本文主要是介绍Educational Codeforces Round 166 (Rated for Div. 2) (A~C)，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！
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;elsecout << "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] << ' ';elsecout << sum2 - b[i] << ' ';}else {if (q[i])cout << sum1 - b[i] << ' ';elsecout << sum2 - a[i] << ' ';}*/
}
int main()
{cin >> t;while (t--) {solve();}
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