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  • [BZOJ3856]Monster

    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


    有一个boss血量为h,每次攻击可以对boss造成a的伤害,boss每轮可以恢复b的血量,连续攻击k轮后必须要休息一轮,问是否能打败boss

    O(1)题。。。具体的判断条件看看代码吧。。。

    /*program from Wolfycz*/
    #include<cmath>
    #include<cstdio>
    #include<cstring>
    #include<iostream>
    #include<algorithm>
    #define inf 0x7f7f7f7f
    using namespace std;
    typedef long long ll;
    typedef unsigned int ui;
    typedef unsigned long long ull;
    inline char gc(){
    	static char buf[1000000],*p1=buf,*p2=buf;
    	return p1==p2&&(p2=(p1=buf)+fread(buf,1,1000000,stdin),p1==p2)?EOF:*p1++;
    }
    inline int frd(){
    	int x=0,f=1;char ch=gc();
    	for (;ch<'0'||ch>'9';ch=gc())	if (ch=='-')    f=-1;
    	for (;ch>='0'&&ch<='9';ch=gc())	x=(x<<1)+(x<<3)+ch-'0';
    	return x*f;
    }
    inline int read(){
    	int x=0,f=1;char ch=getchar();
    	for (;ch<'0'||ch>'9';ch=getchar())	if (ch=='-')	f=-1;
    	for (;ch>='0'&&ch<='9';ch=getchar())	x=(x<<1)+(x<<3)+ch-'0';
    	return x*f;
    }
    inline void print(int x){
    	if (x<0)    putchar('-'),x=-x;
    	if (x>9)	print(x/10);
    	putchar(x%10+'0');
    }
    bool check(int h,int a,int b,int k){
    	if (h<=a||1ll*(a-b)*k-b>0||h<=1ll*(a-b)*(k-1)+a)	return 1;
    	return 0;
    }
    int main(){
    	int Cas=0;
    	while (true){
    		int h=read(),a=read(),b=read(),k=read();
    		if (!(h+a+b+k))	break;
    		printf("Case #%d: ",++Cas);
    		printf(check(h,a,b,k)?"YES
    ":"NO
    ");
    	}
    	return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/Wolfycz/p/9949232.html
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