Codeforces Round #405 (rated, Div. 2, based on VK Cup 2017 Round 1)

来源：互联网发布：centos更新python 编辑：程序博客网时间：2024/04/29 01:33

#include <cstdio>#include <cmath>#include <cstring>#include <cctype>#include <sstream>#include <iostream>#include <algorithm>#include <queue>#include <stack>#include <vector>#include <string>#include <map>#include <set>#include <utility>using namespace std;#define LL long long#define mst(Arr, x)memset(Arr, x, sizeof(Arr));#define pb push_back#define mk make_pair#define REP(i, x, n)for(int i = x; i < n; ++i)const int qq = 1e5 + 10;const int INF = 1e9 + 10;int n;int gcd(int a, int b){return b == 0 ? a : gcd(b, a % b);}int lcm(int a, int b){return a / gcd(a, b) * b;}int main(){LL a, b;scanf("%lld%lld", &a, &b);if(a >= b){printf("1\n");return 0;}for(int i = 2; ; ++i){a = a * 3LL;b = b * 2LL;if(a > b){printf("%d\n", i);return 0;}}return 0;}

對於每一個連通塊，都要求是一個完全字圖

也就是每一個節點的度數都等於這個連通塊頂點的個數 - 1

#include <cstdio>#include <cmath>#include <cstring>#include <cctype>#include <sstream>#include <iostream>#include <algorithm>#include <queue>#include <stack>#include <vector>#include <string>#include <map>#include <set>#include <utility>using namespace std;#define LL long long#define mst(Arr, x)memset(Arr, x, sizeof(Arr));#define pb push_back#define mk make_pair#define REP(i, x, n)for(int i = x; i < n; ++i)const int qq = 3e5 + 10;const int INF = 1e9 + 10;int n, m;int deg[qq];int gcd(int a, int b){return b == 0 ? a : gcd(b, a % b);}int lcm(int a, int b){return a / gcd(a, b) * b;}int fa[qq];int Find(int x){return fa[x] == -1 ? x : fa[x] = Find(fa[x]);}queue<int> Q[qq];int main(){mst(fa, -1);mst(deg, 0);scanf("%d%d", &n, &m);for(int i = 0; i < m; ++i){int a, b;scanf("%d%d", &a, &b);deg[a]++, deg[b]++;int x = Find(a), y = Find(b);if(x != y)fa[y] = x;}for(int i = 1; i <= n; ++i){Q[Find(i)].push(i);}bool flag = true;for(int i = 1; i <= n; ++i){if(Q[i].size() == 0)continue;int k = Q[i].size() - 1;Q[i].pop();while(!Q[i].empty()){int v = Q[i].front();Q[i].pop();if(k != deg[v])flag = false;}}if(flag)printf("YES\n");elseprintf("NO\n");return 0;}

對於每一個假設第i個區間是NO，假設這個區間的最右端是r，讓str[r] = str[r - k + 1] 相等即可，它不影響其他的連續區間

#include <cstdio>#include <cmath>#include <cstring>#include <cctype>#include <sstream>#include <iostream>#include <algorithm>#include <queue>#include <stack>#include <vector>#include <string>#include <map>#include <set>#include <utility>using namespace std;#define LL long long#define mst(Arr, x)memset(Arr, x, sizeof(Arr));#define pb push_back#define mk make_pair#define REP(i, x, n)for(int i = x; i < n; ++i)const int qq = 1e5 + 10;const int INF = 1e9 + 10;int n, k;int gcd(int a, int b){return b == 0 ? a : gcd(b, a % b);}int lcm(int a, int b){return a / gcd(a, b) * b;}string str[105];string name[105];int num[105];string f;map<string, bool> mp;int main(){for(int i = 1; i <= 26; ++i)name[i] = 'A' + (i - 1);for(int i = 27; i <= 26*2; ++i){name[i] = name[i - 26] + "c";}cin >> n >> k;int p = 1;for(int i = 1; i < k; ++i)str[i] = name[p++];for(int i = k; i <= n; ++i){cin >> f;if(f == "YES"){str[i] = name[p++];}else{str[i] = str[i - k + 1];}}for(int i = 1; i <= n; ++i){if(str[i] == "")str[i] = name[p++];}for(int i = 1; i <= n; ++i)cout << str[i] << " ";return 0;}

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