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  • CodeForces Round #560 Div.3

    A. Remainder

    #include <bits/stdc++.h>
    using namespace std;
    
    string s;
    int N, x, y;
    
    int main() {
        scanf("%d%d%d", &N, &x, &y);
        cin >> s;
        for(int i = 0; i < N / 2; i ++)
            swap(s[i], s[N - i - 1]);
    
        int ans = 0;
        for(int i = 0; i <= y; i ++) {
            if(i != y && s[i] != '0') ans ++;
            else if(i == y && s[i] != '1') ans ++;
        }
        
        for(int i = y + 1; i < x; i ++) {
            if(s[i] == '1') ans ++;
        }
        
        printf("%d
    ", ans);
    
        return 0;
    }
    View Code

    B. Polycarp Training

    #include <bits/stdc++.h>
    using namespace std;
    
    const int maxn = 2e5 + 10;
    int N;
    int a[maxn];
    
    int main() {
        scanf("%d", &N);
        for(int i = 1; i <= N; i ++)
            scanf("%d", &a[i]);
    
        sort(a + 1, a + N + 1);
    
        int ans = 0;
        int pos = 1;
        for(int i = 1; i <= N; i ++) {
            if(a[i] >= pos) {
                ans ++;
                pos ++;
            }
        }
    
        printf("%d
    ", ans);
    
        return 0;
    }
    View Code

    C. Good String

    #include <bits/stdc++.h>
    using namespace std;
    
    int N;
    string s, t;
    
    int main() {
        scanf("%d", &N);
        cin >> s;
    
        int pos = 0;
        for(int i = 0; i < N; i ++) {
            if(pos % 2 == 0 || t[pos - 1] != s[i])
                t += s[i], pos ++;
        }
    
        if(t.length() % 2) pos --;
        t = t.substr(0, pos);
    
        printf("%d
    ", N - pos);
        cout << t << endl;
    
        return 0;
    }
    View Code

    D. Almost All Divisors

    #include <bits/stdc++.h>
    using namespace std;
    
    const int maxn = 550;
    int T, N;
    long long a[maxn];
    
    bool IsPrime(long long x) {
        if(x == 1 || x == 0) return false;
        if(x == 2) return true;
        for(long long i = 2; i * i <= x; i ++)
            if(x % i == 0) return false;
        return true;
    }
    
    int Count(long long x) {
        int ans = 0;
        for(long long i = 2; i * i <= x; i ++) {
            if(x % i == 0) {
                if(x / i == i) ans += 1;
                else ans += 2;
            }
        }
        return ans;
    }
    
    int main() {
        scanf("%d", &T);
        while(T --) {
            bool flag = true;
            scanf("%d", &N);
            for(int i = 1; i <= N; i ++)
                cin >> a[i];
    
            sort(a + 1, a + 1 + N);
    
            if(N == 1 && !IsPrime(a[1])) flag = false;
    
            long long sum;
            sum = a[1] * a[N];
            for(int i = 2; i <= N / 2; i ++) {
                if(a[i] * a[N - i + 1] != sum) {
                    flag = false;
                    break;
                }
            }
            if(N % 2 && N != 1 && a[N / 2 + 1] * a[N / 2 + 1] != sum) flag = false;
            if(Count(sum) != N) flag = false;
    
            if(flag) cout << sum << endl;
            else printf("-1
    ");
    
        }
        return 0;
    }
    View Code

    E. Two Arrays and Sum of Functions

    #include <bits/stdc++.h>
    using namespace std;
    
    const int maxn = 2e5 + 10;
    const long long mod = 998244353;
    int N;
    long long a[maxn], b[maxn];
    vector<long long> v;
    
    int main() {
        scanf("%d", &N);
        for(int i = 0; i < N; i ++) {
            scanf("%lld", &a[i]);
            v.push_back(a[i] * (i + 1) * (N - i));
        }
    
        sort(v.rbegin(), v.rend());
        for(int i = 0; i < N; i ++)
            scanf("%lld", &b[i]);
    
        long long ans = 0;
        sort(b, b + N);
        for(int i = 0; i < N; i ++) {
            ans = (ans + b[i] * (v[i] % mod)) % mod;
            ans %= mod;
        }
    
        printf("%lld
    ", ans % mod);
    
        return 0;
    }
    View Code

    明天周一不想上班

     

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  • 原文地址:https://www.cnblogs.com/zlrrrr/p/10890289.html
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