A. Stones
Description
给出3堆物品，个数分别为a,b,c
有两种取数方式，a1b2，b1c2，问最多取多少物品
Solution
$O(n^2)暴力$
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 <queue> 10 #include <set> 11 #include <stack> 12 #if __cplusplus >= 201103L 13 #include <unordered_map> 14 #include <unordered_set> 15 #endif 16 #include <vector> 17 #define lson rt << 1, l, mid 18 #define rson rt << 1 | 1, mid + 1, r 19 #define LONG_LONG_MAX 9223372036854775807LL 20 #define ll LL 21 using namespace std; 22 typedef long long ll; 23 typedef long double ld; 24 typedef unsigned long long ull; 25 typedef pair<int, int> P; 26 int n, m, k; 27 const int maxn = 1e5 + 10; 28 template <class T> 29 inline T read() 30 { 31 int f = 1; 32 T ret = 0; 33 char ch = getchar(); 34 while (!isdigit(ch)) 35 { 36 if (ch == '-') 37 f = -1; 38 ch = getchar(); 39 } 40 while (isdigit(ch)) 41 { 42 ret = (ret << 1) + (ret << 3) + ch - '0'; 43 ch = getchar(); 44 } 45 ret *= f; 46 return ret; 47 } 48 template <class T> 49 inline void write(T n) 50 { 51 if (n < 0) 52 { 53 putchar('-'); 54 n = -n; 55 } 56 if (n >= 10) 57 { 58 write(n / 10); 59 } 60 putchar(n % 10 + '0'); 61 } 62 template <class T> 63 inline void writeln(const T &n) 64 { 65 write(n); 66 puts(""); 67 } 68 int main(int argc, char const *argv[]) 69 { 70 #ifndef ONLINE_JUDGE 71 freopen("in.txt", "r", stdin); 72 freopen("out.txt", "w", stdout); 73 #endif 74 int t = read<int>(); 75 while (t--) 76 { 77 int a = read<int>(), b = read<int>(), c = read<int>(); 78 int res = 0; 79 for (int t1 = 0; t1 <= 100; t1++) 80 for (int t2 = 0; t2 <= 100; t2++) 81 if (t1 <= a && 2 * t1 + t2 <= b && 2 * t2 <= c) 82 res = max(res, 3 * (t1 + t2)); 83 writeln(res); 84 } 85 return 0; 86 }
B. Alice and the List of Presents
Description
给出n种无穷多的球，m个有差别的盒子，要求每种球至少取一个放在某个盒子里的方案数。
Solution
人肉打表找规律
$res={(2^m-1)}^n$
快速幂取模
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 <queue> 10 #include <set> 11 #include <stack> 12 #if __cplusplus >= 201103L 13 #include <unordered_map> 14 #include <unordered_set> 15 #endif 16 #include <vector> 17 #define lson rt << 1, l, mid 18 #define rson rt << 1 | 1, mid + 1, r 19 #define LONG_LONG_MAX 9223372036854775807LL 20 #define ll LL 21 using namespace std; 22 typedef long long ll; 23 typedef long double ld; 24 typedef unsigned long long ull; 25 typedef pair<int, int> P; 26 int n, m, k; 27 const int maxn = 1e5 + 10; 28 const int mod = 1e9 + 7; 29 template <class T> 30 inline T read() 31 { 32 int f = 1; 33 T ret = 0; 34 char ch = getchar(); 35 while (!isdigit(ch)) 36 { 37 if (ch == '-') 38 f = -1; 39 ch = getchar(); 40 } 41 while (isdigit(ch)) 42 { 43 ret = (ret << 1) + (ret << 3) + ch - '0'; 44 ch = getchar(); 45 } 46 ret *= f; 47 return ret; 48 } 49 template <class T> 50 inline void write(T n) 51 { 52 if (n < 0) 53 { 54 putchar('-'); 55 n = -n; 56 } 57 if (n >= 10) 58 { 59 write(n / 10); 60 } 61 putchar(n % 10 + '0'); 62 } 63 template <class T> 64 inline void writeln(const T &n) 65 { 66 write(n); 67 puts(""); 68 } 69 ll qpow(ll a, int b) 70 { 71 ll res = 1; 72 while (b) 73 { 74 if (b & 1) 75 res = res * a % mod; 76 a = a * a % mod; 77 b >>= 1; 78 } 79 return res; 80 } 81 int main(int argc, char const *argv[]) 82 { 83 #ifndef ONLINE_JUDGE 84 freopen("in.txt", "r", stdin); 85 freopen("out.txt", "w", stdout); 86 #endif 87 n = read<int>(), m = read<int>(); 88 writeln(qpow(qpow(2, m) - 1, n)); 89 return 0; 90 }
C. Labs
Description
Solution
将$n^2$个元素按大小分为n组，大小大小...组合
不知道咋证明
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 <queue> 10 #include <set> 11 #include <stack> 12 #if __cplusplus >= 201103L 13 #include <unordered_map> 14 #include <unordered_set> 15 #endif 16 #include <vector> 17 #define lson rt << 1, l, mid 18 #define rson rt << 1 | 1, mid + 1, r 19 #define LONG_LONG_MAX 9223372036854775807LL 20 #define ll LL 21 using namespace std; 22 typedef long long ll; 23 typedef long double ld; 24 typedef unsigned long long ull; 25 typedef pair<int, int> P; 26 int n, m, k; 27 const int maxn = 1e5 + 10; 28 template <class T> 29 inline T read() 30 { 31 int f = 1; 32 T ret = 0; 33 char ch = getchar(); 34 while (!isdigit(ch)) 35 { 36 if (ch == '-') 37 f = -1; 38 ch = getchar(); 39 } 40 while (isdigit(ch)) 41 { 42 ret = (ret << 1) + (ret << 3) + ch - '0'; 43 ch = getchar(); 44 } 45 ret *= f; 46 return ret; 47 } 48 template <class T> 49 inline void write(T n) 50 { 51 if (n < 0) 52 { 53 putchar('-'); 54 n = -n; 55 } 56 if (n >= 10) 57 { 58 write(n / 10); 59 } 60 putchar(n % 10 + '0'); 61 } 62 template <class T> 63 inline void writeln(const T &n) 64 { 65 write(n); 66 puts(""); 67 } 68 int a[maxn]; 69 set<int> s[305]; 70 int main(int argc, char const *argv[]) 71 { 72 #ifndef ONLINE_JUDGE 73 freopen("in.txt", "r", stdin); 74 freopen("out.txt", "w", stdout); 75 #endif 76 n = read<int>(); 77 for (int i = 1; i <= n; i++) 78 for (int j = 1; j <= n; j++) 79 s[i].emplace((i - 1) * n + j); 80 for (int i = 1; i <= n; i++) 81 { 82 for (int j = 1; j <= n; j++) 83 { 84 if (!(j & 1)) 85 { 86 write(*s[j].begin()); 87 putchar(' '); 88 s[j].erase(s[j].begin()); 89 } 90 else 91 { 92 write(*s[j].rbegin()); 93 putchar(' '); 94 s[j].erase(*s[j].rbegin()); 95 } 96 } 97 puts(""); 98 } 99 return 0; 100 }
D. Alice and the Doll
Description
Solution
赛后补题，模拟题意。
每次直走到不能走为止，然后右转继续，蛇形走位。
记录四个边界，最上一行，最下一行，最左一列，最右一列。每次更新坐标需要在边界以内。
注意开始可以直接右转往下走。
1 /* 2 蛇形走位 3 */ 4 #include <algorithm> 5 #include <cctype> 6 #include <cmath> 7 #include <cstdio> 8 #include <cstdlib> 9 #include <cstring> 10 #include <iostream> 11 #include <map> 12 #include <queue> 13 #include <set> 14 #include <stack> 15 #if __cplusplus >= 201103L 16 #include <unordered_map> 17 #include <unordered_set> 18 #endif 19 #include <vector> 20 #define lson rt << 1, l, mid 21 #define rson rt << 1 | 1, mid + 1, r 22 #define LONG_LONG_MAX 9223372036854775807LL 23 #define ll LL 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 set<int> xi[maxn], yi[maxn], xd[maxn], yd[maxn]; 72 int x, y; 73 int l, r, u, d, dic; 74 void init() 75 { 76 x = y = 1; 77 dic = 0; 78 l = u = 0; 79 r = m + 1, d = n + 1; 80 } 81 ll solve() 82 { 83 ll cnt = 1; 84 while (1) 85 { 86 if (dic == 0) 87 { 88 int cur = *xi[x].upper_bound(y); 89 cur = min(cur, r); 90 if (cur - y == 1) 91 return cnt; 92 cnt += cur - y - 1; 93 y = cur - 1; 94 u = max(x, u); //右是由上转移而来，更新上边界 95 } 96 else if (dic == 1) 97 { 98 int cur = *yi[y].upper_bound(x); 99 cur = min(cur, d); 100 if (cur - x == 1) 101 return cnt; 102 cnt += cur - x - 1; 103 x = cur - 1; 104 r = min(r, y); //下是由右转移而来，更新右边界 105 } 106 else if (dic == 2) 107 { 108 int cur = -*xd[x].upper_bound(-y); 109 cur = max(cur, l); 110 if (y - cur == 1) 111 return cnt; 112 cnt += y - cur - 1; 113 y = cur + 1; 114 d = min(d, x); 115 } 116 else if (dic == 3) 117 { 118 int cur = -*yd[y].upper_bound(-x); 119 cur = max(cur, u); 120 if (x - cur == 1) 121 return cnt; 122 cnt += x - cur - 1; 123 x = cur + 1; 124 l = max(l, y); 125 } 126 dic++; 127 if (dic == 4) 128 dic = 0; 129 } 130 return cnt; 131 } 132 int main(int argc, char const *argv[]) 133 { 134 #ifndef ONLINE_JUDGE 135 freopen("in.txt", "r", stdin); 136 // freopen("out.txt", "w", stdout); 137 #endif 138 n = read<int>(), m = read<int>(); 139 k = read<int>(); 140 for (int i = 1; i <= n; i++) 141 xi[i].emplace(0), xi[i].emplace(m + 1), xd[i].emplace(0), xd[i].emplace(-m - 1); 142 for (int i = 1; i <= m; i++) 143 yi[i].emplace(0), yi[i].emplace(n + 1), yi[i].emplace(0), yi[i].emplace(-n - 1); 144 for (int i = 0; i < k; i++) 145 { 146 int a = read<int>(), b = read<int>(); 147 xi[a].emplace(b); 148 yi[b].emplace(a); 149 xd[a].emplace(-b); 150 yd[b].emplace(-a); 151 } 152 init(); 153 ll cur = solve(); 154 if (cur == 1LL * n * m - k) 155 puts("Yes"); 156 else 157 { 158 init(); 159 dic = 1; 160 cur = solve(); 161 if (cur == 1LL * n * m - k) 162 puts("Yes"); 163 else 164 puts("No"); 165 } 166 return 0; 167 }