1381 - Scientific Experiment

Time Limit: 20 Sec  Memory Limit: 256 MB

题目连接

http://www.lightoj.com/volume_showproblem.php?problem=1381

Description

John wants to be a scientist. A first step of becoming a scientist is to perform experiment. John has decided to experiment with eggs. He wants to compare the hardness of eggs from different species. He has decided to use a nearby large multi-storied building for this purpose. For each species he will try to find the highest floor from which he can drop the egg and it will not break. The building has (n+1) floors numbered from 0 to n. John has a book from which he knows that

1. If an egg is dropped from the topmost floor, it will surely break.
2. If an egg is dropped from floor 0, it will not break.
3. The eggs of same species are of same strength. That means if any egg breaks when dropped from the k^th floor; all the eggs of that species will break if dropped from k^th floor.
4. If an egg is unbroken after dropping it from any floor, it remains unharmed, that means the strength of the egg remains same.

Unfortunately John has a few problems:

1. He can only carry one egg at a time.
2. He can buy eggs from a shop inside the building and an egg costs x cents.
3. To enter the building he has to pay y cents if he has no egg with him and z cents if he carries an egg with him.
4. After dropping an egg, John must go outside the building to check whether it‘s broken or not.
5. He does not want to waste any egg so he will not leave any unbroken egg on the ground. But if an egg is broken, he leaves it there.
6. If he has an intact egg at the end, he can sell it for x/2 cents. He does not need to enter the building to sell the egg.

These problems are not going to tame John‘s curious mind. So he has decided to use an optimal strategy and minimize his cost in worst case. As John is not a programmer, he asked your help.

Input

Input starts with an integer T (≤ 50), denoting the number of test cases.

Each case starts with a line containing four integers n x y z as described in the statement. You may assume that 1 < n ≤ 1000 and 1 ≤ x, y, z ≤ 10⁵ and x is even.

Output

For each test case, print the case number and the minimized worst case cost.

Sample Input

7

4 2 998 1000

16 2 1000 1000

16 1000 1 1

4 1000 1 1

7 2 2 2

9 2 1 100

11 2 100 1

Sample Output

Case 1: 2000

Case 2: 4008

Case 3: 1015

Case 4: 1003

Case 5: 10

Case 6: 24

Case 7: 111

HINT

For case 1, John knows that the egg will break if dropped from 4^th floor, but will not break if dropped from 0^th floor. An optimal solution may be

1. John enters the building without any egg (¢998).
2. John buys an egg (¢2).
3. John drops an egg from 2nd floor. John goes out and checks the egg.
   1. If it breaks,
      1. i.      John again enters the building without any egg (¢998) and buys an egg there ¢2.
      2. ii.      He drops the egg from 1st floor.
         1. If it does not break then answer to his problem is 1 and he can sell the egg for ¢1. So his final cost in ¢1999.
         2. If it breaks then the answer to his problem is 0th floor and his final cost is ¢2000.
4. If it does not break
   1. i.      John enters the building with the egg (¢1000).
   2. ii.      He drops it from 3rd floor.
      1. If it does not break then answer to his problem is 3 and he can sell the egg for ¢1. So his final cost in ¢1999.
      2. If it breaks then the answer to his problem is 2 and final cost is ¢2000.

So, using this strategy, his worst case cost is ¢2000.

题意

一个评估蛋的硬度方法是测量蛋从多高摔下来会碎。现在佳佳想以楼层高度作为考量依据评估蛋的 硬度。如果蛋从i楼上掉下来会碎，而i-1不会，那么蛋的硬度为i。高为n层的实验楼里面有蛋卖，一个X元。佳佳开始没有蛋，并且他只能随身携带一个蛋， 不带蛋进楼需要Y元，带蛋需要Z元，做完试验之后如果还有一个蛋，可以卖掉得X/2元（卖蛋不需要进楼）。佳佳把鸡蛋扔出去后，会出楼检查蛋的情况。如果 蛋扔下后没有碎掉，佳佳一定会把蛋捡起，然后进楼，如蛋碎掉了，佳佳就不会管它。 佳佳想知道在最糟糕的情况下，测出蛋的硬度最少需要花费多少钱。

题解：

dp[i]表示在楼层高度为i的情况下，检测出碎蛋的位置的最小最差花费是多少

然后转移就好，注意这个dp[i]只和楼层高度有关！

代码：

//qscqesze
#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include <fstream>
#include <map>
#include <stack>
typedef long long ll;
using namespace std;
//freopen("D.in","r",stdin);
//freopen("D.out","w",stdout);
#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
#define maxn 200001
#define mod 10007
#define eps 1e-9
int Num;
char CH[20];
//const int inf=0x7fffffff;   //нчоч╢С
const int inf=0x3f3f3f3f;
/*

inline void P(int x)
{
    Num=0;if(!x){putchar(‘0‘);puts("");return;}
    while(x>0)CH[++Num]=x%10,x/=10;
    while(Num)putchar(CH[Num--]+48);
    puts("");
}
*/
inline ll read()
{
    ll x=0,f=1;char ch=getchar();
    while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();}
    while(ch>=‘0‘&&ch<=‘9‘){x=x*10+ch-‘0‘;ch=getchar();}
    return x*f;
}
inline void P(int x)
{
    Num=0;if(!x){putchar(‘0‘);puts("");return;}
    while(x>0)CH[++Num]=x%10,x/=10;
    while(Num)putchar(CH[Num--]+48);
    puts("");
}
//**************************************************************************************
ll dp[maxn];
ll d1,d2;
int n,x,y,z;
int main()
{
    int t=read();
    for(int cas=1;cas<=t;cas++)
    {
        n=read(),x=read(),y=read(),z=read();
        dp[0]=dp[1]=0;
        for(int i=2;i<=n;i++){
            dp[i]=inf;
            for(int j=1;j<i;j++)
            {
                d1=dp[j]+x+y;
                d2=dp[i-j]+z;
                if(i-j==1)
                    d2=y+x/2;
                dp[i]=min(dp[i],max(d1,d2));
            }
        }
        printf("Case %d: %lld\n",cas,dp[n]);
    }
}