AtCoder Beginner Contest 365

A - Leap Year

思路

模拟即可;

AC代码

    #include<bits/stdc++.h>
    #define endl '\n'
    #define int int long long
    #define pb push_back
    #define bs bitset
    using namespace std;
    typedef pair<char,int> PCI;
    typedef pair<int,int> PII;
    typedef priority_queue<int> PQ;

    const int N = 2e5+10, MAX = 1e9, INF = -1e9;

    int n;

    signed main()
    {
        ios::sync_with_stdio(false);
        cin.tie(0);
        cin>>n;
        if(n%4!=0)cout<<365<<endl;
        else{
            if(n%100!=0)cout<<366<<endl;
            else{
                if(n%400!=0)cout<<365<<endl;
                else cout<<366<<endl;
            }
        }
        
        return 0;
    }

B - Second Best

思路

模拟即可;

AC代码

    #include<bits/stdc++.h>
    #define endl '\n'
    #define int int long long
    #define pb push_back
    #define bs bitset
    using namespace std;
    typedef pair<char,int> PCI;
    typedef pair<int,int> PII;
    typedef priority_queue<int> PQ;

    const int N = 2e5+10, MAX = 1e9, INF = -1e9;

    int n;
    PII a[N];
    int e;

    bool cmp(PII a,PII b){
        return a.first>b.first;
    }

    signed main()
    {
        ios::sync_with_stdio(false);
        cin.tie(0);
        cin>>n;

        for(int i=1;i<=n;i++){
            cin>>e;
            a[i].first=e;
            a[i].second=i;
        }
        sort(a+1,a+1+n,cmp);
        cout<<a[2].second<<endl;

        return 0;
    }

C - Transportation Expenses

思路

其实二分\(x\)就可以了，我还使用了前缀和，最后的复杂度是\(O(nlogn+logxlogn)\);

AC代码

    #include<bits/stdc++.h>
    #define endl '\n'
    #define int int long long
    #define pb push_back
    #define bs bitset
    using namespace std;
    typedef pair<char,int> PCI;
    typedef pair<int,int> PII;
    typedef priority_queue<int> PQ;

    const int N = 2e5+10, MAX = 1e9, INF = -1e9;

    int n,m;
    vector<int> v;
    int sum=0;
    int e;
    int s[N];

    int counts(int x){
        int p=upper_bound(v.begin(),v.end(),x)-v.begin();
        return s[p-1]+(n-p+1)*x;
    }

    int f(int l,int r){
        while(l<r){
            int mid=(l+r+1)>>1;
            if(counts(mid)<=m)l=mid;
            else r=mid-1;
        }
        return l;
    }

    signed main()
    {
        ios::sync_with_stdio(false);
        cin.tie(0);
        cin>>n>>m;
        v.pb(0);s[0]=0;
        for(int i=1;i<=n;i++){
            cin>>e;
            sum+=e;
            v.pb(e);
        }

        sort(v.begin(),v.end());
        for(int i=1;i<=n;i++){
            s[i]=s[i-1]+v[i];
        }

        if(sum<=m)cout<<"infinite"<<endl;
        else{
            cout<<f(1,1e9+1)<<endl;
        }

        return 0;
    }

D - AtCoder Janken 3

思路

动态规划，很简单的转移；

AC代码

#include<bits/stdc++.h>
#define endl '\n'
#define int int long long
#define pb push_back
#define bs bitset
using namespace std;
typedef pair<char,int> PCI;
typedef pair<int,int> PII;
typedef priority_queue<int> PQ;

const int N = 2e5+10, MAX = 1e9, INF = -1e9;

int n;
char c;
int dp[N][4];

signed main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cin>>n;
    memset(dp,0,sizeof(dp));
	for(int i=1; i<=n; i++) {
		cin>>c;
		if(c=='R') {
			dp[i][1]=max(dp[i-1][2],dp[i-1][3]);
			dp[i][2]=-10086;
			dp[i][3]=max(dp[i-1][2],dp[i-1][1])+1;
		}
		if(c=='S') {
			dp[i][2]=max(dp[i-1][1],dp[i-1][3]);
			dp[i][3]=-10086;
			dp[i][1]=max(dp[i-1][2],dp[i-1][3])+1;
		}
		if(c=='P') {
			dp[i][3]=max(dp[i-1][1],dp[i-1][2]);
			dp[i][1]=-10086;
			dp[i][2]=max(dp[i-1][1],dp[i-1][3])+1;
		}
	}
	cout<<max(dp[n][1],max(dp[n][2],dp[n][3]))<<endl;
    return 0;
}

t.b.c.