P1095 守望者的逃離: https://www.luogu.org/problemnew/show/P1095

題意：

　　有一個人要在S長度的直線上跑過去，初始有M的魔法值，用10點魔法值可以在一秒內跑60米，而普通跑一秒17米。保持靜止可以恢復4點的魔法值。問能否在T秒前跑完。

思路：

　　分開兩次dp，第一次跑出能用加速就用加速的路程。第二次比較dp【i】和dp【i-1】+17的值即可。

#include <algorithm>
#include  <iterator>
#include  <iostream>
#include   <cstring>
#include   
 
<iomanip>
#include   <cstdlib>
#include    <cstdio>
#include    <string>
#include    <vector>
#include    <bitset>
#include    <cctype>
#include     <queue>
#include     <cmath>
#include      <list>
#include       <map>
#include       
 
<set>
//#include <unordered_map>
//#include <unordered_set>
//#include<ext/pb_ds/assoc_container.hpp>
//#include<ext/pb_ds/hash_policy.hpp>
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<ll ,ll > pll;
typedef pair<int ,int > pii;
typedef pair<int ,pii> p3;
//priority_queue<int> q;//這是一個大根堆q
//priority_queue<int,vector<int>,greater<int> >q;//這是一個小根堆q
//__gnu_pbds::cc_hash_table<int,int>ret[11];    //這是很快的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)  //用來壓行
#define REP(i , j , k)  for(int i = j ; i <  k ; ++i)
//priority_queue<int ,vector<int>, greater<int> >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<typename T>
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;
}

luogu -P1095 守望者的逃離 - 分別DP