codeforces #603 div2 ABCD

A. Sweet Problem

Description

 

 

 

Solution

瞎搞的规律

 

 

 

B. PIN Codes

Description

给出n个4位数字密码，每次操作可以改变一个密码中的一位数字。

求最少的操作次数使得n个密码不同。

Solution

hash一下序列乱搞，代码里的while应该可以去掉，扫一遍就行。

  1 #include <algorithm>
  2 #include <cctype>
  3 #include <cmath>
  4 #include <cstdio>
  5 #include <cstdlib>
  6 #include <cstring>
  7 #include <iostream>
  8 #include <map>
  9 #include <numeric>
 10 #include <queue>
 11 #include <set>
 12 #include <stack>
 13 #if __cplusplus >= 201103L
 14 #include <unordered_map>
 15 #include <unordered_set>
 16 #endif
 17 #include <vector>
 18 #define lson rt << 1, l, mid
 19 #define rson rt << 1 | 1, mid + 1, r
 20 #define LONG_LONG_MAX 9223372036854775807LL
 21 #define pblank putchar(' ')
 22 #define ll LL
 23 #define fastIO ios::sync_with_stdio(false), cin.tie(0), cout.tie(0)
 24 using namespace std;
 25 typedef long long ll;
 26 typedef long double ld;
 27 typedef unsigned long long ull;
 28 typedef pair<int, int> P;
 29 int n, m, k;
 30 const int maxn = 1e5 + 10;
 31 template <class T>
 32 inline T read()
 33 {
 34     int f = 1;
 35     T ret = 0;
 36     char ch = getchar();
 37     while (!isdigit(ch))
 38     {
 39         if (ch == '-')
 40             f = -1;
 41         ch = getchar();
 42     }
 43     while (isdigit(ch))
 44     {
 45         ret = (ret << 1) + (ret << 3) + ch - '0';
 46         ch = getchar();
 47     }
 48     ret *= f;
 49     return ret;
 50 }
 51 template <class T>
 52 inline void write(T n)
 53 {
 54     if (n < 0)
 55     {
 56         putchar('-');
 57         n = -n;
 58     }
 59     if (n >= 10)
 60     {
 61         write(n / 10);
 62     }
 63     putchar(n % 10 + '0');
 64 }
 65 template <class T>
 66 inline void writeln(const T &n)
 67 {
 68     write(n);
 69     puts("");
 70 }
 71 template <typename T>
 72 void _write(const T &t)
 73 {
 74     write(t);
 75 }
 76 template <typename T, typename... Args>
 77 void _write(const T &t, Args... args)
 78 {
 79     write(t), pblank;
 80     _write(args...);
 81 }
 82 template <typename T, typename... Args>
 83 inline void write_line(const T &t, const Args &... data)
 84 {
 85     _write(t, data...);
 86     puts("");
 87 }
 88 vector<int> g[100];
 89 inline int _hash(vector<int> &t)
 90 {
 91     int p = 1;
 92     for (int i = 0; i < 4; i++)
 93         p = (p * 2017 + t[i]) % 10007;
 94     return p;
 95 }
 96 inline int isok()
 97 {
 98     unordered_set<int> s;
 99     for (int i = 0; i < n; i++)
100     {
101         int p = _hash(g[i]);
102         if (s.count(p))
103             return 0;
104         s.emplace(p);
105     }
106     return 1;
107 }
108 int main(int argc, char const *argv[])
109 {
110 #ifndef ONLINE_JUDGE
111     freopen("in.txt", "r", stdin);
112     // freopen("out.txt", "w", stdout);
113 #endif
114     int t = read<int>();
115     while (t--)
116     {
117         n = read<int>();
118         for (int i = 0; i < n; i++)
119             g[i].clear();
120         for (int i = 0; i < n; i++)
121         {
122             char s[10];
123             scanf("%s", s);
124             for (int j = 0; j < 4; j++)
125                 g[i].emplace_back(s[j] - '0');
126         }
127         int res = 0;
128         while (!isok())
129         {
130             unordered_map<int, int> mp;
131             mp.clear();
132             for (int i = 0; i < n; i++)
133                 mp[_hash(g[i])]++;
134             for (int i = 0; i < n; i++)
135             {
136                 int p = _hash(g[i]);
137                 if (mp[p] == 1)
138                     continue;
139                 for (int j = 0; j < 4; j++)
140                 {
141                     int f = 1;
142                     for (int k = 0; k < 10 && f; k++)
143                     {
144                         int tmp = g[i][j];
145                         g[i][j] = k;
146                         if (!mp[_hash(g[i])])
147                         {
148                             mp[_hash(g[i])] = 1;
149                             f = 0;
150                             ++res;
151                         }
152                     }
153                     if (!f)
154                         break;
155                 }
156                 mp[p]--;
157             }
158         }
159         writeln(res);
160         for (int i = 0; i < n; i++)
161         {
162             for (int j : g[i])
163                 write(j);
164             puts("");
165         }
166     }
167     return 0;
168 }
View Code

C. Everyone is a Winner!

Description

 

 

Solution

数论分块儿裸题，加上一个0即是答案。

  1 #include <algorithm>
  2 #include <cctype>
  3 #include <cmath>
  4 #include <cstdio>
  5 #include <cstdlib>
  6 #include <cstring>
  7 #include <iostream>
  8 #include <map>
  9 #include <numeric>
 10 #include <queue>
 11 #include <set>
 12 #include <stack>
 13 #if __cplusplus >= 201103L
 14 #include <unordered_map>
 15 #include <unordered_set>
 16 #endif
 17 #include <vector>
 18 #define lson rt << 1, l, mid
 19 #define rson rt << 1 | 1, mid + 1, r
 20 #define LONG_LONG_MAX 9223372036854775807LL
 21 #define pblank putchar(' ')
 22 #define ll LL
 23 #define fastIO ios::sync_with_stdio(false), cin.tie(0), cout.tie(0)
 24 using namespace std;
 25 typedef long long ll;
 26 typedef long double ld;
 27 typedef unsigned long long ull;
 28 typedef pair<int, int> P;
 29 int n, m, k;
 30 const int maxn = 1e5 + 10;
 31 template <class T>
 32 inline T read()
 33 {
 34     int f = 1;
 35     T ret = 0;
 36     char ch = getchar();
 37     while (!isdigit(ch))
 38     {
 39         if (ch == '-')
 40             f = -1;
 41         ch = getchar();
 42     }
 43     while (isdigit(ch))
 44     {
 45         ret = (ret << 1) + (ret << 3) + ch - '0';
 46         ch = getchar();
 47     }
 48     ret *= f;
 49     return ret;
 50 }
 51 template <class T>
 52 inline void write(T n)
 53 {
 54     if (n < 0)
 55     {
 56         putchar('-');
 57         n = -n;
 58     }
 59     if (n >= 10)
 60     {
 61         write(n / 10);
 62     }
 63     putchar(n % 10 + '0');
 64 }
 65 template <class T>
 66 inline void writeln(const T &n)
 67 {
 68     write(n);
 69     puts("");
 70 }
 71 template <typename T>
 72 void _write(const T &t)
 73 {
 74     write(t);
 75 }
 76 template <typename T, typename... Args>
 77 void _write(const T &t, Args... args)
 78 {
 79     write(t), pblank;
 80     _write(args...);
 81 }
 82 template <typename T, typename... Args>
 83 inline void write_line(const T &t, const Args &... data)
 84 {
 85     _write(t, data...);
 86     puts("");
 87 }
 88 int main(int argc, char const *argv[])
 89 {
 90 #ifndef ONLINE_JUDGE
 91     freopen("in.txt", "r", stdin);
 92     // freopen("out.txt", "w", stdout);
 93 #endif
 94     int t = read<int>();
 95     while (t--)
 96     {
 97         n = read<int>();
 98         vector<int> res;
 99         res.clear();
100         res.emplace_back(0);
101         for (int i = 1, j; i <= n; i = j + 1)
102         {
103             res.emplace_back(n / i);
104             j = n / (n / i);
105         }
106         writeln(res.size());
107         sort(res.begin(), res.end());
108         for (int x : res)
109             write(x), pblank;
110         puts("");
111     }
112 
113     return 0;
114 }
View Code

D. Secret Passwords

Description

盗贼剽窃到了n个由小写字母构成的密码串。

密码串T,S被认为相同效果的条件：

1，存在一个字符同时出现在T,S中。

2，存在第三个密码串P，满足P=S&&P=T

两者满足其一即可。

问盗贼最少要试多少个密码才能打开保险柜，保险柜仅有一个正确密码。

Solution

并查集维护连通块。

  1 #include <algorithm>
  2 #include <cctype>
  3 #include <cmath>
  4 #include <cstdio>
  5 #include <cstdlib>
  6 #include <cstring>
  7 #include <iostream>
  8 #include <map>
  9 #include <numeric>
 10 #include <queue>
 11 #include <set>
 12 #include <stack>
 13 #if __cplusplus >= 201103L
 14 #include <unordered_map>
 15 #include <unordered_set>
 16 #endif
 17 #include <vector>
 18 #define lson rt << 1, l, mid
 19 #define rson rt << 1 | 1, mid + 1, r
 20 #define LONG_LONG_MAX 9223372036854775807LL
 21 #define pblank putchar(' ')
 22 #define ll LL
 23 #define fastIO ios::sync_with_stdio(false), cin.tie(0), cout.tie(0)
 24 using namespace std;
 25 typedef long long ll;
 26 typedef long double ld;
 27 typedef unsigned long long ull;
 28 typedef pair<int, int> P;
 29 int n, m, k;
 30 const int maxn = 2e5 + 10;
 31 template <class T>
 32 inline T read()
 33 {
 34     int f = 1;
 35     T ret = 0;
 36     char ch = getchar();
 37     while (!isdigit(ch))
 38     {
 39         if (ch == '-')
 40             f = -1;
 41         ch = getchar();
 42     }
 43     while (isdigit(ch))
 44     {
 45         ret = (ret << 1) + (ret << 3) + ch - '0';
 46         ch = getchar();
 47     }
 48     ret *= f;
 49     return ret;
 50 }
 51 template <class T>
 52 inline void write(T n)
 53 {
 54     if (n < 0)
 55     {
 56         putchar('-');
 57         n = -n;
 58     }
 59     if (n >= 10)
 60     {
 61         write(n / 10);
 62     }
 63     putchar(n % 10 + '0');
 64 }
 65 template <class T>
 66 inline void writeln(const T &n)
 67 {
 68     write(n);
 69     puts("");
 70 }
 71 template <typename T>
 72 void _write(const T &t)
 73 {
 74     write(t);
 75 }
 76 template <typename T, typename... Args>
 77 void _write(const T &t, Args... args)
 78 {
 79     write(t), pblank;
 80     _write(args...);
 81 }
 82 template <typename T, typename... Args>
 83 inline void write_line(const T &t, const Args &... data)
 84 {
 85     _write(t, data...);
 86     puts("");
 87 }
 88 int fa[maxn];
 89 int ch[26];
 90 inline int getfa(int x)
 91 {
 92     if (!fa[x])
 93         return x;
 94     return fa[x] = getfa(fa[x]);
 95 }
 96 inline void merge(int x, int y)
 97 {
 98     x = getfa(x);
 99     y = getfa(y);
100     if (x != y)
101         fa[y] = x;
102 }
103 char s[55];
104 int main(int argc, char const *argv[])
105 {
106 #ifndef ONLINE_JUDGE
107     freopen("in.txt", "r", stdin);
108     // freopen("out.txt", "w", stdout);
109 #endif
110     n = read<int>();
111     memset(ch, -1, sizeof(ch));
112     for (int i = 0; i < n; i++)
113     {
114         scanf("%s", s);
115         int len = strlen(s);
116         for (int j = 0; j < len; j++)
117         {
118             int idx = s[j] - 'a';
119             if (ch[idx] == -1)
120                 ch[idx] = i;
121             else
122                 merge(i, ch[idx]);
123         }
124     }
125     int res = 0;
126     for (int i = 0; i < n; i++)
127         if (!fa[i])
128             ++res;
129     writeln(res);
130     return 0;
131 }
View Code

 

posted @ 2019-11-30 13:13  mool  阅读(189)  评论(0)  编辑  收藏  举报