poj 3783

Balls

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 1196		Accepted: 783

Description

The classic Two Glass Balls brain-teaser is often posed as: 

"Given two identical glass spheres, you would like to determine the lowest floor in a 100-story building from which they will break when dropped. Assume the spheres are undamaged when dropped below this point. What is the strategy that will minimize the worst-case scenario for number of drops?"

Suppose that we had only one ball. We'd have to drop from each floor from 1 to 100 in sequence, requiring 100 drops in the worst case. 

Now consider the case where we have two balls. Suppose we drop the first ball from floor n. If it breaks we're in the case where we have one ball remaining and we need to drop from floors 1 to n-1 in sequence, yielding n drops in the worst case (the first ball is dropped once, the second at most n-1 times). However, if it does not break when dropped from floor n, we have reduced the problem to dropping from floors n+1 to 100. In either case we must keep in mind that we've already used one drop. So the minimum number of drops, in the worst case, is the minimum over all n. 

You will write a program to determine the minimum number of drops required, in the worst case, given B balls and an M-story building.

Input

The first line of input contains a single integer P, (1 ≤ P ≤ 1000), which is the number of data sets that follow. Each data set consists of a single line containing three(3) decimal integer values: the problem number, followed by a space, followed by the number of balls B, (1 ≤ B ≤ 50), followed by a space and the number of floors in the building M, (1 ≤ M ≤ 1000).

Output

For each data set, generate one line of output with the following values: The data set number as a decimal integer, a space, and the minimum number of drops needed for the corresponding values of B and M.

Sample Input

4 
1 2 10 
2 2 100 
3 2 300 
4 25 900

Sample Output

1 4
2 14
3 24
4 10

Source

Greater New York Regional 2009

 

 

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstdlib>
#include <cstring>
#include <string>
#include <deque>
using namespace std;
#define ll long long     
#define  N 1000009
#define  gep(i,a,b)  for(int  i=a;i<=b;i++)
#define  gepp(i,a,b) for(int  i=a;i>=b;i--)
#define  gep1(i,a,b)  for(ll i=a;i<=b;i++)
#define  gepp1(i,a,b) for(ll i=a;i>=b;i--)    
#define  mem(a,b)  memset(a,b,sizeof(a))
#define  mod  1000000007
#define  lowbit(x)  x&(-x)
#define  inf 100000
int t,a,b,m;
int dp[1009][60];
/*
dp[i][j]：i层楼，J个球在最坏的情况下需要的次数
枚举前面的k ： 1 ~ i
没碎  dp[i][j]=dp[i-k][j]+1//还有i-k层楼，下面的楼肯定不需要了，还有j个球
碎了 dp[i][j]=dp[k-1][j-1]+1//往下k-1层，上面的楼肯定不用查了，还有j-1个球
dp[i][j]=min(dp[i][j],max(dp[i-k][j],dp[k-1][j-1])+1);//+1因为 k 层楼需要一次
*/
void init(){
    gep(i,0,1001){
        gep(j,0,51){
            dp[i][j]=inf;
        }
    }
    gep(i,0,51) dp[0][i]=0;
    //从1开始
    gep(i,1,1001){
        gep(j,1,51){
            gep(k,1,i){
                dp[i][j]=min(dp[i][j],max(dp[i-k][j],dp[k-1][j-1])+1);
            }
        }
    }
}
int main()
{
     init();
    scanf("%d",&t);
    while(t--){
        scanf("%d%d%d",&a,&b,&m);
        printf("%d %d\n",a,dp[m][b]);
    }
    return 0;
}