A. Verify Password

思路:按照ASCLL值进行比较就行(因为字母的ASCLL本来就在数字后面),所以,只要找到前面比后面的数大就输出NO,反之YES

代码实现:

#include<bits/stdc++.h>
using namespace std;
#define N 100005
typedef long long ll;
ll n, m, num, sum, t;
ll a[N], dp[N];
char s[N];
int main()
{
	cin >> t;
	while (t--)
	{
		int flag = 0;
		cin >> n;
		cin >> s + 1;
		for (int i = 1; i < n; i++) {
			if (s[i] > s[i + 1]) {
				flag = 1;
			}
		}
		if (flag == 1)
			cout << "NO" << endl;
		else
			cout << "YES" << endl;
	}
	return 0;
}

B. Increase/Decrease/Copy

思路:

因为对于前n项,我们并不会改变其位置,只有第n+1项,我们进行特判

对于a[i]->b[i],，转变的最小代价就是它们的差值。

对于b[n+1]，得到它的最小代价就是a[i]->b[i]过程中与它最小的差值+1

实现代码:

#include<bits/stdc++.h>
using namespace std;
#define N 200005
typedef long long ll;
ll n, m, num, sum, t;
ll a[N], b[N];
char s[N];
int main()
{
	cin >> t;
	while (t--)
	{
		int flag=0;
		num = 0;
		sum = 1e9;
		cin >> n;
		for (int i = 1; i <= n; i++) {
			cin >> a[i];
		}
		for (int i = 1; i <= n + 1; i++) {
			cin >> b[i];
		}
		for (int i = 1; i <= n; i++) {
			num += abs(a[i] - b[i]);
			sum = min(sum, abs(a[i] - b[n + 1]));
			sum = min(sum, abs(b[i] - b[n + 1]));
		}
		for (int i = 1; i <= n; i++) {
			if (a[i] > b[n + 1] && b[i] > b[n + 1] || a[i] < b[n + 1] && b[i] < b[n + 1])
				continue;
			else {
				flag = 1;
			}
		}
		if (flag == 1) {
			num += 1;
		}
		else {
			num += 1 + sum;
		}
		cout << num << endl;
	}
	return 0;
}

C. Job Interview

 

 思路:

从前往后遍历，检查一下是a能力值浪费了还是b能力值浪费了，然后从后往前枚举，开一个数组维护一下最近后缀损失能力值。输出答案的时候，如果当前的人的站的职位刚好是能力值被浪费的职位，输出总和减去当前的人的能力值加上最近损失能力值

实现代码:

#include<bits/stdc++.h>
using namespace std;
#define N 1000005
#define inf 1e9+7
typedef long long ll;
ll n, m, sum, ans, t, sum1, sum2, num1, num2, f;
ll q[N], p[N], a[N], b[N];
void solve()
{
	cin >> n >> m;
	sum1 = sum2 = num1 = num2 = f=0;
	for (int i = 0; i <= n+m; i++)
		cin >> a[i], q[i] = p[i] = 0;
	for (int i = 0; i <= n+m; i++) {
		cin >> b[i], f += a[i] > b[i];
		if (a[i] > b[i] && num1 <= n || m == i - num1)
			num1++, sum1 += a[i], p[i] = 1; 
		else sum1 += b[i];
		if (a[i] < b[i] && num2 <= m || n == i - num2)
			num2++, sum2 += b[i], q[i] = 1; 
		else sum2 += a[i];
	}
	for (int i = 0; i <= n+m; i++, cout << ' ')
		cout << (f > n ? (p[i] ? sum1 - a[i] : sum2 - b[i]) : (q[i] ? sum2 - b[i] : sum1 - a[i]));
		/*if (f > n) {
			if (p[i])
				cout << sum1 - a[i] << ' ';
			else
				cout << sum2 - b[i] << ' ';
		}
		else {
			if (q[i])
				cout << sum1 - b[i] << ' ';
			else
				cout << sum2 - a[i] << ' ';
		}*/
}
int main()
{
	cin >> t;
	while (t--) {
		solve();
	}
}