A. Catch the Coin
解题思路:
最终𝑥一定会相等，我们考虑直接到下面接住他。
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define N 1000005
ll dp[N], w[N], v[N], h[N];
ll dis[1005][1005];
ll a, b, c, n, m, t;
ll ans, sum, num, minn = 1e9 + 7, maxx = 0;
struct node {ll a, b, c;
}f[N];
int main()
{cin >> t;while (t--){cin >> a >> b;if (b>=-1) {cout << "YES" << endl;}else {cout << "NO" << endl;}}return 0;
}B. Substring and Subsequence
解题思路:
由于 a 是肯定全部要出现的，因此只要找出在 a 中最多能匹配上 b 中的多少个字符
不难发现 b 中匹配的一定是连续的一段，因此可以暴力枚举区间然后暴力检验
代码:
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define N 1000005
ll dp[N], w[N], v[N], h[N];
ll dis[1005][1005];
ll a, b, c, n, m, t;
ll ans, sum, num, minn = 1e9 + 7, maxx = 0;
struct node {ll a, b, c;
}f[N];
string s1, s2;
int main()
{cin >> t;while (t--){cin >> s1 >> s2;ans = 0;for (int i = 0; i < s2.size(); i++) {ll num1 = i, num2 = 0;for (int j = 0; j < s1.size(); j++) {if (s1[j] == s2[num1]) {num1++;num2++;}}ans = max(ans, num2);}cout << s1.size() + s2.size() - ans << endl;}return 0;
}C. Two Movies
解题思路:
首先如果 ai,bi不同，则直接把评分加到较大的那个数上去一定是最优的，这个分析一下会发现对于所有情形成立
在此基础上可以先得到两部电影的评分，分别记为 x, y，然后考虑将剩下的1 1数量记为 X，-1 -1数量记为 Y
分类讨论一下加入 X 个 1 和 Y 个 −1 的影响即可，每次先把两个数做到平均一定是最优的
代码:
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define N 1000005
ll dp[N], w[N], v[N], h[N];
ll dis[1005][1005];
ll a, b, c, n, m, t;
ll ans, sum, num, minn = 1e9 + 7, maxx = 0;
struct node {ll a, b, c;
}f[N];
string s1, s2;
int main()
{cin >> t;while (t--){cin >> n;int x=0, y=0, X=0, Y=0;for (int i = 1; i <= n; i++)cin >> w[i];for (int j = 1; j <= n; j++)cin >> v[j];for (int i = 1; i <= n; i++) {if (w[i] != v[i]) {if (w[i] > v[i]) x+=w[i];else y+=v[i];}else {if (w[i] == 1) {X++;}else if (w[i]==-1) {Y++;}}}if (X <= abs(x - y)) {if (x <= y) x += X;else y += X;}else {int k = X - abs(x - y);maxx = max(x, y);x = y = maxx;x += k / 2, y += (k + 1) / 2;}if (Y <= abs(x - y)) {if (x <= y) y -= Y;else x -= Y;}else {int k = Y - abs(x - y);minn = min(x, y);x = y = minn;x -= k / 2, y -= (k + 1) / 2;}cout << min(x, y) << endl;}return 0;
}
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