【持续更新】2024牛客寒假算法基础集训营4 题解 | JorbanS

string solve() {int a, b, k; cin >> a >> b >> k;return a >= k * b ? "good" : "bad";
}

每次将一堆石子分成两堆，必然最后结果会多一个 $1$，因此只要看能多几个 $1$，即为可以操作的次数

string solve() {cin >> n;ll sum = 0;for (int i = 0; i < n; i ++) cin >> a[i], sum += a[i];return sum - n & 1 ? "gui" : "sweet";
}

暴力模拟

char solve() {cin >> n >> m >> x >> y;for (int i = 0; i < n; i ++) cin >> s[i];cin >> p >> q;for (int i = 0; i < q; i ++) {cin >> e[i].aa >> e[i].bb;e[i].bb --;}while (p --) {for (int i = 0; i < q; i ++) {if (e[i].aa == 1) {char c = s[e[i].bb][m - 1];for (int j = m - 1; j; j --) s[e[i].bb][j] = s[e[i].bb][j - 1];s[e[i].bb][0] = c;} else {char c = s[n - 1][e[i].bb];for (int j = n - 1; j; j --) s[j][e[i].bb] = s[j - 1][e[i].bb];s[0][e[i].bb] = c;}}}return s[-- x][-- y];
}

重新分配数，由于 $gcd\times k=sum,k\ge n$，故 $gcd|n$

注意特判 $n=1$ 时

ll solve() {cin >> n;ll sum = 0, res = 0;for (int i = 0; i < n; i ++) cin >> a[i], sum += a[i];for (int i = 1; i <= sum / n; i ++)if (sum % i == 0) res ++;if (n == 1) return 1;return res;
}	

只许看前缀和相同的区间点，每次找到后 $clear$ 前面的已经被选择过的

int solve() {cin >> n >> k;int idx = 0;map<int, int> mp;for (int i = 1; i <= n; i ++) {cin >> a[i];a[i] += a[i - 1];a[i] %= k;}set<int> s;s.insert(0);int res = 0;for (int i = 1; i <= n; i ++) {if (s.count(a[i])) {res ++;s.clear();}s.insert(a[i]);}return res;
}

前缀和查询底边是否存在，总复杂度 $O(n^3)$

ll solve() {cin >> n >> m;for (int i = 0; i < n; i ++) cin >> s[i];for (int i = 0; i < n; i ++)for (int j = 0; j < m; j ++)a[i][j + 1] = a[i][j] + (s[i][j] == '*');ll res = 0;for (int i = 0; i < n; i ++)for (int j = 1; j < m - 1; j ++) {if (s[i][j] == '.') continue;for (int k = 1; k <= min(j, min(m - 1 - j, n - 1 - i)); k ++) {if (s[i + k][j - k] == '*' && s[i + k][j + k] == '*') {if (a[i + k][j + k + 1] - a[i + k][j - k] >= k * 2 + 1) res ++;} else break;}}return res;
}