Codeforces Round 923 (Div. 3)
Codeforces Round 923 (Div. 3)
A. Make it White
题意:略
思路:找最小和最大的‘B’下标即可
AC code:
void solve() {cin >>n;string s; cin>> s;int mn = INF, mx = 0;for (int i = 0; i < n; i ++) {if (s[i] == 'B') {mn = min(mn, i);mx = max(mx, i);}}cout << mx - mn + 1 << endl;
}
B. Following the String
题意:略
思路:map记录26个字母的出现次数,0即是用新的字母
AC code:
void solve() {cin >> n;for (int i = 1; i <= n; i ++) cin >> a[i];map<char, int> mp;string s = "";for (int i = 1; i <= n; i ++) {for (char c = 'a'; c <= 'z'; c ++) {if (mp[c] == a[i]) {s += c;mp[c] ++;break;}}} cout << s << endl;
}
C. Choose the Different Ones!
题意:给出n个a数组元素和m个b数组元素,是否可以分别从a和b数组中取k/2个元素来组成1到k的每个元素
思路:用map分别记录ab数组元素,然后枚举1到k的元素,若有未出现的直接返回NO,然后记录ab数组的交集元素,最后检查仅存在于其中一个集合的元素是否大于k/2,若不存在则YES
AC code:
void solve() {cin >> n >> m >> k;map<int, int> ma, mb;for (int i = 1; i <= n; i ++) {int x; cin >> x;if (x <= k && x >= 1) ma[x] ++;}for (int i = 1; i <= m; i ++) {int x; cin >> x;if (x <= k && x >= 1) mb[x] ++;}int ca = ma.size(), cb = mb.size();int cnt = 0;for (int i = 1; i <= k; i ++) {if (!ma[i] && !mb[i]) {cout << "NO" << endl;return;}if (ma[i] && mb[i]) cnt ++;}if (ca - cnt > k / 2 || cb - cnt > k / 2) {cout << "NO" << endl;return;}cout << "YES" << endl;
}
D. Find the Different Ones!
题意:查询任意区间元素是否存在两个不同的元素
思路:从后往前记录最近的一个不同元素的下标,依次向前迭代
AC code:
void solve() {cin >> n;for (int i = 1; i <= n; i ++) cin >> a[i];nex[n] = -1;for (int i = n - 1; i >= 1; i --) {if (a[i] != a[i + 1]) nex[i] = i + 1;else nex[i] = nex[i + 1];}cin >> q;while (q --) {int l, r; cin >> l >> r;if (nex[l] != -1 && nex[l] <= r) {cout << l << " " << nex[l] << endl;} else {cout << "-1 -1" << endl;}}
}
E. Klever Permutation
题意:将1到n的数排列组合,使得每组相邻的k个元素的和相差不超过1
思路:
可以发现,每组向前迭代的过程都是先去掉一个头元素,再添一个尾元素;
那么只要去掉的和新的元素差不超过1即可,而两元素下标差为k;
所以我们可以枚举答案序列的前k个起始元素,每个元素以下标+k的方式迭代到最后;
这样可以满足答案序列每组删除和新增的元素差为1,但是不能保证每一组的和都是相差小于1;
所以可以从小到大一组,从大到小一组,通过一个标记点可以实现;
AC code:
void solve() {cin >> n >> k;vector<int> ans(n + 10, 0);int st = 1;bool flag = true;for (int i = 1; i <= k; i ++) {int x = i;if (flag) {while (x <= n) {ans[x] = st;st ++;x += k;}flag = false;} else {while (x <= n) {st ++;x += k;}int now;if (x > n) x -= k, now = st - 1;x = i;while (x <= n) {ans[x] = now;now --;x += k;}flag = true;}}for (int i = 1; i <= n; i ++) cout << ans[i] << " ";cout << endl;
}
F待补。。。。。。