A. Distinct Digits
Description
Solution
B. Filling the Grid
Description
Solution
模拟题意，找出没有被固定的方格个数，快速幂。
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 const ll mod = 1000000007; 32 template <class T> 33 inline T read() 34 { 35 int f = 1; 36 T ret = 0; 37 char ch = getchar(); 38 while (!isdigit(ch)) 39 { 40 if (ch == '-') 41 f = -1; 42 ch = getchar(); 43 } 44 while (isdigit(ch)) 45 { 46 ret = (ret << 1) + (ret << 3) + ch - '0'; 47 ch = getchar(); 48 } 49 ret *= f; 50 return ret; 51 } 52 template <class T> 53 inline void write(T n) 54 { 55 if (n < 0) 56 { 57 putchar('-'); 58 n = -n; 59 } 60 if (n >= 10) 61 { 62 write(n / 10); 63 } 64 putchar(n % 10 + '0'); 65 } 66 template <class T> 67 inline void writeln(const T &n) 68 { 69 write(n); 70 puts(""); 71 } 72 template <typename T> 73 void _write(const T &t) 74 { 75 write(t); 76 } 77 template <typename T, typename... Args> 78 void _write(const T &t, Args... args) 79 { 80 write(t), pblank; 81 _write(args...); 82 } 83 template <typename T, typename... Args> 84 inline void write_line(const T &t, const Args &... data) 85 { 86 _write(t, data...); 87 } 88 int h, w; 89 int r[maxn], c[maxn]; 90 int mp[1010][1010]; 91 int main(int argc, char const *argv[]) 92 { 93 #ifndef ONLINE_JUDGE 94 freopen("in.txt", "r", stdin); 95 // freopen("out.txt","w", stdout); 96 #endif 97 h = read<int>(), w = read<int>(); 98 for (int i = 1; i <= h; i++) 99 r[i] = read<int>(); 100 for (int i = 1; i <= w; i++) 101 c[i] = read<int>(); 102 for (int i = 1; i <= h; i++) 103 if (r[i]) 104 { 105 for (int j = 1; j <= r[i]; j++) 106 mp[i][j] = 1; 107 mp[i][r[i] + 1] = -1; 108 } 109 else 110 mp[i][1] = -1; 111 int f = 1; 112 for (int i = 1; i <= w; i++) 113 if (c[i]) 114 { 115 for (int j = 1; j <= c[i]; j++) 116 { 117 if (mp[j][i] == -1) 118 { 119 f = 0; 120 break; 121 } 122 mp[j][i] = 1; 123 } 124 if (mp[c[i] + 1][i] == 1) 125 { 126 f = 0; 127 break; 128 } 129 else 130 mp[c[i] + 1][i] = -1; 131 } 132 else 133 { 134 if (mp[1][i] == 1) 135 { 136 f = 0; 137 break; 138 } 139 else 140 mp[1][i] = -1; 141 } 142 ll res = 0; 143 for (int i = 1; i <= h; i++) 144 for (int j = 1; j <= w; j++) 145 if (!mp[i][j]) 146 ++res; 147 ll ans = 1; 148 ll base = 2; 149 for (; res; res >>= 1) 150 { 151 if (res & 1) 152 ans = ans * base % mod; 153 base = base * base % mod; 154 } 155 if (f) 156 writeln(ans); 157 else 158 puts("0"); 159 return 0; 160 }
C. Primes and Multiplication
Description
Solution
注意到g(x,y)是计算x里y的最大幂次约数。
那么答案可以转换为$n!$里包含的x素因子的最大幂次约数之积。
唯一分解+阶乘素因子分解。
！！！由于n数量级在1e18，其阶乘的素因子指数有可能爆int，注意开ll。（wa了一发）
1 #include <algorithm> 2 #include <numeric> 3 #include <cctype> 4 #include <cmath> 5 #include <cstdio> 6 #include <cstdlib> 7 #include <cstring> 8 #include <iostream> 9 #include <map> 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 ll n, m, k; 30 const int maxn = 1e6 + 10; 31 const ll mod = 1e9 + 7; 32 template <class T> 33 inline T read() 34 { 35 int f = 1; 36 T ret = 0; 37 char ch = getchar(); 38 while (!isdigit(ch)) 39 { 40 if (ch == '-') 41 f = -1; 42 ch = getchar(); 43 } 44 while (isdigit(ch)) 45 { 46 ret = (ret << 1) + (ret << 3) + ch - '0'; 47 ch = getchar(); 48 } 49 ret *= f; 50 return ret; 51 } 52 template <class T> 53 inline void write(T n) 54 { 55 if (n < 0) 56 { 57 putchar('-'); 58 n = -n; 59 } 60 if (n >= 10) 61 { 62 write(n / 10); 63 } 64 putchar(n % 10 + '0'); 65 } 66 template <class T> 67 inline void writeln(const T &n) 68 { 69 write(n); 70 puts(""); 71 } 72 template <typename T> 73 void _write(const T &t) 74 { 75 write(t); 76 } 77 template <typename T, typename... Args> 78 void _write(const T &t, Args... args) 79 { 80 write(t), pblank; 81 _write(args...); 82 } 83 template <typename T, typename... Args> 84 inline void write_line(const T &t, const Args &... data) 85 { 86 _write(t, data...); 87 puts(""); 88 } 89 int prime[maxn], vis[maxn], pcnt; 90 void init(){ 91 for (int i = 2; i < maxn;i++){ 92 if (!vis[i]) 93 prime[pcnt++] = i; 94 for (int j = 0; j < pcnt && i * prime[j] < maxn;j++){ 95 vis[i * prime[j]] = 1; 96 if (i%prime[j]==0) 97 break; 98 } 99 } 100 } 101 ll fac[maxn], tot; 102 inline ll qpow(ll base,ll n){ 103 ll res = 1; 104 while(n){ 105 if(n&1) 106 res = res * base % mod; 107 base = base * base % mod; 108 n >>= 1; 109 } 110 return res; 111 } 112 ll f(ll n, int p) 113 { 114 if (n == 0) 115 return 0; 116 return f(n / p, p) + n / p; 117 } 118 int main(int argc, char const *argv[]) 119 { 120 #ifndef ONLINE_JUDGE 121 freopen("in.txt","r", stdin); 122 // freopen("out.txt","w", stdout); 123 #endif 124 init(); 125 ll x = read<ll>(); 126 n = read<ll>(); 127 for (int i = 0; i < pcnt;i++){ 128 if (x%prime[i]==0){ 129 fac[tot++] = prime[i]; 130 while(x%prime[i]==0) 131 x /= prime[i]; 132 } 133 } 134 if (x!=1) 135 fac[tot++] = x; 136 ll res = 1; 137 for (int i = 0; i < tot;i++){ 138 ll t = f(n, fac[i]); 139 res = res * qpow(fac[i], t) % mod; 140 } 141 writeln(res); 142 return 0; 143 }
D. Complete Tripartite
Description
给出一个可能不连通的无自环无向图，问是否能划分出三个点集。
任意两个点集应该满足每一个顶点到另一个集合的任意点有边，自己点集内任意两点无边相连。
Solution
自己思路补全，看了题解才会。
首先想到dfs一遍判断图是否连通，不连通直接输出-1。
首先将所有点染色为1。
遍历所有染色1的点的邻接点v，如果v的颜色依然为1，则将其标为2。
遍历所有染色2的点的邻接点v，如果v的颜色依然为2，则将其标为3。
做完以上步骤，可以保证三个点集内部点不互相连。
接下来使用点的度数判断是否满足点集间互相有边。
再者是判断是否有一个集合为空。
判断完输出结果。
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 } 87 int col[maxn], deg[maxn]; 88 vector<int> g[maxn]; 89 void dfs(int u, int p) 90 { 91 col[u] = 1; 92 int sz = g[u].size(); 93 for (int i = 0; i < sz; i++) 94 { 95 int v = g[u][i]; 96 if (col[v]) 97 continue; 98 dfs(v, u); 99 } 100 } 101 int main(int argc, char const *argv[]) 102 { 103 #ifndef ONLINE_JUDGE 104 freopen("in.txt", "r", stdin); 105 // freopen("out.txt","w", stdout); 106 #endif 107 n = read<int>(), m = read<int>(); 108 for (int i = 0; i < m; i++) 109 { 110 int x = read<int>(), y = read<int>(); 111 g[x].emplace_back(y), g[y].emplace_back(x); 112 deg[x]++, deg[y]++; 113 } 114 dfs(1, 0); 115 int p = accumulate(col + 1, col + 1 + n, 0); 116 if (p != n) 117 { 118 puts("-1"); 119 return 0; 120 } 121 for (int i = 1; i <= n; i++) 122 { 123 if (col[i] == 1) 124 { 125 int sz = g[i].size(); 126 for (int j = 0; j < sz; j++) 127 { 128 int k = g[i][j]; 129 if (col[k] == 1) 130 col[k] = 2; 131 } 132 } 133 } 134 for (int i = 1; i <= n; i++) 135 { 136 if (col[i] == 2) 137 { 138 int sz = g[i].size(); 139 for (int j = 0; j < sz; j++) 140 { 141 int k = g[i][j]; 142 if (col[k] == 2) 143 col[k] = 3; 144 } 145 } 146 } 147 int s[4] = {0}; 148 for (int i = 1; i <= n; i++) 149 ++s[col[i]]; 150 int f = 1; 151 if (!s[1]||!s[2]||!s[3]) 152 f = 0; 153 for (int i = 1; i <= n&&f;i++){ 154 int cur = 0; 155 for (int j = 1; j <= 3;j++) 156 if (j!=col[i]) 157 cur += s[j]; 158 if (cur!=deg[i]) 159 f = 0; 160 } 161 if (f) 162 for (int i = 1;i<=n;i++) 163 write(col[i]), pblank; 164 else 165 puts("-1"); 166 return 0; 167 }