标签：组成 注意 scanf 背景 bin ios alt mic dynamic
运
【问题背景】
zhx和妹子们玩数数游戏。
【问题描述】
仅包含4或7的数被称为幸运数。
一个序列的子序列被定义为从序列中删去若干个数，剩下的数组成的新序列。两个子序列被定义为不同的当且仅当其中的元素在原始序列中的下标的集合不相等。对于一个长度为N的序列，共有2N个不同的子序列。（包含一个空序列）。
一个子序列被称为不幸运的，当且仅当其中不包含两个相同的幸运数。
对于一个给定序列，求其中长度恰好为K的不幸运子序列的个数，答案mod 109+7输出。【输入格式】
第一行两个正整数N，K，表示原始序列的长度和题目中的K。
接下来一行N个整数ai，表示序列中第i个元素的值。【输出格式】
仅一个数，表示不幸运子序列的个数。（mod 10^9+7）
【样例输入】
3 2
1 1 14 2
4 7 4 7【样例输出】
3
4【样例解释】
对于样例1，每个长度为2的子序列都是符合条件的。
对于样例2，4个不幸运子序列元素下标分别为：{1, 2}, {3, 4}, {1, 4}, {2, 3}。注意下标集{1, 3}对应的子序列不是“不幸运”的，因为它包含两个相同的幸运数4.【数据范围与规定】
对于50%的数据，1≤N≤16。
对于70%的数据，1≤N≤1000,ai≤10000。
对于100%的数据，1≤N≤100000,K≤N,1≤ai≤10^9。
此题比较水但是又略鬼畜...
首先我们可以发现一个幸运数一股子0/1串的味道...所以很显然对于 $10^9$ 之内的数据, 幸运数的数量只有 $1 \times 10^3$ 左右个. 所以我们第一可以选择打个表上去(重度懒癌患者表示还是直接打个数组表上去w), 第二可以选择DFS预处理一下, 然后将这些幸运数离散化. 离散化之后进行计数, 计算每个幸运数出现的次数 $cnt_i$ , 同时处理出非幸运数的个数 $d$ . 处理到这里我们已经成功将输入分成了两类, 然后需要分别来处理.
对于非幸运数随便瞎**挑够数就行, 然后非幸运数的部分就变成了一个组合问题. 我们枚举子序列中幸运数的个数 $i$ , 则非幸运数的个数为 $k-i$ , 与对应的幸运数部分相乘即可求得答案的一部分. 然后一直枚举 $i$ 即可得出答案. 公式如下:
\[ans=\sum_{i=1}^k {\binom{d}{k-i} \times f_i}\]
其中 $f_i$ 为从所有幸运数中取出 $i$ 个的方案数. 由于如果要保证子序列的非幸运性(???)的话, 幸运数的个数不能达到 $2$ , 也就是说每种要么取一个要么就不取, 然后我们似乎闻到了一股子0/1背包的味道所以我们可以将它转化成 $DP$ 来求. 对于 $C^d_{k-i}$ , 因为 $d$ 是固定的, 所以我们可以递推出一个组合数表出来. 当 $f_i$ 和 $\binom{d}{i}$ 两张表打好之后就可以算总答案辣w
参考代码
1 #include <cstdio> 2 #include <cstring> 3 #include <cstdlib> 4 #include <iostream> 5 #include <algorithm> 6 7 const int MAXM=1024; 8 const int MAXN=100010; 9 const long long p=1e9+7; 10 const int 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11 12 int n; 13 int k; 14 int unlucky; 15 long long c[MAXN]; 16 long long dp[MAXN]; 17 long long cnt[MAXN]; 18 19 void Initialize(); 20 void Combination(); 21 void DynamicProgramming(); 22 long long Pow(long long,int); 23 24 int main(){ 25 Initialize(); 26 DynamicProgramming(); 27 Combination(); 28 long long ans=0; 29 for(int i=1;i<=k;i++) 30 ans=(ans+c[i]*dp[k-i]%p)%p; 31 printf("%lld\n",ans); 32 return 0; 33 } 34 35 void DynamicProgramming(){ 36 dp[0]=1; 37 for(int i=1;i<=MAXM;i++){ 38 if(cnt[i]>=2){ 39 for(int j=i;j>0;j--){ 40 dp[j]=(dp[j]+dp[j-1]*cnt[i])%p; 41 } 42 } 43 } 44 } 45 46 void Combination(){ 47 c[0]=1; 48 for(int i=1;i<=unlucky;i++) 49 c[i]=c[i-1]*(unlucky-i+1)%p*Pow(i,p-2)%p; 50 } 51 52 void Initialize(){ 53 int tmp; 54 scanf("%d%d",&n,&k); 55 unlucky=n; 56 for(int i=0;i<n;i++){ 57 scanf("%d",&tmp); 58 int pos=std::lower_bound(lucky,lucky+MAXM,tmp)-lucky; 59 if(lucky[pos]==tmp){ 60 cnt[pos]++; 61 if(cnt[pos]==2) 62 unlucky-=2; 63 else if(cnt[pos]>2) 64 unlucky--; 65 } 66 } 67 } 68 69 long long Pow(long long a,int n){ 70 long long ans=1; 71 while(n>0){ 72 if((n&1)!=0){ 73 ans=ans*a%p; 74 } 75 a=a*a%p; 76 n>>=1; 77 } 78 return ans; 79 }
标签：组成 注意 scanf 背景 bin ios alt mic dynamic
原文地址：http://www.cnblogs.com/rvalue/p/7286715.html
