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 << " "; 81 else if (a == 9 && b == 1) 82 cout << 9 << " " << 10 << " "; 83 else if (a == b - 1) 84 cout << a << " " << b << " "; 85 else 86 cout << "-1 "; 87 return 0; 88 }
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 }
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 }
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] << " "; 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 }