|洛谷|NOIP2007|动态规划|P1095 守望者的逃离

http://www.luogu.org/problem/show?pid=1095

第一次循环让守望者只能闪烁，然后第二次循环让守望者只能跑步，比较两次的长短即可

#include
#include
#include
#include
#define ms(i,j) memset(i,j, sizeof i);
using namespace std;
int f[300005];
int main()
{
	int m,s,t;
	scanf("%d%d%d", &m, &s, &t);
	f[0] = 0;
	for (int i=1;i<=t;i++)
	{
		if (m<10)
		{
			f[i] = f[i-1];
			m += 4;
		}else {m -= 10; f[i] = f[i-1]+60;}
	}
	for (int i=1;i<=t;i++)
	{
		
		f[i] = max(f[i], f[i-1]+17);
		if (f[i]>=s)
		{
			printf("Yes\n%d", i);
			return 0;
		}
	}
	printf("No\n%d", f[t]);
    return 0;
}

转载于:https://www.cnblogs.com/flyinthesky1/p/6384252.html