夺宝网站还可以做吗,网络服务营销,火星建站和八亿建站,网站权重怎么做Codeforces Round #260 (Div. 2) 题目链接 A#xff1a;水题#xff0c;事实上仅仅要推断有没有一个ai ! bi就可以#xff0c;由于都保证是1 - n的不相等数字 B#xff1a;找到2 3 4的循环节#xff0c;发现仅仅有4和2#xff0c;于是把大数%4#xff0c;%2#xff0c;… Codeforces Round #260 (Div. 2) 题目链接 A水题事实上仅仅要推断有没有一个ai ! bi就可以由于都保证是1 - n的不相等数字 B找到2 3 4的循环节发现仅仅有4和2于是把大数%4%2在依据循环节去计算就可以 Cdpdp[i][0]表示不拿i数字。dp[i][1]表示拿i数字。状态转移为dp(i,0)max(dp(i−1,0),dp(i−1,1)),dp(i,1)dp(i−1,0)val[i] DTrie博弈。依据字符串建Trie然后两遍dfs找出能控制自己必胜和能控制自己必败的状态假设能控制必胜又能控制必败就赢了假设仅仅能控制必胜那么假设k为奇数也是赢剩下都是输 E并查集贪心对于每一个集合两遍DFS能找出最长链然后每次合并操作的时候肯定是拿两遍的最长链中点去合并是最优的这样带个权值len表示集合最长长度就可以 代码 A: #include cstdio
#include cstring
#include algorithm
using namespace std;const int N 100005;
int n, a[N], b[N];bool judge() {for (int i 0; i n; i)if (a[i] ! b[i]) return true;return false;
}int main() {scanf(%d, n);for (int i 0; i n; i)scanf(%d%d, a[i], b[i]);if (judge()) printf(Happy Alex\n);else printf(Poor Alex\n);return 0;
} B: #include cstdio
#include cstringconst int N 100005;
char str[N];int num[5][10];int main() {num[2][0] 1; num[2][1] 2; num[2][2] 4; num[2][3] 3;num[3][0] 1; num[3][1] 3; num[3][2] 4; num[3][3] 2;num[4][0] 1; num[4][1] 4;scanf(%s, str);if (strcmp(str, 0) 0) {printf(4\n);return 0;}int yu 0;for (int i 0; i strlen(str); i) {yu (yu * 10 str[i] - 0) % 4;}int sb yu % 2;int ans (1 num[2][yu] num[3][yu] num[4][sb]) % 5;printf(%d\n, ans);return 0;
} C: #include cstdio
#include cstring
#include algorithm
using namespace std;const int N 100005;
int n;
long long vis[N], dp[N][2];int main() {int a;scanf(%d, n);for (int i 1; i n; i) {scanf(%d, a);vis[a];}for (long long i 1; i 100000; i) {dp[i][0] max(dp[i - 1][0], dp[i - 1][1]);dp[i][1] max(dp[i][1], dp[i - 1][0] vis[i] * i);}printf(%lld\n, max(dp[100000][0], dp[100000][1]));return 0;
} D: #include cstdio
#include cstringconst int MAXNODE 100005;
const int SIGMA_SIZE 26;struct Trie {int ch[MAXNODE][SIGMA_SIZE];int win[MAXNODE], lose[MAXNODE];int sz;void init() {sz 1;memset(ch[0], 0, sizeof(ch[0]));}int idx(char c) {return c - a;}void insert(char *str) {int n strlen(str);int u 0;for (int i 0; i n; i) {int c idx(str[i]);if (!ch[u][c]) {memset(ch[sz], 0, sizeof(ch[sz]));ch[u][c] sz;}u ch[u][c];}}int dfs1(int u) {win[u] 0;for (int i 0; i SIGMA_SIZE; i) {int v ch[u][i];if (!v) continue;int tmp dfs1(v);if (!tmp) win[u] 1;}return win[u];}int dfs2(int u) {lose[u] 0;int bo 1;for (int i 0; i SIGMA_SIZE; i) {int v ch[u][i];if (!v) continue;bo 0;int tmp dfs2(v);if (!tmp) lose[u] 1;}if (bo) return lose[u] 1;return lose[u];}void getsg() {dfs1(0);dfs2(0);}
};const int N 100005;
int n, k;
char str[N];
Trie gao;bool judge() {gao.init();scanf(%d%d, n, k);for (int i 0; i n; i) {scanf(%s, str);gao.insert(str);}gao.getsg();if (gao.win[0] gao.lose[0])return true;if (gao.win[0]) {if (k1) return true;return false;}return false;
}int main() {if (judge()) printf(First\n);else printf(Second\n);return 0;
} E: #include cstdio
#include cstring
#include vector
using namespace std;const int N 300005;
int n, m, q, parent[N], len[N], vis[N], Maxu, Maxl;
vectorint g[N];void dfs(int u, int h, int p) {vis[u] 1;int flag 1;for (int i 0; i g[u].size(); i) {int v g[u][i];if (v p) continue;flag 0;dfs(v, h 1, u);}if (flag) {if (h Maxl) {Maxl h;Maxu u; }}
}int find(int x) {return x parent[x] ? x : parent[x] find(parent[x]);
}int main() {scanf(%d%d%d, n, m, q);for (int i 1; i n; i)parent[i] i;int u, v;while (m--) {scanf(%d%d, u, v);int pu find(u);int pv find(v);parent[pu] pv;g[u].push_back(v);g[v].push_back(u);}for (int i 1; i n; i) {if (vis[i]) continue;Maxl -1;dfs(i, 0, 0);Maxl -1;dfs(Maxu, 0, 0);len[find(i)] Maxl;}int c, a, b;while (q--) {scanf(%d, c);if (c 1) {scanf(%d, a);printf(%d\n, len[find(a)]);}else {scanf(%d%d, a, b);int pa find(a);int pb find(b);if (pa ! pb) {parent[pa] pb;len[pb] max(len[pa], max(len[pb], (len[pb] 1) / 2 (len[pa] 1) / 2 1));}}}return 0;
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