UVa 10081 tight Words (DP)

10081-tight Words

Time limit:3.000 seconds

Http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=115&page=show_ problem&problem=1022

Train of thought: a string of length I, if the end number is J, then the number of such strings must be long as i-1 string, the mantissa is j-1,j,j+1 the sum of the number of three kinds of cases (within the boundary)

Complete code:

/*0.019s*/
  
#include <bits/stdc++.h>
using namespace std;
  
Double dp[111][11];
  
int main ()
{
    int k, n, I, J;
    Double ans;
    while (~SCANF ("%d%d", &k, &n))
    {for
        (i = 0; I <= k; ++i)  dp[1][i] = 1.0/(k + 1);
        for (i = 2; I <= n; ++i) for
            (j = 0; j <= K; ++j)
            {
                dp[i][j] = Dp[i-1][j]/(k + 1);
                if (J > 0) dp[i][j] + = Dp[i-1][j-1]/(k + 1);
                if (J < K) Dp[i][j] + = dp[i-1][j + 1]/(k + 1);
            }
        Ans = 0;
        for (i = 0; I <= K; i++) ans + = dp[n][i];
        printf ("%.5f\n", ans *);
    return 0;
}

Author: csdn Blog Synapse7

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