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  • LuoguP2150 [NOI2015]寿司晚宴

    LuoguP2150 [NOI2015]寿司晚宴

    题意:

    链接

    题解:

    考虑暴力状压,发现显然过不了

    然后套路根号分治

    #include<bits/stdc++.h>
    
    using namespace std;
    
    inline int read()
    {
        int f = 1,x = 0;
        char ch;
        do
        {
            ch = getchar();
            if(ch == '-') f = -1;
        }while(ch < '0'||ch > '9');
        do
        {
            x = (x<<3) +(x<<1) + ch - '0';
            ch = getchar();
        }while(ch >= '0'&&ch <= '9');
        return f*x;
    }
    
    const int MAXN = 500 + 2;
    
    const int Prime[10] = {0,2,3,5,7,11,13,17,19};
    int n,p;
    
    struct node
    {
        int sta;
        int P;
        friend bool operator < (node a1,node a2)
        {
            return a1.P < a2.P;
        }
    }a[MAXN];
    
    int dp[2][(1<<8) + 1][(1<<8) + 1];
    int f[(1<<8) + 1][(1<<8) + 1];
    
    int main()
    {
        n = read(),p = read();
        for(int i=2;i<=n;i++)
        {
            int cur = i;
            for(int j=1;j<=8;j++) 
            {
                if(cur % Prime[j]) continue;
                while(cur % Prime[j] == 0) cur /= Prime[j];
                a[i-1].sta |= (1<<j-1);
            }
            a[i-1].P = cur == 1 ? -1 : cur;
        }
        sort(a+1,a+n);
        f[0][0] = 1;
        for(int i=1;i<n;i++)
        {
            if(a[i].P == -1 || a[i].P != a[i-1].P) 
            {
                memcpy(dp[0],f,sizeof(f));
                memcpy(dp[1],f,sizeof(f));
            }
            for(int sta1 = (1<<8)-1;sta1>=0;sta1--)
            {
                for(int sta2 = (1<<8)-1;sta2>=0;sta2--)
                {
                    if(sta1 & sta2) continue;
                    if((a[i].sta & sta2) == 0) dp[0][sta1|a[i].sta][sta2] = (dp[0][sta1|a[i].sta][sta2] + dp[0][sta1][sta2]) % p; 
                    if((a[i].sta & sta1) == 0) dp[1][sta1][sta2|a[i].sta] = (dp[1][sta1][sta2|a[i].sta] + dp[1][sta1][sta2]) % p;            
                }
            }
            if(i == n - 1 || a[i].P != a[i+1].P || a[i].P == -1)
            {
                for(int sta1 = (1<<8)-1;sta1>=0;sta1--)
                {
                    for(int sta2 = (1<<8)-1;sta2>=0;sta2--)
                    {
                        if(sta1 & sta2) continue;
                        f[sta1][sta2] = ((dp[0][sta1][sta2] + dp[1][sta1][sta2] - f[sta1][sta2]) % p + p) % p;
                    }
                }            
            }
        }
        long long ans = 0;
        for(int sta1 = (1<<8)-1;sta1>=0;sta1--)
        {
            for(int sta2 = (1<<8)-1;sta2>=0;sta2--)
            {
                if(sta1 & sta2) continue;
                ans += f[sta1][sta2]; 
                ans %= p;
            }
        }
        cout << ans << endl;
    }
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  • 原文地址:https://www.cnblogs.com/wlzs1432/p/13996854.html
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