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  • 【洛谷 P1707】 刷题比赛 (矩阵加速)

    题目连接
    很久没写矩阵加速了,复习一下,没想到是一道小毒瘤题。

    状态矩阵(a[k],b[k],c[k],a[k+1],b[k+1],c[k+1],k,k^2,w^k,z^k,1)
    转移矩阵

      0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
      0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0,
      0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0,
      0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0,
      0, q, 0, 0, p, 1, 1, t, r, 0, 0, 1,
      0, 0, v, 0, 1, u, 1, 0, 0, 1, 0, 0,
      0, 0, 0, y, 1, 1, x, 1, 0, 0, 1, 2,
      0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1,
      0, 0, 0, 0, 0, 0, 0, 2, 1, 0, 0, 1,
      0, 0, 0, 0, 0, 0, 0, 0, 0, w, 0, 0,
      0, 0, 0, 0, 0, 0, 0, 0, 0, 0, z, 0,
      0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1
    

    手写不易。

    #include <cstdio>
    #include <cstdlib>
    #include <cstring>
    // a[k],b[k],c[k],a[k+1],b[k+1],c[k+1],k,k^2,w^k,z^k,1
    int u, v, p, q, r, t, x, w, z, y;
    typedef long long ll;
    ll n, k;
    ll f[12], tmp[12], temp[12][12], dp[12][12];
    inline ll Slow_Mul(ll a, ll b){
        ll ans = 0;
        while(b){
          if(b & 1) ans = (ans + a) % k;
          b >>= 1;
          a = (a + a) % k;
        }
        return ans;
    }
    inline void Mult(){
        for(int i = 1; i <= 11; ++i){
           tmp[i] = 0;
           for(int j = 1; j <= 11; ++j)
              (tmp[i] += Slow_Mul(f[j], dp[i][j])) %= k;
        }
        for(int i = 1; i <= 11; ++i)
           f[i] = tmp[i];
    }
    inline void Self(){
        for(int i = 1; i <= 11; ++i)
           for(int j = 1; j <= 11; ++j){
              temp[i][j] = 0;
              for(int l = 1; l <= 11; ++l)
                 (temp[i][j] += Slow_Mul(dp[i][l], dp[l][j])) %= k;
           }
        for(int i = 1; i <= 11; ++i)
           for(int j = 1; j <= 11; ++j)
              dp[i][j] = temp[i][j];
    }
    inline void Fast_Pow(){
        n -= 2;
        while(n){
          if(n & 1) Mult();
          n >>= 1;
          Self();
        }
    }
    int main(){
        scanf("%lld%lld%d%d%d%d%d%d%d%d%d%d", &n, &k, &p, &q, &r, &t, &u, &v, &w, &x, &y, &z);
        ll xs[12][12] = 
    { 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
      0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0,
      0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0,
      0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0,
      0, q, 0, 0, p, 1, 1, t, r, 0, 0, 1,
      0, 0, v, 0, 1, u, 1, 0, 0, 1, 0, 0,
      0, 0, 0, y, 1, 1, x, 1, 0, 0, 1, 2,
      0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1,
      0, 0, 0, 0, 0, 0, 0, 2, 1, 0, 0, 1,
      0, 0, 0, 0, 0, 0, 0, 0, 0, w, 0, 0,
      0, 0, 0, 0, 0, 0, 0, 0, 0, 0, z, 0,
      0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1
    };
        memcpy(dp, xs, sizeof dp);
        f[1] = f[2] = f[3] = f[11] = 1; f[4] = f[5] = f[6] = 3; f[9] = w; f[10] = z; f[7] = 1; f[8] = 1;
        Fast_Pow();
        printf("nodgd %lld
    Ciocio %lld
    Nicole %lld
    ", f[4], f[5], f[6]);
        return 0;
    }
    
    
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  • 原文地址:https://www.cnblogs.com/Qihoo360/p/9790216.html
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