hey_left 11 Codeforces Round 859 (Div. 4)

题目链接

A.

直接判断输出

#include <bits/stdc++.h>
using namespace std;

void solve(){
    int a,b,c;cin>>a>>b>>c;
    if(a+b==c)cout<<'+'<<'\n';
    else if(a-b==c)cout<<"-"<<'\n';
}

signed main(){
    int hey_left=1;
    cin>>hey_left;
    while(hey_left--){
        solve();
    }
}

B.

把偶数全部放前面
再维护偶数和与奇数和
遍历所有数，是偶数就+到偶数和，是奇数就加到奇数和，若某一个时刻奇数和大等于偶数和了，就是no

#include <bits/stdc++.h>
using namespace std;

void solve(){
    int n;cin>>n;
    vector<int>a(n+1);
    for(int i=1;i<=n;i++)cin>>a[i];
    vector<int>ou,ji;
    for(int i=1;i<=n;i++){
        if(a[i]%2==0)ou.push_back(a[i]);
        else ji.push_back(a[i]);
    }
    for(int i=0;i<ji.size();i++){
        ou.push_back(ji[i]);
    }
    int sum_ou=0,sum_ji=0;
    bool f=0;
    for(int i=0;i<ou.size();i++){
        if(ou[i]%2==0)sum_ou+=ou[i];
        else sum_ji+=ou[i];
        if(sum_ji>=sum_ou){
            f=1;break;
        }
    }
    if(f)cout<<"NO"<<'\n';
    else cout<<"YES"<<'\n';
}

signed main(){
    int hey_left=1;
    cin>>hey_left;
    while(hey_left--){
        solve();
    }
}

C.

直接按题意模拟
需要注意的是，mp[]=0相当于没标记
这里换成1和2即可

#include <bits/stdc++.h>
using namespace std;

void solve(){
    int n;cin>>n;
    string s;cin>>s;
    map<char,int>mp;
    mp[s[0]]=1;
   // cout<<"s[0]="<<s[0]<<' '<<"mp[]="<<mp[s[0]]<<'\n';
    bool f=0;
    for(int i=1;i<s.size();i++){
        if(s[i]==s[i-1]){
            f=1;
            break;
        }
        if(mp[s[i]]){
            if(mp[s[i-1]]==mp[s[i]]){
                f=1;
                break;
            }
        }else {
            if(mp[s[i-1]]==1)mp[s[i]]=2;
            else if(mp[s[i-1]]==2)mp[s[i]]=1;
        }
        //cout<<"s[i]="<<s[i]<<' '<<"mp[]="<<mp[s[i]]<<'\n';
    }
    if(f)cout<<"NO"<<'\n';
    else cout<<"YES"<<'\n';
}

signed main(){
    int hey_left=1;
    cin>>hey_left;
    while(hey_left--){
        solve();
    }
}

D.

记录下前缀和与后缀和，再加上改变的和
判奇偶性

#include <bits/stdc++.h>
using namespace std;

void solve(){
    int n,q;cin>>n>>q;
    vector<int>a(n+1),pre(n+5),post(n+5);
    for(int i=1;i<=n;i++){
        cin>>a[i];
        pre[i]=pre[i-1]+a[i];
    }
    for(int i=n;i>=1;i--)post[i]=post[i+1]+a[i];
    while(q--){
        int l,r,k;cin>>l>>r>>k;
        int sum=0;
        sum+=pre[l-1]+post[r+1]+(r-l+1)*k;
        if(sum&1){
            cout<<"YES"<<'\n';
        }else cout<<"NO"<<'\n';
    }
}

signed main(){
    int hey_left=1;
    cin>>hey_left;
    while(hey_left--){
        solve();
    }
}

E.

二分特殊石头所在的区间
注意要特判边界，详情见代码

#include <bits/stdc++.h>
using namespace std;

void solve(){
    int n;cin>>n;
    vector<int>a(n+1),pre(n+1);
    for(int i=1;i<=n;i++){
        cin>>a[i];
        pre[i]=pre[i-1]+a[i];
    }
    int l=1,r=n,mid;
    while(l<=r){
        mid=(l+r)/2;
        cout<<"? "<<(mid-l+1)<<' ';
        for(int i=l;i<=mid;i++){
            cout<<i;
            if(i==mid)cout<<endl;
            else cout<<' ';
        }
        int x;cin>>x;
        if(x==pre[mid]-pre[l-1])l=mid+1;
        else {
            cout<<"? "<<1<<' '<<mid<<endl;
            cin>>x;
            if(x==a[mid])r=mid-1;
            else {
                cout<<"! "<<mid<<endl;
                return ;
            }
        }
    }
    cout<<"! "<<r<<endl;
}

signed main(){
    int hey_left=1;
    cin>>hey_left;
    while(hey_left--){
        solve();
    }
}