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hdu4950 Monster

編輯：關於C++

Monster

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1782 Accepted Submission(s): 550

Problem Description Teacher Mai has a kingdom. A monster has invaded this kingdom, and Teacher Mai wants to kill it.

Monster initially has h HP. And it will die if HP is less than 1.

Teacher Mai and monster take turns to do their action. In one round, Teacher Mai can attack the monster so that the HP of the monster will be reduced by a. At the end of this round, the HP of monster will be increased by b.

After k consecutive round's attack, Teacher Mai must take a rest in this round. However, he can also choose to take a rest in any round.

Output YES if Teacher Mai can kill this monster, else output NO.
Input There are multiple test cases, terminated by a line 0 0 0 0.

For each test case, the first line contains four integers h,a,b,k(1<=h,a,b,k <=10^9).
Output For each case, output Case #k: first, where k is the case number counting from 1. Then output YES if Teacher Mai can kill this monster, else output NO.

Sample Input

5 3 2 2
0 0 0 0

Sample Output

Case #1: NO

Author xudyh
題意：已知你的攻擊力a、怪物血量h、每回合怪物的回血量b和每k回合你必須休息一回合；你也可以不到k回合就休息。分析：只需要判斷每k回合怪物掉的血量比回的血量多就YES，否則就NO。

#include 
#include 
#include 
#include 
#include 
#include
#include 
#include 
#include 
#include 
using namespace std;
const double eps = 1e-6;
const double pi = acos(-1.0);
const int INF = 0x3f3f3f3f;
const int MOD = 1000000007;
#define ll long long
#define CL(a) memset(a,0,sizeof(a))

int main()  
{  
    ll h,a,b,k,cas=1;  
    while(~scanf(%I64d%I64d%I64d%I64d,&h,&a,&b,&k),h+a+b+k)  
    {  
        if(a>=h||(a-b)*k-b>0||(a-b)*(k-1)+a>=h)//只要判斷每個k回合怪物掉的血比回的血多久行
            printf(Case #%lld: YES
,cas++);  
        else  
            printf(Case #%lld: NO
,cas++);  
    }  
    return 0;  
}