Codeforces Round #104 (Div. 2) E DP(01背包模型) +组和+除法取模求逆元
题意:
规定只包含4或7的数为幸运数字,给定n个数的序列,求他的子序列,使得该子序列的长度为k并且满足该子序列中不存在相同的两个幸运数字。问一共寻在多少种可能。(只要该数的下标不同则认为是不同的序列)
思路:
记录每个幸运数字的个数,枚举从非幸运数中取出的个数i,那么在幸运数字中取k - i。这里C(no,i)好算,直接带公式算除法取模,而在幸运数字中取k-i个的可能需要同过dp来算,这里类似于01背包模型,dp[i +1][j + 1] = dp[i][j]*a[i] + dp[i][j + 1]; 表示前i个取了j个的可能数。
这里对dp进行了空间优化
View Code
//#pragma comment(linker,"/STACK:327680000,327680000") #include <iostream> #include <cstdio> #include <cmath> #include <vector> #include <cstring> #include <algorithm> #include <string> #include <set> #include <functional> #include <numeric> #include <sstream> #include <stack> #include <map> #include <queue> #define CL(arr, val) memset(arr, val, sizeof(arr)) #define lc l,m,rt<<1 #define rc m + 1,r,rt<<1|1 #define ll __int64 #define L(x) (x) << 1 #define R(x) (x) << 1 | 1 #define MID(l, r) (l + r) >> 1 #define Min(x, y) (x) < (y) ? (x) : (y) #define Max(x, y) (x) < (y) ? (y) : (x) #define E(x) (1 << (x)) #define iabs(x) (x) < 0 ? -(x) : (x) #define OUT(x) printf("%I64d\n", x) #define lowbit(x) (x)&(-x) #define Read() freopen("din.txt", "r", stdin) #define Write() freopen("dout.txt", "w", stdout); #define M 26 #define N 100007 using namespace std; const int inf = 0x1F1F1F1F; const int mod = 1000000007; const int X = 1000000005; ll fac[N]; ll a[N]; ll dp[1230]; set<int> iset; map<int,int> imap; void init() { int i; fac[0] = 1; for (i = 1; i < N; ++i) { fac[i] = fac[i - 1]*i%mod; } } bool isL(int x) { while (x) { if (x % 10 != 4 && x % 10 != 7) return false; x /= 10; } return true; } ll modexp(ll a,ll b,ll c) { ll rs = 1; ll tmp = a; while (b) { if (b&1) rs = rs*tmp%c; tmp = tmp*tmp%c; b >>= 1; } return rs; } ll cal(int x,int y) { ll ans = 0; ans = fac[x]*modexp(fac[x - y],X,mod)%mod; ans = ans*modexp(fac[y],X,mod)%mod; return ans; } int main() { // Read(); int n,m; int i,x,j; scanf("%d%d",&n,&m); init(); iset.clear(); imap.clear(); int no = 0; for (i = 0; i < n; ++i) { scanf("%d",&x); if (isL(x)) { iset.insert(x); imap[x]++; } else no++; } set<int>::iterator it; int la = 0; for (it = iset.begin(); it != iset.end(); ++it) a[++la] = imap[*it]; for (i = 0; i <= la; ++i) dp[0] = 1; for (i = 0; i < la; ++i) { for (j = i; j >= 0; --j) { dp[j + 1] += dp[j]*a[i + 1]; dp[j + 1] %= mod; } } ll ans = 0; for (i = 0; i <= m; ++i) { if (no >= i && la >= m - i) { ll tmp = cal(no,i)*dp[m - i]%mod; ans = (ans + tmp)%mod; } } cout<<ans<<endl; return 0; }