思路:一共三种情况,分开处理,先处理用能量的,再处理跑步的
#include<bits/stdc++.h>
#define fi first
#define se second
#define INF 0x3f3f3f3f
#define ll long long
#define ld long double
#define mem(ar,num) memset(ar,num,sizeof(ar))
#define me(ar) memset(ar,0,sizeof(ar))
#define lowbit(x) (x&(-x))
#define IOS ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define lcm(a,b) ((a)*(b)/(__gcd((a),(b))))
#define Max 300010
#define mod 1000000007
using namespace std;
int dp[300001];
int main() {
int m, s, t;
scanf("%d%d%d", &m, &s, &t);
for(int i = 1; i <= t; i++) { //处理闪烁法术
if(m >= 10)
dp[i] = dp[i - 1] + 60, m -= 10; //如果能用,就用
else
dp[i] = dp[i - 1], m += 4; //否则休息
}
for(int i = 1; i <= t; i++) {
dp[i] = max(dp[i], dp[i - 1] + 17); //处理跑步,dp[i]为使用法术和跑步的最大值(最优)
if(dp[i] >= s) {
printf("Yes\n%d", i); //如果超过了距离s,就成功了,输出yes
return 0;
}
}
printf("No\n%d", dp[t]); //没成功,输出no
return 0;
}