守望者的逃离 题解

通过计算，我们可以发现闪现+等CD会比一直跑快，所以能闪现就闪现。走路(plan1)和闪现(plan2)同时进行，如果plan2比plan1快，就用plan2更新plan1。如果plan1比plan2快，就继续进行。如果plan1超过了目标距离，输出"Yes"和plan1的值。否则输出"No"和plan1的值。

AC代码

#include<set>
#include<map>
#include<stack>
#include<queue>
#include<math.h>
#include<stdio.h>
#include<limits.h>
#include<iostream>
#include<algorithm>
#define N 1005
#define int long long
using namespace std;
inline int read(void){
    int x=0,f=1;
    char c=getchar();
    while(c<'0' || c>'9'){if(c == '-') f=-1;c=getchar();}
    while(c>='0' && c<='9') x=x*10+c-'0',c=getchar();
    return x*f;
}
 
inline void write(int x) {
    if(x<0) putchar('-'),x=-x;
    if(x>9) write(x/10);
    putchar(x%10+'0');
    return;
}
 
class node{
private:
	long long plan_1,plan_2;
	unsigned long long m,s,t,now;
public:
	explicit node(register long long K){
		plan_1 = K,plan_2 = K,now = K;
		return;
	}
	inline void get_num(void){
		cin >> m >> s >> t;
		_start(); 
	}
	inline void _start(void){
		for(register int i=1; i<=t; i++){
			plan_1+=17;
			if(m >= 10){
				plan_2+=60;
				m-=10;
			}else m+=4;
			if(plan_2 > plan_1)	plan_1=plan_2;
			if(plan_1 > s){
				write_num_1(i);
				return;
			}
		}
		write_num_2(plan_1);
		return;
	}
	inline void write_num_1(register int answer){
		printf("Yes\n");
		write(answer);
		return;
	}
	inline void write_num_2(register int answer){
		printf("No\n");
		write(answer);
		return;
	}
};
 
signed main(void){
	node a(0);
	a.get_num();
	return 0;
}

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