A. Doggo Recoloring

ps：注意 n == 1

B. Weakened Common Divisor

題解：WCD出現中的數必然是 < a, b >中某個數的公約數。而 < a, b > 的貢獻和 a * b 的貢獻是等價，所以gcd就好了。

const int N = 150005;struct node {LL sum, x, y; }a[N];LL gcd(LL x, LL y) {return (!y ? x : gcd(y, x % y)); }void ca(LL x) {for (int i = 2; i <= (int)sqrt(x); ++i) if (x % i == 0) {cout << i << endl;return ;}cout << x << endl; }LL n;int main() {sc(n);for (int i = 1; i <= n; ++i) {sc(a[i].x), sc(a[i].y);a[i].sum = a[i].x * a[i].y;}if (n == 1) {cout << a[1].x << endl;return 0;}LL res = gcd(a[1].x, a[2].sum);for (int i = 3; i <= n; ++i) {res = gcd(res, a[i].sum);}if (res != 1) {ca(res);return 0;}res = gcd(a[1].y, a[2].sum);for (int i = 3; i <= n; ++i) {res = gcd(res, a[i].sum);}if (res != 1){ca(res);return 0;}cout << "-1" << endl;return 0;
}

?C. Plasticine zebra

題解：觀察可得，分成幾部分考慮，中間的最大值，以及頭和尾的拼接。（細節啊啊啊啊啊，沒想清楚魯代碼是很危險的！！！！），頭部如果是奇數，那么尾部和頭部是相反的。

const int N = 150005;string s;int main() {cin >> s;int n = s.size();if (n == 1) {cout << 1 << endl;return 0;}int res = 1, ans = 0;char last = s[0];for (int i = 1; i < n; ++i) {if (last != s[i]) res++;else {upd(ans, res);res = 1;}last = s[i];}if (res) upd(ans, res);last = s[0];int l = 1, r = 1;for (int i = 1; i < n; ++i) {if (last != s[i]) l++;else break;last = s[i];}if (l & 1) {if (last == 'b') last = 'w';else last = 'b';}if (s[n - 1] != last) r = 0;else {for (int i = n - 2; ~i; --i) {if (last != s[i]) r++;else break;last = s[i];}}if ((l != 1 || r != 1) && l + r <= n) {upd(ans, l + r);}cout << ans << endl;return 0; }

?

轉載于:https://www.cnblogs.com/zgglj-com/p/9504505.html

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