题目传送门

A. Floor Number

签到

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define rep(i, a, b) for (register int i = a; i <= b; i++)
 
int n, x;
 
inline void solve(int T)
{
    cin >> n >> x;
    if(n <= 2) cout << "1" << endl;
    else {
        n -= 2;
        cout << ceil(1.0 * n / x) + 1 << endl;
    }
}
 
 
int main()
{
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
 
    // freopen("in.txt", "r", stdin);
    // freopen("ans.txt", "w", stdans);
 
    int T = 1;
    cin >> T;
    rep(i, 1, T) solve(i);
}

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B. Symmetric Matrix

容易发现只需要一个右上角等于左下角的矩阵就可以构造任意偶数大小的矩阵

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define rep(i, a, b) for (register int i = a; i <= b; i++)
 
int n, m;
int a[4], flag, cnt;
 
inline void solve(int T)
{
    cin >> n >> m;
    flag = 0;
    rep(i, 1, n) {
        rep(j, 0, 3) cin >> a[j];
        if(a[1] == a[2]) flag = 1;
        
    }
    if((m & 1) || (!flag)) cout << "NO\n";
    else cout << "YES\n";
}
 
 
int main()
{
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
 
    // freopen("in.txt", "r", stdin);
    // freopen("ans.txt", "w", stdans);
 
    int T = 1;
    cin >> T;
    rep(i, 1, T) solve(i);
}

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C. Increase and Copy

贪心的想法：先将1加x，再把$x+1$复制y份，答案要满足$(x+1)*y>=n$

答案最多$63244$，可以直接枚举x

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define rep(i, a, b) for (register int i = a; i <= b; i++)
 
 
inline void solve(int T)
{
    int n, ans = 63244;
    cin >> n;
    rep(i, 0, 63244) if(i - 1 + ceil(1.0 * n / (i + 1)) < ans) ans = i - 1 + ceil(1.0 * n / (i + 1));
    cout << ans << endl;
}
 
 
int main()
{
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
 
    // freopen("in.txt", "r", stdin);
    // freopen("ans.txt", "w", stdans);
 
    int T = 1;
    cin >> T;
    rep(i, 1, T) solve(i);
}

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D. Non-zero Segments

如果$1$至$i-1$没有和为$0$的子段，且$1$至$i$存在和为$0$的子段，那就在$i-1$和$i$之间插入一个值，值是什么不重要

大概就是这样贪

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define rep(i, a, b) for (register int i = a; i <= b; i++)
 
map<ll, ll> mp;
ll n;
ll tmp, sum, ans;
 
inline void solve(int T)
{
    mp.clear();
    sum = ans = 0;
    cin >> n;
    rep(i, 1, n) {
        cin >> tmp;
        sum += tmp;
        if(mp[sum] || sum == 0) ans++, mp.clear(), sum = tmp;
        mp[sum] = 1;
    }
    cout << ans << endl;
}
 
 
int main()
{
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
 
    // freopen("in.txt", "r", stdin);
    // freopen("ans.txt", "w", stdans);
 
    int T = 1;
    // cin >> T;
    rep(i, 1, T) solve(i);
}

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