2013多校联合训练第五场_2013多校训练5-优快云博客

。。。被05卡了全场

保存前缀和的模m值，对于每一个值找它第一次出现的地方即可。

const int N = 1e5 + 5;
int n , m , a[N];
int s[N] , p[N];
int main()
{
  int  i , j;
  while (~scanf("%d%d",&n,&m))
  {
    for (i = 1 ; i <= n ; ++ i)
      scanf("%d",&a[i]) , s[i] = (s[i - 1] + a[i] + 20000 * m) , s[i] %= m;
    int ans = 0;
    memset(p , -1 , sizeof(p));
    p[0] = 0;
    for (i = 1 ; i <= n ; ++ i)
    {
      if (!~p[s[i]])
        p[s[i]] = i;
      else ans = max(ans , i - p[s[i]]);
    }
    cout << ans << endl;
  }
  return 0;
}

每一位是独立的，就可以分别进行一个DP，找出每一位是1的概率再累加。

const int N = 205;
int a[N] , n , ca;
char o[N][2];
double f[20][N][2] , p[N];
int main()
{
  int i , j , k;
  while (~scanf("%d",&n))
  {
    printf("Case %d:\n" , ++ ca);
    for (i = 0 ; i <= n ; ++ i)
      scanf("%d",&a[i]);
    for (i = 1 ; i <= n ; ++ i)
      scanf("%s", o[i]);
    for (i = 1 ; i <= n ; ++ i){
      scanf("%lf", &p[i]);
      p[i] = 1. - p[i];
    }

    memset(f , 0 , sizeof(f));

    for (i = 0 ; i < 20 ; ++ i)
      if (a[0] & (1 << i))
        f[i][0][1] = 1;
      else f[i][0][0] = 1;
    double ans = 0;
    for (i = 19 ; i >= 0 ; -- i)
    {
      for (j = 1 ; j <= n ; ++ j)
      {
        if (o[j][0] == '&')
        {
          if (a[j] & (1 << i))
            f[i][j][0] = f[i][j - 1][0] * (1.0 - p[j]) + f[i][j - 1][0] * p[j],
            f[i][j][1] = f[i][j - 1][1] * (1.0 - p[j]) + f[i][j - 1][1] * p[j];
          else
            f[i][j][0] = f[i][j - 1][0] * (1.0 - p[j]) + (f[i][j - 1][0] + f[i][j - 1][1]) * p[j],
            f[i][j][1] = f[i][j - 1][1] * (1.0 - p[j]);
        }
        if (o[j][0] == '|')
        {
          if (a[j] & (1 << i))
            f[i][j][1] = f[i][j - 1][1] * (1.0 - p[j]) + (f[i][j - 1][0] + f[i][j - 1][1]) * p[j],
            f[i][j][0] = f[i][j - 1][0] * (1.0 - p[j]);
          else
            f[i][j][0] = f[i][j - 1][0] * (1.0 - p[j]) + f[i][j - 1][0] * p[j],
            f[i][j][1] = f[i][j - 1][1] * (1.0 - p[j]) + f[i][j - 1][1] * p[j];
        }
        if (o[j][0] == '^')
        {
          if (a[j] & (1 << i))
            f[i][j][1] = f[i][j - 1][1] * (1.0 - p[j]) + f[i][j - 1][0] * p[j],
            f[i][j][0] = f[i][j - 1][0] * (1.0 - p[j]) + f[i][j - 1][1] * p[j];
          else
            f[i][j][0] = f[i][j - 1][0] ,
            f[i][j][1] = f[i][j - 1][1] ;
        }
      //   printf("0:%lf 1:%lf\n" , f[i][j][0] , f[i][j][1]);
      }
//      printf("%lf\n" , f[i][n][1]);
      ans *= 2.0 , ans += f[i][n][1];
    }
    printf("%.6lf\n" , ans + 1e-9);
  }
  return 0;
}

这个平均值只有两种可能，一个是不走圈，就是一条密度最小的路径。一个是把一个圈走无穷次，就是这个圈的密度。

所以只需用floyd求得包含某点的最小密度圈以及两天之间的最小密度路径，就可以得到答案了。

using namespace std;
typedef long long LL;
#define N 105
#define M 200005

int n , m , g[N][N] , f[N][N][N];
double h[N] , w[N][N];
bool p[N][N];
void work()
{
  int i , j , k , l , x , y , z;
  for (i = 1 ; i <= n ; ++ i)
    for (j = 1 ; j <= n ; ++ j)
      g[i][j] = f[1][i][j] = 1 << 29 , w[i][j] = 1e30 , p[i][j] = 0;
  for (i = 1 ; i <= n ; ++ i) h[i] = 1e30;
  while (m --)
  {
    scanf("%d%d%d",&x,&y,&z);
    g[x][y] = f[1][x][y] = z;
    p[x][y] = 1;
  }
  for (k = 1 ; k <= n ; ++ k)
    for (i = 1 ; i <= n ; ++ i)
      for (j = 1 ; j <= n ; ++ j)
        p[i][j] |= (p[i][k] && p[k][j]);
  for (k = 2 ; k <= n ; ++ k)
  {

    for (i = 1 ; i <= n ; ++ i)
      for (j = 1 ; j <= n ; ++ j)
      {
        f[k][i][j] = 1 << 29;
        for (l = 1 ; l <= n ; ++ l)
          f[k][i][j] = min(f[k][i][j] , f[k - 1][i][l] + g[l][j]);
      }
  /*  for (i = 1 ; i <= n ; ++ i)
      for (j = 1 ; j <= n ; ++ j)
        printf("%d%c" , f[k][i][j] , " \n"[j == n]);
*/
  }
  for (i = 1 ; i <= n ; ++ i)
    for (j = 1 ; j <= n ; ++ j)
      if (f[j][i][i] < 1 << 28)
        h[i] = min(h[i] , 1.0 * f[j][i][i] / j);
  for (i = 1 ; i <= n ; ++ i)
    for (j = 1 ; j <= n ; ++ j) if (p[i][j])
      w[i][j] = min(h[i] , h[j]);

  for (k = 1 ; k <= n ; ++ k)
    for (i = 1 ; i <= n ; ++ i) if (p[i][k])
      for (j = 1 ; j <= n ; ++ j) if (p[k][j])
        w[i][j] = min(w[i][j] , h[k]);
  for (i = 1 ; i <= n ; ++ i)
    for (j = 1 ; j <= n ; ++ j)
      for (k = 1 ; k <= n ; ++ k)
        if (f[k][i][j] < 1 << 28)
          w[i][j] = min(w[i][j] , 1. * f[k][i][j] / k);
  for (i = 1 ; i <= n ; ++ i)
    for (j = 1 ; j <= n ; ++ j)
    {
      if (w[i][j] == 1e30)
        printf("NO");
      else printf("%.3lf" , w[i][j]);
      if (j == n) puts(""); else printf(" ");
    }
}

int main()
{
  freopen("~input.txt","r",stdin);
  freopen("~output.txt","w",stdout);

  while (~scanf("%d%d",&n,&m))
  //int _; scanf("%d\n" , &_) ; while (_ --)
    work();
  return 0;
}