Codeforces round 810 (Div. 2) [competition record]

A. Perfect Permutation

n Even numbers ,i~i+1 Just switch positions in pairs ;

n For an odd number , Exchange anyway , There must be one that satisfies the divisible condition , So the first number doesn't move , Then even numbers can be exchanged in pairs ！！！

#include <bits/stdc++.h>
using namespace std;
#define sc(x) scanf("%d",&x)
#define sl(x) scanf("%lld",&x)
#define ll long long
#define pb push_back
#define INF 0x3f3f3f3f
#define eps 1e-8
const int Max=2e5+5;
int main(){
	int t;sc(t);
	while(t--){
		int n;sc(n);
		if(n%2==1){
			printf("1 ");
			for(int i=2;i<=n;i+=2){
				printf("%d %d ",i+1,i);
			}
		}else{
			for(int i=1;i<=n;i+=2){
				printf("%d %d ",i+1,i);
			}
		}
		printf("\n");
	}
}

B. Party

Watch this question for half an hour during the competition , Sit for half an hour , What a strange solution , See you for the first time ！！！ Drop points ,T_T

m If it is an even number , Just choose all , The answer for 0！！！

m If it is an odd number , The more people you invite, the better , So don't invite someone to have an even number of relationships , All the rest are invited , At this time, there may be an odd number of relationships , So this is not the optimal situation , Therefore, we also need to consider not inviting a pair of friends , There are even relationships in addition to each other ！！！

#include <bits/stdc++.h>
using namespace std;
#define sc(x) scanf("%d",&x)
#define sl(x) scanf("%lld",&x)
#define ll long long
#define pb push_back
#define INF 0x3f3f3f3f
#define eps 1e-8
const int Max=2e5+5;
int a[Max],sum[Max];
pair<int,int> mp[Max];
int main(){
    int t;
    sc(t);
    while(t--){
        int n;int m;
        sc(n);sc(m);
        for(int i=1;i<=n;i++) sc(a[i]),sum[i]=0;
        int ans=1e9+5;
        for(int i=1;i<=m;i++){
            int u,v;sc(u);sc(v);
            sum[u]++;
            sum[v]++;
            mp[i]=make_pair(u,v);
        }
        if(m%2==0) printf("0\n");
        else{
            for(int i=1;i<=n;i++){
                if(sum[i]%2) ans=min(ans,a[i]);
            }
            for(int i=1;i<=m;i++){
                if((sum[mp[i].first]+sum[mp[i].second])%2==0){
                    ans=min(ans,a[mp[i].first]+a[mp[i].second]);
                }
            }
            printf("%d\n",ans);
        }
    }
}

C. Color the Picture

This problem could have been solved , It's all wrong ！！！

There are two situations , One is to put it vertically , For example , Another is to put it horizontally , Every time you put a color, pay attention to the number of rows or columns >=2, From small to large, just be greedy ！！！

#include <bits/stdc++.h>
using namespace std;
#define sc(x) scanf("%d",&x)
#define sl(x) scanf("%lld",&x)
#define ll long long
#define pb push_back
#define INF 0x3f3f3f3f
#define eps 1e-8
const int Max=2e5+5;
ll a[Max];
int main(){
	int t;sc(t);
	while(t--){
		ll n,m,k;sl(n);sl(m);sl(k);
		for(int i=1;i<=k;i++){
			sl(a[i]);
		}
		bool flag=false;
		ll ans=0;
		sort(a+1,a+1+k);
		for(int i=1;i<=k;i++){
			if(a[i]/n>=2){
				if((m-(ans+a[i]/n))==1){
					if(a[i]/n>2) ans+=(a[i]/n-1);
				}
				else ans+=a[i]/n;
			}
		}
		if(ans>=m){
			printf("YES\n");continue;
		}
		ans=0;
		for(int i=1;i<=k;i++){
			if(a[i]/m>=2){
				if((n-(ans+a[i]/m))==1){
					if(a[i]/m>2) ans+=(a[i]/m-1);
				}
				else ans+=a[i]/m;
			}
		}
		if(ans>=n){
			printf("YES\n");continue;
		}else printf("NO\n");
	}
}

Terrible , Drop field , It seems that I haven't played well once ！！！！！！！ What can go out doesn't go out