P1095 守望者的逃离: https://www.luogu.org/problemnew/show/P1095
有一个人要在S长度的直线上跑过去，初始有M的魔法值，用10点魔法值可以在一秒内跑60米，而普通跑一秒17米。保持静止可以恢复4点的魔法值。问能否在T秒前跑完。
分开两次dp，第一次跑出能用加速就用加速的路程。第二次比较dp【i】和dp【i-1】+17的值即可。
#include #include #include #include #include #include #include #include <string> #include #include #include #include #include #include #include #include <set> //#include //#include //#include //#include using namespace std; //#pragma GCC optimize(3) //#pragma comment(linker, "/STACK:102400000,102400000") //c++ #define lson (l , mid , rt << 1) #define rson (mid + 1 , r , rt << 1 | 1) #define debug(x) cerr << #x << " = " << x << "\n"; #define pb push_back #define pq priority_queue typedef long long ll; typedef unsigned long long ull; typedef pair pll; typedef pair<int ,int > pii; typedef pair<int ,pii> p3; //priority_queue q;//这是一个大根堆q //priority_queue,greater >q;//这是一个小根堆q //__gnu_pbds::cc_hash_tableret[11]; //这是很快的hash_map #define fi first #define se second //#define endl '\n' #define OKC ios::sync_with_stdio(false);cin.tie(0) #define FT(A,B,C) for(int A=B;A <= C;++A) //用来压行 #define REP(i , j , k) for(int i = j ; i < k ; ++i) //priority_queue, greater >que; const ll mos = 0x7FFFFFFFLL; //2147483647 const ll nmos = 0x80000000LL; //-2147483648 const int inf = 0x3f3f3f3f; const ll inff = 0x3f3f3f3f3f3f3f3fLL; //18 const double PI=acos(-1.0); template inline T read(T&x){ x=0;int f=0;char ch=getchar(); while (ch<'0'||ch>'9') f|=(ch=='-'),ch=getchar(); while (ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar(); return x=f?-x:x; } /*-----------------------showtime----------------------*/ const int maxn = 300009; int dp[maxn],m,s,t; int main(){ cin>>m>>s>>t; for(int i=1; i<=t; i++){ if(m >= 10){ dp[i] = dp[i-1]+60; m -= 10; } else { dp[i] = dp[i-1]; m += 4; } } for(int i=1; i<=t; i++){ dp[i] = max(dp[i], dp[i-1]+17); if(dp[i]>=s){ puts("Yes"); printf("%d\n", i); return 0; } } puts("No"); printf("%d\n", dp[t]); return 0; }
转载于:https://www.cnblogs.com/ckxkexing/p/9734244.html
