HUAS 1476 不等数列(DP)

淡淡的烟草味﹌ 2021-10-01 02:08 177阅读 0赞

考虑DP。

如果把转移看出当前位填什么数的话，这样是有后效性的。

如果考虑当前的序列是将1至n依次插入序列中的话。

考虑将i插入1到i-1的序列中，如果插入到<号中或者首部，那么最后就会多出一个大于号。

如果插入到>号中或者尾部，那么最后就会多出一个小于号。

所以定义状态dp\[i\]\[j\]表示1到i组成的序列中，小于号的数目为j的方法数。转移方程即为所求。

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# include <cstdio>
    # include <cstring>
    # include <cstdlib>
    # include <iostream>
    # include <vector>
    # include <queue>
    # include <stack>
    # include <map>
    # include <set>
    # include <cmath>
    # include <algorithm>
    using namespace std;
    # define lowbit(x) ((x)&(-x))
    # define pi 3.1415926535
    # define eps 1e-9
    # define MOD 100000007
    # define INF 1000000000
    # define mem(a,b) memset(a,b,sizeof(a))
    # define FOR(i,a,n) for(int i=a; i<=n; ++i)
    # define FO(i,a,n) for(int i=a; i<n; ++i)
    # define bug puts("H");
    # define lch p<<1,l,mid
    # define rch p<<1|1,mid+1,r
    # define mp make_pair
    # define pb push_back
    typedef pair<int,int> PII;
    typedef vector<int> VI;
    # pragma comment(linker, "/STACK:1024000000,1024000000")
    typedef long long LL;
    int Scan() {
        int res=0, flag=0;
        char ch;
        if((ch=getchar())=='-') flag=1;
        else if(ch>='0'&&ch<='9') res=ch-'0';
        while((ch=getchar())>='0'&&ch<='9')  res=res*10+(ch-'0');
        return flag?-res:res;
    }
    void Out(int a) {
        if(a<0) {putchar('-'); a=-a;}
        if(a>=10) Out(a/10);
        putchar(a%10+'0');
    }
    const int N=1005;
    //Code begin...
     
    int dp[N][N];
     
    int dfs(int x, int y){
        if (~dp[x][y]) return dp[x][y];
        if (y==0||y==x-1) return dp[x][y]=1;
        if (y>=x) return 0;
        int ans=(dfs(x-1,y)*(y+1)%2012+dfs(x-1,y-1)*(x-y)%2012)%2012;
        return dp[x][y]=ans;
    }
    int main ()
    {
        int n, k;
        mem(dp,-1);
        scanf("%d%d",&n,&k);
        printf("%d\n",dfs(n,k));
        return 0;
    }

转载于:https://www.cnblogs.com/lishiyao/p/6623394.html

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