A. Forgetting Things
Description
给出a和b的最高位，输出一组可能的a+1=b
Solution
判断a=b,a=b-1,b=9&a=1三种情况
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 int main(int argc, char const *argv[])
72 {
73 #ifndef ONLINE_JUDGE
74 freopen("in.txt", "r", stdin);
75 // freopen("out.txt", "w", stdout);
76 #endif
77 ll a, b;
78 cin >> a >> b;
79 if (a == b)
80 cout << a * 10 << " " << a * 10 + 1 << "\n";
81 else if (a == 9 && b == 1)
82 cout << 9 << " " << 10 << "\n";
83 else if (a == b - 1)
84 cout << a << " " << b << "\n";
85 else
86 cout << "-1\n";
87 return 0;
88 }
View Code
B. TV Subscriptions
找一段连续序列满足长度大于等于d且其内的不同元素最少
Solution
滑动窗口
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, d;
30 const int maxn = 1e6 + 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 unordered_map<int, int> mp;
72 unordered_set<int> s;
73 int a[maxn];
74 int main(int argc, char const *argv[])
75 {
76 #ifndef ONLINE_JUDGE
77 freopen("in.txt", "r", stdin);
78 // freopen("out.txt", "w", stdout);
79 #endif
80 int t = read<int>();
81 while (t--)
82 {
83 mp.clear();
84 s.clear();
85 n = read<int>(), k = read<int>(), d = read<int>();
86 for (int i = 1; i <= n; i++)
87 a[i] = read<int>();
88 int head;
89 int res = 1e9;
90 for (int i = 1, j = 1; i <= n; i++)
91 {
92 head = a[i];
93 while (j - i + 1 <= d && j <= n)
94 s.emplace(a[j]), ++mp[a[j]], j++;
95 if (j - i + 1 > d)
96 res = min(res, (int)s.size());
97 int t = mp[head];
98 if (t == 1)
99 s.erase(head), mp.erase(head);
100 else
101 --mp[head];
102 }
103 writeln(res);
104 }
105 return 0;
106 }
View Code
C. p-binary
定义p-二进制数如上，给定n和p，求最少的p-二进制数之和等于n
Solution
注意到如果一个数最少被表示成二进制数的数位可以贪心得到,假设为minx，那么如果当前判断的个数满足$ minx \leq x \leq newn$,其中$newn=n-x*k$
则当前x满足条件
暴力枚举
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 ll p[62];
72 void init()
73 {
74 p[0] = 1;
75 for (int i = 1; i < 62; i++)
76 p[i] = 1LL << i;
77 }
78 int main(int argc, char const *argv[])
79 {
80 #ifndef ONLINE_JUDGE
81 freopen("in.txt", "r", stdin);
82 // freopen("out.txt", "w", stdout);
83 #endif
84 init();
85 ll n = read<ll>(), k = read<ll>();
86 ll res = -1;
87 for (ll i = 1;; i++)
88 {
89 ll tmp = n - 1LL * i * k;
90 if (tmp <= 0)
91 break;
92 ll t1 = 0, t2 = tmp;
93 for (int j = 61; j >= 0 && tmp; j--)
94 if (tmp >= p[j])
95 tmp -= p[j], ++t1;
96 if (i >= t1 && i <= t2)
97 {
98 res = i;
99 break;
100 }
101 }
102 writeln(res);
103 return 0;
104 }
View Code
D. Power Products
Description
给一个长为n的序列，求满足$1 \leq i \lt j \leq n,a[i]*a[j]=x^k$的数对个数
Solution
赛后A，思路来自nb群友
对于每一个数唯一分解，且指数q对k取模，那么当$q_i \gt 0$时可以得到其a[j]需要满足$q_j=k-q_i$
对于大于1e5的数据直接舍弃，注意开ll，从右往左扫一遍即为答案
1 #include <algorithm>
2 #include <cctype>
3 #include <cmath>
4 #include <cstdio>
5 #include <cstdlib>
6 #include <cstring>
7 #include <ext/rope>
8 #include <iostream>
9 #include <map>
10 #include <numeric>
11 #include <queue>
12 #include <set>
13 #include <stack>
14 #if __cplusplus >= 201103L
15 #include <unordered_map>
16 #include <unordered_set>
17 #endif
18 #include <vector>
19 #define lson rt << 1, l, mid
20 #define rson rt << 1 | 1, mid + 1, r
21 #define LONG_LONG_MAX 9223372036854775807LL
22 #define pblank putchar(' ')
23 #define ll LL
24 #define fastIO ios::sync_with_stdio(false), cin.tie(0), cout.tie(0)
25 using namespace std;
26 using namespace __gnu_cxx;
27 typedef long long ll;
28 typedef long double ld;
29 typedef unsigned long long ull;
30 typedef pair<int, int> P;
31 int n, m, k;
32 const int maxn = 1e5 + 10;
33 template <class T>
34 inline T read()
35 {
36 int f = 1;
37 T ret = 0;
38 char ch = getchar();
39 while (!isdigit(ch))
40 {
41 if (ch == '-')
42 f = -1;
43 ch = getchar();
44 }
45 while (isdigit(ch))
46 {
47 ret = (ret << 1) + (ret << 3) + ch - '0';
48 ch = getchar();
49 }
50 ret *= f;
51 return ret;
52 }
53 template <class T>
54 inline void write(T n)
55 {
56 if (n < 0)
57 {
58 putchar('-');
59 n = -n;
60 }
61 if (n >= 10)
62 {
63 write(n / 10);
64 }
65 putchar(n % 10 + '0');
66 }
67 template <class T>
68 inline void writeln(const T &n)
69 {
70 write(n);
71 puts("");
72 }
73 unordered_map<int, int> mp;
74 int prime[maxn], vis[maxn], pcnt;
75 int a[maxn], b[maxn], c[maxn], d[maxn];
76 void init()
77 {
78 for (int i = 2; i < maxn; i++)
79 {
80 if (!vis[i])
81 prime[pcnt++] = i;
82 for (int j = 0; j < pcnt && prime[j] * i < maxn; j++)
83 {
84 vis[i * prime[j]] = 1;
85 if (i % prime[j] == 0)
86 break;
87 }
88 }
89 }
90 ll qpow(ll a, int n)
91 {
92 ll res = 1;
93 while (n)
94 {
95 if (n & 1)
96 res = res * a;
97 a = a * a;
98 n >>= 1;
99 }
100 return res;
101 }
102 int main(int argc, char const *argv[])
103 {
104 #ifndef ONLINE_JUDGE
105 freopen("in.txt", "r", stdin);
106 // freopen("out.txt", "w", stdout);
107 #endif
108 init();
109 n = read<int>(), k = read<int>();
110 for (int i = 1; i <= n; i++)
111 a[i] = read<int>();
112 for (int i = 1; i <= n; i++)
113 {
114 if (mp.count(a[i]))
115 {
116 b[i] = b[mp[a[i]]];
117 c[i] = c[mp[a[i]]];
118 continue;
119 }
120 ll newcur = 1, tocur = 1;
121 int tmp = a[i];
122 int f = 1;
123 for (int i = 0; i < pcnt && prime[pcnt] < tmp; i++)
124 if (tmp % prime[i] == 0)
125 {
126 int now = 0;
127 while (tmp % prime[i] == 0)
128 {
129 tmp /= prime[i];
130 ++now;
131 }
132 now %= k;
133 if (now)
134 {
135 newcur *= qpow(prime[i], now);
136 ll p = 1, flag = 1;
137 for (int j = 1; j <= k - now; j++)
138 {
139 p = p * prime[i];
140 if (p > 1e5)
141 {
142 flag = 0;
143 break;
144 }
145 }
146 if (flag && tocur * p <= 1e5)
147 tocur *= p;
148 else
149 f = 0;
150 }
151 }
152 if (tmp > 1)
153 {
154 newcur *= tmp;
155 ll p = 1, flag = 1;
156 for (int j = 1; j <= k - 1; j++)
157 {
158 p = p * tmp;
159 if (p > 1e5)
160 {
161 flag = 0;
162 break;
163 }
164 }
165 if (flag && tocur * p <= 1e5)
166 tocur *= p;
167 else
168 f = 0;
169 }
170 b[i] = newcur;
171 if (f && tocur <= 1e5)
172 c[i] = tocur;
173 else
174 c[i] = -1;
175 mp[a[i]] = i;
176 }
177 // for (int i = 1; i <= n; i++)
178 // cout << b[i] << " " << c[i] << "\n";
179 ll res = 0;
180 for (int i = n; i >= 1; i--)
181 {
182 res += d[c[i]];
183 d[b[i]]++;
184 }
185 writeln(res);
186 return 0;
187 }
View Code