数对

题目传送门

第一次不记得了，用的暴力，居然拿了 $76$ 分 (奇迹)！！！

$Wrong$ $Code:$

#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 7;
int a[maxn];
int main()
{
	ios::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0); 
	int n, c;
	cin >> n >> c;
	for (int i = 1; i <= n; ++i)
	{
		cin >> a[i];
	}
	int ans = 0;
	for (int i = 1; i <= n; ++i)
	{
		for (int j = 1; j <= n; ++j)
		{
			if (a[i] - a[j] == c)
			{
				ans++;
			}
		}
	}
	cout << ans << "\n";
	return 0;
}

实现方法

lower_bound 和 upper_bound 的另外一个作用就是直接获取相等的个数

lower_bound 找到相等的第一个位置 $l$

upper_bound 找到不相等的第一个位置

那么相等的最后一个位置 $r = $ upper_bound $-1$

相等的个数就是 $r - l + 1$。

$AC$ $Code:$

#include <bits/stdc++.h>
using namespace std;
const int maxn = 2e5 + 7;
int a[maxn];
int n, c;
int main()
{
    cin >> n >> c;
    for (int i = 1; i <= n; ++i)
    {
    	cin >> a[i];
	}
	sort(a + 1, a + 1 + n);
	long long cnt = 0;
	for (int i = 1; i <= n; ++i)
	{
		int b = a[i] - c;
		int geshu = upper_bound(a + 1, a + 1 + n, b) - lower_bound(a + 1, a + 1 + n, b);
		cnt += geshu;
	}
	cout << cnt;
	return 0;
}

T6：B3799 [NICA #1] 序列

题目传送门

看到题后：

眼睛：我会了

脑子：不，你不会

引用一下陈老师的AC代码 $↓$

$AC$ $Code:$

#include <bits/stdc++.h>
#define int long long
using namespace std;
const int maxn = 5e5 + 7;
int a[maxn];
int sum[maxn];
int n, m;
signed main()
{
    cin >> n >> m;
    for (int i = 1; i <= n; ++i)
    {
    	cin >> a[i];
	}
	sort(a + 1, a + 1 + n);
	for (int i = 1; i <= n; ++i)
	{
		sum[i] = sum[i - 1] + a[i];
	}
	int num = 0;
	while (m--)
	{
		int op;
		cin >> op;
		if (op == 1)
		{
			int k;
			cin >> k;
			num += k;
		}
		else
		{
			int pos = lower_bound(a + 1, a + 1 + n, -num) - a;
			int l = pos;
			int r = n;
			cout <<  (sum[r] - sum[l - 1]) + (r - l + 1) * num << "\n";
		}
	}
	return 0;
}

$The$ $end$

posted @ 2024-08-13 21:15 yucheng0630 阅读(6) 评论(0) 收藏举报

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8.13日测试内容

8.13日测试内容

T1：P1571 眼红的Medusa

题目传送门

实现方法

\(AC\) \(Code:\)

T2：P2249 【深基13.例1】查找

题目传送门

\(Wrong\) \(Code:\)

错误原因

\(AC\) \(Code:\)

T3：P1678 烦恼的高考志愿

题目传送门

实现方法

\(AC\) \(Code:\)

T4：P1918 保龄球

题目传送门

实现方法

\(AC\) \(Code:\)

T5：P1102 A-B 数对

题目传送门

\(Wrong\) \(Code:\)

实现方法

\(AC\) \(Code:\)

T6：B3799 [NICA #1] 序列

题目传送门

\(AC\) \(Code:\)

\(The\) \(end\)

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