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
;
ll a
;
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;
}