【题解】Educational Codeforces Round 130 (Rated for Div. 2)ABC

A. Parkway Walk

题意：

给你初始体力k，从一个座位跳到相邻的作为需要花费体力$ai$，可以在任意作为回复任意体力，求从起点到终点至少恢复多少体力

思路：

greedy，体力恰好足够即可，没必要多休息

code：

#include 
using namespace std;
const double pi = acos(-1.0);
#define x first
#define y second
#define LL long long 
#define int LL
#define pb push_back
#define all(v) (v).begin(),(v).end()
#define PII pair<int,int>
#define ll_INF 0x7f7f7f7f7f7f7f7f
#define INF 0x3f3f3f3f
#define debug(x) cerr << #x << ": " << x << endl
#define io ios_base::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr)
LL Mod(LL a,LL mod){return (a%mod+mod)%mod;}
LL lowbit(LL x){return x&-x;}//最低位1及其后面的0构成的数值
LL qmi(LL a,LL b,LL mod) {LL ans = 1; while(b){ if(b & 1) ans = ans * (a % mod) % mod; a = a % mod * (a % mod) % mod; b >>= 1;} return ans; }
int _;
int n,m;
const int N=110;
int a[N];
void solve()
{
	cin>>n>>m;
	for(int i=1;i<=n;i++)cin>>a[i];
	int res=0;
	int k=m;
	for(int i=1;i<=n;i++)
	{
		if(k>=a[i])k-=a[i];
		else res+=a[i]-k,k=0;
	} 
	cout<<res<<endl;
}
signed main()
{
	io;
	cin>>_;
	while(_--)
	solve();
	return 0;
}

B. Promo

题意：

对一个数组排序，求最大的 $x$ 个的前 $y$ 个最小的和

思路：

sort排序，前缀和优化一下

code:

#include 
using namespace std;
const double pi = acos(-1.0);
#define x first
#define y second
#define LL long long 
#define int LL
#define pb push_back
#define all(v) (v).begin(),(v).end()
#define PII pair<int,int>
#define ll_INF 0x7f7f7f7f7f7f7f7f
#define INF 0x3f3f3f3f
#define debug(x) cerr << #x << ": " << x << endl
#define io ios_base::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr)
LL Mod(LL a,LL mod){return (a%mod+mod)%mod;}
LL lowbit(LL x){return x&-x;}//最低位1及其后面的0构成的数值
LL qmi(LL a,LL b,LL mod) {LL ans = 1; while(b){ if(b & 1) ans = ans * (a % mod) % mod; a = a % mod * (a % mod) % mod; b >>= 1;} return ans; }
int _;
int n,q;
const int N=2e5+10;
int a[N];
int s[N];
void solve()
{
	scanf("%lld%lld",&n,&q);
	for(int i=1;i<=n;i++)scanf("%lld",&a[i]);
	sort(a+1,a+1+n);
	for(int i=1;i<=n;i++)s[i]=s[i-1]+a[i];
	while(q--)
	{
		int x,y;
		scanf("%lld%lld",&x,&y);
		printf("%lld\n",s[n-x+y]-s[n-x]);
	}
	
}
signed main()
{
//	io;
	//cin>>_;
//	while(_--)
	solve();
	return 0;
}

C. awoo’s Favorite Problem

题意：

给定两个字符串$s,t$，给定两种操作，问$s$ 是否能转换到$t$

1. $ab->ba$
2. $bc->cb$

思路：

发现以 $b$ 进行交换，$b$ 的位置是不受印象的，$a$只能换到比原先下标大的，$c$只能换到比原先下标小的。我们只存储 $a,c$的下标，判断字典序即可（a,b,c三种字符的数量首先得相同）

code:

#include 
using namespace std;
const double pi = acos(-1.0);
#define x first
#define y second
#define LL long long 
#define int LL
#define pb push_back
#define all(v) (v).begin(),(v).end()
#define PII pair<int,int>
#define ll_INF 0x7f7f7f7f7f7f7f7f
#define INF 0x3f3f3f3f
#define debug(x) cerr << #x << ": " << x << endl
#define io ios_base::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr)
LL Mod(LL a,LL mod){return (a%mod+mod)%mod;}
LL lowbit(LL x){return x&-x;}//最低位1及其后面的0构成的数值
LL qmi(LL a,LL b,LL mod) {LL ans = 1; while(b){ if(b & 1) ans = ans * (a % mod) % mod; a = a % mod * (a % mod) % mod; b >>= 1;} return ans; }
int _;
int n;
void solve()
{
	cin>>n;
	string s,t;
	cin>>s>>t;
    for (auto c : { 'a', 'b', 'c' }) 
	{
        if (std::count(s.begin(), s.end(), c) != std::count(t.begin(), t.end(), c)) {
            std::cout << "NO\n";
            return;
        }
    }
    string a,b;
    vector<int>A,B;
    for(int i=0;i<n;i++)
    {
    	if(s[i]!='b')
    	{
    		a+=s[i];
    		A.pb(i);
    	}
    }
    for(int i=0;i<n;i++)
    {
    	if(t[i]!='b')
    	{
    		b+=t[i];
    		B.pb(i);
    	}
    }
    if(a!=b)
    {
    	cout<<"NO"<<endl;
    	return;
    }
    for(int i=0;i<A.size();i++)
    {
    	if(a[i]=='a'&&A[i]>B[i])
    	{
    		cout<<"NO"<<endl;
    		return;
    	}
    	if(a[i]=='c'&&A[i]<B[i])
    	{
    		cout<<"NO"<<endl;
    		return;
    	}
    }
    cout<<"YES"<<endl;
}
signed main()
{
	io;
	cin>>_;	
	while(_--)
	solve();
	return 0;
}