Codeforces Round #425 (Div. 2)

A

#include <cstdio>#include <cstring>#include <cmath>#include <iostream>#include <algorithm>#include <sstream>#include <map>#include <set>#include <vector>#include <utility>#include <queue>#include <stack>#include <string>using namespace std;#define LL long long#define pb push_back#define mk make_pair#define pill pair<int, int>#define mst(a, b)memset(a, b, sizeof a)#define REP(i, x, n)for(int i = x; i <= n; ++i)const int MOD = 1e9;const int qq = 2e5 + 10;int main(){LL n, k;cin >> n >> k;LL tmp = n / k;if(tmp % 2 == 1) {puts("YES");} else {puts("NO");}return 0;}

B

题意：给出好字符，给出串st，st中'?'可以被好字符替代，‘*’可以被坏字符组成的字符串替代，然后给出q个询问，给出字符ct问是否能使得st等于ct

思路：关键其实就是处理*。

#include <cstdio>#include <cstring>#include <cmath>#include <iostream>#include <algorithm>#include <sstream>#include <map>#include <set>#include <vector>#include <utility>#include <queue>#include <stack>#include <string>using namespace std;#define LL long long#define pb push_back#define mk make_pair#define pill pair<int, int>#define mst(a, b)memset(a, b, sizeof a)#define REP(i, x, n)for(int i = x; i <= n; ++i)const int MOD = 1e9;const int qq = 1e5 + 10;int mp[30];char st[qq], qt[qq], ct[qq];int main(){scanf("%s", ct);int len = strlen(ct);for(int i = 0; i < len; ++i) {mp[ct[i]] = 1;}scanf("%s", st);int lens = strlen(st);int q;scanf("%d", &q);bool isexit = false;for(int i = 0; i < lens; ++i) {if(st[i] == '*') {isexit = true;break;}}int lenq = 0;if(isexit) {for(int i = 0; i < lens; ++i) {if(st[i] != '*') {qt[lenq++] = st[i];}}}while(q--) {scanf("%s", ct);int lenc = strlen(ct);bool f = true;if(isexit) {if(lens - 1 > lenc) {f = false;} else if(lens - 1 ==  lenc) {for(int i = 0; i < lenc; ++i) {if(qt[i] == '?') {if(!mp[ct[i]])f = false;} else if(qt[i] != ct[i]) {f = false;}}} else {int t = lenc - (lens - 1);for(int i = 0, j = 0; i < lens && j < lenc; ++i, ++j) {if(st[i] == '?') {if(!mp[ct[j]])f = false;} else if(st[i] == '*') {int c = 0;while(c < t) {if(mp[ct[j]]) f = false;++j, c++;}--j;} else if(st[i] != ct[j]) {f = false;} }}} else {if(lenc == lens) {for(int i = 0; i < lenc; ++i) {if(st[i] == '?') {if(!mp[ct[i]])f = false;} else if(st[i] != ct[i]) {f = false;}}} else {f = false;}}if(f)puts("YES");elseputs("NO");}return 0;}

大牛的代码就是不一样、

#include <cstdio>#include <cstring>#include <cmath>#include <iostream>#include <algorithm>#include <sstream>#include <map>#include <set>#include <vector>#include <utility>#include <queue>#include <stack>#include <string>using namespace std;#define LL long long#define pb push_back#define mk make_pair#define pill pair<int, int>#define mst(a, b)memset(a, b, sizeof a)#define REP(i, x, n)for(int i = x; i <= n; ++i)const int MOD = 1e9;const int qq = 1e5 + 10, z = 60;bool good[z];string goods, pat;int n;bool mat(string pat, string s) {if(pat.size() != s.size()) {return 0;}for(int i = 0; i < s.size(); ++i) {if(pat[i] != '?') {if(s[i] != pat[i])return 0;} else {if(!good[s[i] - 'a'])return 0;}}return 1;}int main(){ios::sync_with_stdio(0), cin.tie(0);cin >> goods >> pat >> n;for(auto c : goods) {good[c - 'a'] = 1;}auto p = pat.find('*');string s;while(n--) {cin >> s;if(s.size() < pat.size() - 1) {cout << "NO\n";} else if(p == string::npos) {cout << (mat(pat, s) ? "YES" : "NO") << endl;} else {bool ok = 1;ok &= mat(pat.substr(0, p), s.substr(0, p));reverse(pat.begin(), pat.end());reverse(s.begin(), s.end());p = pat.size() - p - 1;ok &= mat(pat.substr(0, p), s.substr(0, p));reverse(pat.begin(), pat.end());reverse(s.begin(), s.end());p = pat.size() - p - 1;for(int i = p; i < s.size() - (pat.size() - p - 1); ++i) {ok &= !good[s[i] - 'a'];}cout << (ok ? "YES" : "NO") << endl;}}return 0;}