首页 > 代码库 > CCF 201312-4 有趣的数 (数位DP, 状压DP, 组合数学+暴力枚举, 推公式, 矩阵快速幂)
CCF 201312-4 有趣的数 (数位DP, 状压DP, 组合数学+暴力枚举, 推公式, 矩阵快速幂)
问题描述
我们把一个数称为有趣的，当且仅当：
1. 它的数字只包含0, 1, 2, 3，且这四个数字都出现过至少一次。
2. 所有的0都出现在所有的1之前，而所有的2都出现在所有的3之前。
3. 最高位数字不为0。
因此，符合我们定义的最小的有趣的数是2013。除此以外，4位的有趣的数还有两个：2031和2301。
请计算恰好有n位的有趣的数的个数。由于答案可能非常大，只需要输出答案除以1000000007的余数。
1. 它的数字只包含0, 1, 2, 3，且这四个数字都出现过至少一次。
2. 所有的0都出现在所有的1之前，而所有的2都出现在所有的3之前。
3. 最高位数字不为0。
因此，符合我们定义的最小的有趣的数是2013。除此以外，4位的有趣的数还有两个：2031和2301。
请计算恰好有n位的有趣的数的个数。由于答案可能非常大，只需要输出答案除以1000000007的余数。
输入格式
输入只有一行，包括恰好一个正整数n (4 ≤ n ≤ 1000)。
输出格式
输出只有一行，包括恰好n 位的整数中有趣的数的个数除以1000000007的余数。
样例输入
4
样例输出
3
析：先说第一种方法，数位DP：
首先只有0123，并且有约束条件，那么就只有6种状态，{2},{20},{201},{23},{203},{2031}，然后依次对每一位进行计算即可。
第二种，状压DP：
因为只有4种数字，所以可能用状压DP，1表示0， 2表示1， 4表示2，8表示3.剩下的就简单了。
第三种，组合数学+暴力枚举：
先枚举01的数量，那么最少2个，最多n-2个，由于0是不能放在首位，所以就有c(n-1, i)，i 表示有多少个01，然后再考虑有多种方法，
应该有 n-i 种方法，因为0的数量是从1个到 i-1 个，同理，23就有 n-i-1 种，然后乘起来就好。
第四种，推公式：
这个方法和第三种一样的，只不过是把第三种简化了，然后成一个递推公式，an=（n*n-5n+4）*2^（n-3）+n-1。
第五种，矩阵快速幂：
对每一个组成数字进行分析，字符串必定以2开头，把字符串中字符放入一个集合，我们可以统计长度为N，字符集合为S的合法字符串个数，
然后把它们的关系表示成一个矩阵，就OK了。
代码如下：
数位DP：
#pragma comment(linker, "/STACK:1024000000,1024000000")#include <cstdio>#include <string>#include <cstdlib>#include <cmath>#include <iostream>#include <cstring>#include <set>#include <queue>#include <algorithm>#include <vector>#include <map>#include <cctype>#include <stack>#define freopenr freopen("in.txt", "r", stdin)#define freopenw freopen("out.txt", "w", stdout)using namespace std;typedef long long LL;typedef pair<int, int> P;const int INF = 0x3f3f3f3f;const double inf = 0x3f3f3f3f3f3f;const LL LNF = 100000000000000000;const double PI = acos(-1.0);const double eps = 1e-8;const int maxn = 1e3 + 5;const int mod = 1e9 + 7;const char *mark = "+-*";const int dr[] = {-1, 0, 1, 0};const int dc[] = {0, 1, 0, -1};int n, m;inline bool is_in(int r, int c){ return r >= 0 && r < n && c >= 0 && c < m;}inline LL Max(LL a, LL b){ return a < b ? b : a; }inline LL Min(LL a, LL b){ return a > b ? b : a; }inline int Max(int a, int b){ return a < b ? b : a; }inline int Min(int a, int b){ return a > b ? b : a; }LL dp[maxn][5];int main(){ while(cin >> n){ dp[4][0] = 7; dp[4][1] = 5; dp[4][2] = 3; dp[4][3] = 9; dp[4][4] = 3; for(int i = 5; i <= n; ++i){ dp[i][0] = (1 + 2 * dp[i-1][0]) % mod; dp[i][1] = (dp[i-1][0] + 2 * dp[i-1][1]) % mod; dp[i][2] = (1 + dp[i-1][2]) % mod; dp[i][3] = (dp[i-1][0] + dp[i-1][2] + 2 * dp[i-1][3]) % mod; dp[i][4] = (dp[i-1][1] + dp[i-1][3] + 2 * dp[i-1][4]) % mod; } cout << dp[n][4] << endl; }}
状压DP：
#pragma comment(linker, "/STACK:1024000000,1024000000")#include <cstdio>#include <string>#include <cstdlib>#include <cmath>#include <iostream>#include <cstring>#include <set>#include <queue>#include <algorithm>#include <vector>#include <map>#include <cctype>#include <cmath>#include <stack>#define frer freopen("in.txt", "r", stdin)#define frew freopen("out.txt", "w", stdout)using namespace std;typedef long long LL;typedef pair<int, int> P;const int INF = 0x3f3f3f3f;const double inf = 0x3f3f3f3f3f3f;const double PI = acos(-1.0);const double eps = 1e-8;const int maxn = 1e3 + 5;const int mod = 1e9 + 7;const int dr[] = {-1, 1, 0, 0};const int dc[] = {0, 0, 1, -1};const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};int n, m;const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};inline int Min(int a, int b){ return a < b ? a : b; }inline int Max(int a, int b){ return a > b ? a : b; }inline LL Min(LL a, LL b){ return a < b ? a : b; }inline LL Max(LL a, LL b){ return a > b ? a : b; }inline bool is_in(int r, int c){ return r >= 0 && r < n && c >= 0 && c < m;}LL dp[maxn][(1<<4)+1];int main(){ while(cin >> n){ memset(dp, 0, sizeof dp); dp[0][0] = 1; for(int i = 0; i < n; ++i){ for(int j = 0; j < 16; ++j){ if(!dp[i][j]) continue; if(!(j & 2) && i) dp[i+1][j|1] = (dp[i+1][j|1] + dp[i][j]) % mod; dp[i+1][j|2] = (dp[i+1][j|2] + dp[i][j]) % mod; if(!(j & 8)) dp[i+1][j|4] = (dp[i+1][j|4] + dp[i][j]) % mod; dp[i+1][j|8] = (dp[i+1][j|8] + dp[i][j]) % mod; } } cout << dp[n][15] << endl; } return 0;}
组合数学+暴力枚举：
#pragma comment(linker, "/STACK:1024000000,1024000000")#include <cstdio>#include <string>#include <cstdlib>#include <cmath>#include <iostream>#include <cstring>#include <set>#include <queue>#include <algorithm>#include <vector>#include <map>#include <cctype>#include <cmath>#include <stack>#define frer freopen("in.txt", "r", stdin)#define frew freopen("out.txt", "w", stdout)using namespace std;typedef long long LL;typedef pair<int, int> P;const int INF = 0x3f3f3f3f;const double inf = 0x3f3f3f3f3f3f;const double PI = acos(-1.0);const double eps = 1e-8;const int maxn = 1e3 + 5;const int mod = 1e9 + 7;const int dr[] = {-1, 1, 0, 0};const int dc[] = {0, 0, 1, -1};const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};int n, m;const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};inline int Min(int a, int b){ return a < b ? a : b; }inline int Max(int a, int b){ return a > b ? a : b; }inline LL Min(LL a, LL b){ return a < b ? a : b; }inline LL Max(LL a, LL b){ return a > b ? a : b; }inline bool is_in(int r, int c){ return r >= 0 && r < n && c >= 0 && c < m;}LL c[maxn][maxn];void init(){ for(int i = 0; i < n; ++i) c[i][0] = c[i][i] = 1; for(int i = 2; i < n; ++i) for(int j = 1; j < i; ++j) c[i][j] = (c[i-1][j] + c[i-1][j-1]) % mod;}int main(){ while(cin >> n){ init(); LL ans = 0; for(int i = 2; i < n-1; ++i) ans = (ans + (c[n-1][i] * (i-1) * (n-i-1)) % mod) % mod; cout << ans << endl; } return 0;}
推公式：
#pragma comment(linker, "/STACK:1024000000,1024000000")#include <cstdio>#include <string>#include <cstdlib>#include <cmath>#include <iostream>#include <cstring>#include <set>#include <queue>#include <algorithm>#include <vector>#include <map>#include <cctype>#include <cmath>#include <stack>#define frer freopen("in.txt", "r", stdin)#define frew freopen("out.txt", "w", stdout)using namespace std;typedef long long LL;typedef pair<int, int> P;const int INF = 0x3f3f3f3f;const double inf = 0x3f3f3f3f3f3f;const double PI = acos(-1.0);const double eps = 1e-8;const int maxn = 1e3 + 5;const int mod = 1e9 + 7;const int dr[] = {-1, 1, 0, 0};const int dc[] = {0, 0, 1, -1};const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};int n, m;const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};inline int Min(int a, int b){ return a < b ? a : b; }inline int Max(int a, int b){ return a > b ? a : b; }inline LL Min(LL a, LL b){ return a < b ? a : b; }inline LL Max(LL a, LL b){ return a > b ? a : b; }inline bool is_in(int r, int c){ return r >= 0 && r < n && c >= 0 && c < m;}LL quick_pow(LL a, LL b){ LL ans = 1; while(b){ if(b & 1) ans = (ans * a) % mod; b >>= 1; a = (a * a) % mod; } return ans;}int main(){ while(cin >> n){ LL ans = (n * n - 5 * n + 4) % mod; ans = (ans * quick_pow(2LL, n-3) + n - 1) % mod; cout << ans << endl; } return 0;}
矩阵快速幂：
#pragma comment(linker, "/STACK:1024000000,1024000000")#include <cstdio>#include <string>#include <cstdlib>#include <cmath>#include <iostream>#include <cstring>#include <set>#include <queue>#include <algorithm>#include <vector>#include <map>#include <cctype>#include <cmath>#include <stack>#define frer freopen("in.txt", "r", stdin)#define frew freopen("out.txt", "w", stdout)using namespace std;typedef long long LL;typedef pair<int, int> P;const int INF = 0x3f3f3f3f;const double inf = 0x3f3f3f3f3f3f;const double PI = acos(-1.0);const double eps = 1e-8;const int maxn = 1e3 + 5;const int mod = 1e9 + 7;const int dr[] = {-1, 1, 0, 0};const int dc[] = {0, 0, 1, -1};const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};int n, m;const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};inline int Min(int a, int b){ return a < b ? a : b; }inline int Max(int a, int b){ return a > b ? a : b; }inline LL Min(LL a, LL b){ return a < b ? a : b; }inline LL Max(LL a, LL b){ return a > b ? a : b; }inline bool is_in(int r, int c){ return r >= 0 && r < n && c >= 0 && c < m;}struct Matrix{ LL a[6][6];};Matrix mul(Matrix x, Matrix y){ Matrix ans; memset(ans.a, 0, sizeof ans.a); for(int i = 0; i < 6; ++i){ for(int j = 0; j < 6; ++j){ for(int k = 0; k < 6; ++k){ ans.a[i][j] += (x.a[i][k] * y.a[k][j]) % mod; } ans.a[i][j] %= mod; } } return ans;}Matrix quick_pow(Matrix x, int b){ Matrix ans; memset(ans.a, 0, sizeof ans.a); for(int i = 0; i < 6; ++i) ans.a[i][i] = 1; while(b){ if(b & 1) ans = mul(ans, x); b >>= 1; x = mul(x, x); } return ans;}int main(){ Matrix x, y; memset(x.a, 0, sizeof x.a); memset(y.a, 0, sizeof y.a); x.a[3][3] = x.a[0][0] = x.a[1][0] = x.a[2][1] = x.a[3][0] = x.a[4][1] = x.a[4][3] = x.a[5][2] = x.a[5][4] = x.a[5][5] = 1; x.a[1][1] = x.a[4][4] = x.a[5][5] = x.a[2][2] = 2; y.a[0][0] = 1; while(cin >> n){ Matrix ans = mul(quick_pow(x, n-1), y); cout << ans.a[5][0] << endl; } return 0;}
CCF 201312-4 有趣的数 (数位DP, 状压DP, 组合数学+暴力枚举, 推公式, 矩阵快速幂)
声明：以上内容来自用户投稿及互联网公开渠道收集整理发布，本网站不拥有所有权，未作人工编辑处理，也不承担相关法律责任，若内容有误或涉及侵权可进行投诉： 投诉/举报 工作人员会在5个工作日内联系你，一经查实，本站将立刻删除涉嫌侵权内容。