"URAL 1223" Chernobyl ' Eagle on a Roof (legendary DP God optimized Eagle Egg)

"URAL 1223" Chernobyl ' Eagle on a Roof (legendary DP God optimized Eagle Egg)

Remember that the Jin Gian said this problem in many school races.

Today finally met the Buddha.

The worst is not the topic itself, but Chen Ben's five-weight optimization.

Now the realm only reaches the second level ...

The third, the monotony of the linear programming, and the subsequent diagram of the transfer complexity down to the LOG2 (n) part ....

I would like to say a little bit about the way I n2logn N^2logn. Stuck with the 1s past ... Good Konjac

That is, the initial DP sets an optimization, the optimization is good understanding, because the second test is the least number of times, then when the Eagle Egg number m allows two points to do, the output can be.

The rest of the bare-running DP.
DP[I][J] Represents the minimum number of times I have an eagle egg in a J-floor experiment.

Enumeration 1 <= m <= J,
Dp[i][j]=min1<=m<=j[max (Dp[i][m−1],dp[i−1][j−m]) +1] dp[i][j] = min_{1

I don't see the PPT. Even if there is a picture.

The idea is to throw the first Eagle egg on level m,
If it's broken, then it's the minimum number of experiments in the m-1 layer with i-1 eggs. +1 (dp[i-1][m-1]+1)
If it's not broken, it's using the I egg and then the minimum number of experiments in the M+1~J layer +1 (dp[i][j-m]+1)

And then it's gone ...

The code is as follows:

#include <iostream> #include <cmath> #include <vector> #include <cstdlib> #include <cstdio > #include <climits> #include <ctime> #include <cstring> #include <queue> #include <stack&
Gt  #include <list> #include <algorithm> #include <map> #include <set> #define LL Long Long #define Pr pair<int,int> #define FREAD (CH) freopen (CH, "R", stdin) #define FWRITE (CH) freopen (CH, "w", stdout) using namespace s
td
const int INF = 0X3F3F3F3F;
const int mod = 1E9+7;
Const double EPS = 1e-8;

const int MAXN = 1123;

int DP[10][MAXN];
    int main () {//fread ("");

    Fwrite ("");

    int n,m; 
            while (~SCANF ("%d%d", &m,&n) && (n+m)) {if (M >= ceil (log ((n+1) *1.0)/log (2.0)))
        printf ("%d\n", (int) ceil (log ((n+1) *1.0)/log (2.0)));

            else {memset (dp,0,sizeof (DP));

            for (int i = 1; I <= n; ++i) dp[1][i] = i; for (int i = 2; I <= M ++i) {for (int j = 1; j <= N; ++j) {Dp[i][j] = Dp[i][j
                    -1]+1; for (int k = 1; k < J; ++k) {Dp[i][j] = min (Dp[i][j],max (dp[i][k-1],dp[i-1) [J-k])
                    +1);
        }}} printf ("%d\n", Dp[m][n]);
}} return 0; }