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  • CodeForces Round #563 Div.2

    A. Ehab Fails to Be Thanos

    #include <bits/stdc++.h>
    using namespace std;
     
    const int maxn = 1e5 + 10;
    int N;
    int a[maxn];
     
    int main() {
       scanf("%d", &N);
        for(int i = 0; i < N * 2; i ++)
            scanf("%d", &a[i]);
        
        sort(a, a + N * 2);
        
        long long sum1 = 0, sum2 = 0;
        for(int i = 0; i < N; i ++)
            sum1 += a[i];
        for(int i = N; i < N * 2; i ++)
            sum2 += a[i];
        
        if(sum1 != sum2) {
            for(int  i = 0; i < N * 2; i ++)
                printf("%d%s", a[i], i != 2 * N - 1 ? " " : "
    ");
        } else printf("-1
    ");
        
        
        return 0;
    }
    View Code

    B. Ehab Is an Odd Person

    #include <bits/stdc++.h>
    using namespace std;
     
    const int maxn = 1e5 + 10;
    int N;
    int odd = 0, even = 0;
    long long a[maxn];
     
    int main() {
        scanf("%d", &N);
        for(int i = 0; i < N; i ++) {
            scanf("%lld", &a[i]);
            if(a[i] % 2 == 0) even ++;
            else odd ++;
        }
        if(odd && even) sort(a, a + N);
        
        for(int i = 0; i < N; i ++)
            printf("%lld%s", a[i], i != N - 1 ? " " : "
    ");
        
        return 0;
    }
    View Code

    C. Ehab and a Special Coloring Problem

    #include <bits/stdc++.h>
    using namespace std;
     
    const int maxn = 1e5 + 10;
    int N;
    int pos[maxn];
     
    bool isprime(int x) {
        if(x <= 1) return false;
        if(x == 2) return true;
        
        for(int i = 2; i * i <= x; i ++) {
            if(x % i == 0) return false;
        }
        
        return true;
        
    }
     
    void solve() {
        int p = 1;
        
        for(int i = 2; i <= N; i ++) {
            if(isprime(i)) {
                pos[i] = p;
                for(int j = i; j < maxn; j += i)
                    pos[j] = p;
                
                p ++;
                
            }
        }
        
    }
     
    int main() {
        scanf("%d", &N);
        solve();
        
        for(int i = 2; i <= N; i ++) 
            printf("%d%s", pos[i], i != N ? " " : "
    ");
        
        return 0;
    } 
    View Code

    D. Ehab and the Expected XOR Problem

    #include <bits/stdc++.h>
    using namespace std;
     
    const int maxn = (1 << 18) + 10;
    int N, x;
    int vis[maxn];
    vector<int> ans;
     
    int main() {
        scanf("%d%d", &N, &x);    
        
        for(int i = 0; i < 1 << N; i ++) {
            if(!vis[i]) vis[i ^ x] = 1;
        }
        
        int t = 0;
        for(int i = 1; i < 1 << N; i ++) {
            if(!vis[i]) {
                ans.push_back(t ^ i);
                t = i;
            }
        }
        
        printf("%d
    ", (int)ans.size());
        for(int i = 0; i < (int)ans.size(); i ++)
            printf("%d%s", ans[i], i != (int)ans.size() - 1 ? " " : "
    ");
        
        return 0;
    }
    View Code

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  • 原文地址:https://www.cnblogs.com/zlrrrr/p/11221911.html
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