Codeforces Round 1044 (Div. 2)——D. Chicken Jockey（跌落伤害，最少攻击次数）

思路讲解

其实这个dp的思路非常简单，被AI带偏了。

$dp[i]$ 表示消除到第 i 个序列所需的攻击次数。

$dp[i-1]+A[i]-1$ 这个表示不做任何操作。

$dp[i-2]+A[i-1]+A[i]-\min(A[i],i-1)$ 这个表示做操作，$A[i-1]$ 需要消掉，$A[i]$可以跌落，跌落伤害是 $\min(A[i],i-1)$，$A[i]-\min(A[i],i-1)$ 这个就是剩余所需操作数。

inline void __(){
    cin>>N;
    vll A(N+5);
    FOR(i,1,N){
        cin>>A[i];
    }
    vll dp(N+5);
    dp[1]=A[1];
    FOR(i,2,N){
        dp[i]=min(dp[i-1]+A[i]-1,dp[i-2]+A[i-1]+A[i]-min(A[i],i-1));
    }
    cout<<dp[N]<<"\n";
}

AC代码

https://codeforces.com/contest/2133/submission/335405306

源代码

// by Gospel_rock
#include <bits/stdc++.h>
#define all(vec) vec.begin(),vec.end()
#define PB push_back
#define EB emplace_back
#define EM emplace
#define FI first
#define SE second
#define pct __builtin_popcountll
#define ctz __builtin_ctzll
#define SZ(a) ((long long) a.size())
#define FOR(i, a, b) for (long long i = (a); i <= (b); ++i)
#define ROF(i, a, b) for (long long i = (a); i >= (b); --i)
#define debug(var) cerr << #var <<":"<<var<<"\n";
#define cend cerr<<"\n-----------\n"
#define fsp(x) fixed<<setprecision(x)
#define minn(vec) *min_element(vec.begin(),vec.end())
#define maxx(vec) *max_element(vec.begin(),vec.end())
using namespace std;
#ifdef LOCAL
template<typename T> void _print(T x) { cerr << x; }
// 可以打印多层数组，但是格式不是很好
template<typename T> void _print(vector<T> v) { cerr << "[ "; for (T i : v) _print(i), cerr << " "; cerr << "]"; }
template<typename T> void _print(set<T> s) { cerr << "{ "; for (T i : s) _print(i), cerr << " "; cerr << "}"; }
template<typename T, typename U> void _print(pair<T, U> p) { cerr << "{"; _print(p.first); cerr << ", "; _print(p.second); cerr << "}"; }
template<typename T, typename U> void _print(map<T, U> m) { cerr << "{ "; for (auto &p : m) _print(p), cerr << " "; cerr << "}"; }
// 打印两层数组，格式好一些
template<typename T> void _print(vector<vector<T>> v){cerr<<"{\n";for (auto i : v) _print(i), cerr << "\n";cerr<<" }";}
#endif

using  ll = long long;using ull = unsigned long long;
using DB=double;using LD=long double;
using i128 = __int128;
using pdd = pair<DB,DB>;using plb = pair<ll,bool>;
using pll = pair<ll,ll>;using vll=vector<ll>;using vb=vector<bool>;
using tll = tuple<ll,ll,ll>;using vii=vector<int>;
using vpll = vector<pll>;using vtll = vector<tll>;
using vdb = vector<DB>;using vc=vector<char>;
using arr3 = array<ll,3> ;using arr2 = array<ll,2>;
using CD=complex<double>;
static constexpr ll MAXN=(ll)1e6+10,INF=(ll)1e17+3; // 1e17+3是素数 1e18+9是素数
static constexpr ll mod=(ll)1e9+7; // 998244353;
static constexpr double eps=1e-8;const double pi=acos(-1.0);
ll N,M,Q,X,K,T,lT;

inline void __();
signed main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);cout.tie(0);
    
    cin>>lT;
    while(lT--)
        __();
    return 0;
}
/*

*/
inline void __(){
    cin>>N;
    vll A(N+5);
    FOR(i,1,N){
        cin>>A[i];
    }
    vll dp(N+5);
    dp[1]=A[1];
    FOR(i,2,N){
        dp[i]=min(dp[i-1]+A[i]-1,dp[i-2]+A[i-1]+A[i]-min(A[i],i-1));
    }
    cout<<dp[N]<<"\n";
}
/*
AC
https://codeforces.com/contest/2133/submission/335405306
*/

心路历程（WA，TLE，MLE……）