A題

簽到模擬

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<int,int> pll;
const int inf=0x3f3f3f3f;
const int N=1e5+10;
const int mod=1e9+7;
int a[N];
int main(){
    ios::sync_with_stdio(false);
    int n;
    cin>>n;
    int i;
    for(i=1;i<=n;i++){
        cin
 
>>a[i];
    }
    if(n==1){
        cout<<"-1"<<endl;
    }
    else if(n==2){
        if(a[1]==a[2]){
            cout<<-1<<endl;
        }
        else{
            cout<<1<<endl;
            cout<<1<<endl;
        }
    }
    else{
        int
 
 mx=0x3f3f3f3f;
        cout<<1<<endl;
        int id;
        for(i=1;i<=n;i++){
            if(mx>a[i]){
                mx=a[i];
                id=i;
            }
        }
        cout<<id<<endl;
 
    }
    return 0;
}

B題

按奇數偶數分，然後排序貪心求解

#include<bits/stdc++.h>
using
 
 namespace std;
typedef long long ll;
typedef pair<int,int> pll;
const int inf=0x3f3f3f3f;
const int N=1e5+10;
const int mod=1e9+7;
vector<int> num;
int a[N];
int main(){
    ios::sync_with_stdio(false);
    int n,b;
    cin>>n>>b;
    int i;
    int cnt1=0;
    int cnt2=0;
    for(i=1;i<=n;i++)
        cin>>a[i];
    for(i=1;i<=n;i++){
        if(a[i]%2){
            cnt1++;
        }
        else{
            cnt2++;
        }
        if(cnt1==cnt2){
            if(i+1<=n){
                num.push_back(abs(a[i+1]-a[i]));
            }
        }
    }
    sort(num.begin(),num.end());
    int ans=0;
    int cnt=0;
    while(b){
        if(cnt==(int)num.size())
            break;
        if(b-num[cnt]<0)
            break;
        ans++;
        b-=num[cnt];
        cnt++;
    }
    cout<<ans<<endl;
    return 0;
}

C題

我們發現每次反轉都會少了全0子串，所以貪心判斷即可

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<int,int> pll;
const int inf=0x3f3f3f3f;
const int N=1e5+10;
const int mod=1e9+7;
int main(){
    ios::sync_with_stdio(false);
    int n,x,y;
    cin>>n>>x>>y;
    string s;
    cin>>s;
    int i;
    if(x>y){
        ll ans=0;
        for(i=0;i<(int)s.size();i++){
            if(s[i]=='0'){
                ans+=y;
                while(i<(int)s.size()-1&&s[i+1]=='0')
                    i++;
            }
        }
        cout<<ans<<endl;
    }
    else{
        ll ans=y;
        int cnt=0;
        for(i=0;i<(int)s.size();i++){
            if(s[i]=='0'){
                cnt++;
                while(i<(int)s.size()-1&&s[i+1]=='0')
                    i++;
            }
        }
        if(cnt==0){
            cout<<0<<endl;
        }
        else
        cout<<ans+(ll)x*(cnt-1)<<endl;
    }
    return 0;
}

D題

這種題顯然就是找規律，找到前12個後，後面的規律就比較明顯了

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N=5e5+10;
int d[]={0,4,10,20,35,56,83,116,155,198,244,292};
int main(){
    ios::sync_with_stdio(false);
    ll n;
    cin>>n;
    if(n<=11){
        cout<<d[n]<<endl;
    }
    else{
        ll ans=292;
        cout<<ans+(ll)(n-11)*49<<endl;
    }
}