Codeforces-Round-886-(Div

[[Dashboard - Codeforces Round 886 (Div. 4) - Codeforces)](比赛链接)

新年快乐！ 做个简单的

A. To My Critics

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

void solve() {
    int a, b, c;
    cin >> a >> b >> c;
    if(a + b >= 10 || b + c >= 10 || a + c >= 10) {
        cout << "YES\n";
    }else{
        cout << "NO\n";
    }
}

int main() {
    int t;
    cin >> t;
    while(t--) solve();
    return 0;
}

B. Ten Words of Wisdom

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

void solve() {
    int n;
    cin >> n;
    int ans = 0, t = 0;
    for(int i = 1; i <= n; i++) {
        int a, b;
        cin >> a >> b;
        if(a <= 10 && b > ans) {
            ans = b;
            t = i;
        }
    }
    cout << t << endl;
}

int main() {
    int t;
    cin >> t;
    while(t--) solve();
    return 0;
}

C. Word on the Paper

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

void solve() {
    string s, ans;
    for(int i = 0; i < 8; i++) {
        cin >> s;
        for(auto ch : s) {
            if(ch != '.') ans += ch;
        }
    }
    cout << ans << endl;
}

int main() {
    int t;
    cin >> t;
    while(t--) solve();
    return 0;
}

D. Balanced Round

思路：排序，遍历记录最大值即可，见代码：

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

void solve() {
    int n, k;
    cin >> n >> k;
    vector<int> a(n);
    for(int i = 0; i < n; i++) cin >> a[i];
    int ans = 1, b = 1;
    sort(a.begin(), a.end());
    for(int i = 1; i < n; i++) {
        if(a[i] - a[i - 1] <= k) {
            b++;
            ans = max(ans, b);
        }else{
            b = 1;
        }
    }
    cout << n - ans << endl;
}

int main() {
    int t;
    cin >> t;
    while(t--) solve();
    return 0;
}

E. Cardboard for Pictures

思路：解一下二次函数， 注意开long double

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

void solve() {
    long double n, c, b = 0;
    cin >> n >> c;
    vector<long double> a(n);
    for(int i = 0; i < n; i++) {
        cin >> a[i];
        c -= a[i] * a[i];
        b += a[i];
    }
    b *= 4;
    long double ta = 4 * n;
    // cout << ta << ' ' << b << ' ' << c << endl;
    long double ans = (sqrt(b * b + 4 * ta * c) - b) / (2 * ta);
    cout << (ll)ans << endl;
}

int main() {
    int t;
    cin >> t;
    while(t--) solve();
    return 0;
}

F. We Were Both Children

思路：把出现过的数的小于n的倍数出现的次数用一个数组 cnt 记录下来， 输出max(cnt[i])即可。 先用map记录一下，去一下重，防止复杂度过大。 复杂度是n * ln(n) (shi调和级数

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

void solve() {
    int n;
    cin >> n;
    vector<int> cnt(n + 1);
    map<int, int> mp;
    int x, ans = 0;
    for(int i = 0; i < n; i++) cin >> x, mp[x]++;
    for(auto [k, v] : mp) {
        x = k;
        int b = x;
        while(x <= n) {
            cnt[x] += v;
            x += b;
            ans = max(ans, cnt[x - b]);
        }
    }
    cout << ans << endl;
}

int main() {
    int t;
    cin >> t;
    while(t--) solve();
    return 0;
}

G. The Morning Star

思路：组合数学，哈希记录一下即可

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

void solve() {
    int n;
    cin >> n;
    ll ans = 0;
    map<ll, ll> mp1, mp2, mpx, mpy;
    for(int i = 0; i < n; i++) {
        ll a, b;
        cin >> a >> b;
        ans += mp1[b - a] + mp2[a + b] + mpx[a] + mpy[b];
        mp1[b - a] ++;
        mp2[a + b] ++;
        mpx[a] ++;
        mpy[b] ++;
    }
    cout << ans * 2 << endl;
}

int main() {
    int t;
    cin >> t;
    while(t--) solve();
    return 0;
}

谢谢观看