A. Coins

#include <bits/stdc++.h>

using namespace std;

#define int long long

int read() {
    int x = 0, f = 1, ch = getchar();
    while ((ch < '0' || ch > '9') && ch != '-') ch = getchar();
    if (ch == '-') f = -1, ch = getchar();
    while (ch >= '0' && ch <= '9') x = (x << 3) + (x << 1) + ch - '0', ch = getchar();
    return x * f;
}


void solve() {
    int n = read() , k = read();
    if( n % 2 == 0 ) printf("YES\n");
    else if( k & 1 ) printf("YES\n");
    else printf("NO\n");
    return;
}

int32_t main() {
    for( int t = read() ; t ; t -- )
        solve();
    return 0;
}

B. Long Legs

枚举$m$，$m$确定了答案是$m-1+\left \lceil \frac x m \right \rceil + \left \lceil \frac y m \right \rceil $

至于枚举的范围，感觉是玄学，大家随便搞的都能过

#include <bits/stdc++.h>

using namespace std;

#define int long long

int read() {
    int x = 0, f = 1, ch = getchar();
    while ((ch < '0' || ch > '9') && ch != '-') ch = getchar();
    if (ch == '-') f = -1, ch = getchar();
    while (ch >= '0' && ch <= '9') x = (x << 3) + (x << 1) + ch - '0', ch = getchar();
    return x * f;
}


void solve() {
    int x = read(), y = read() , res = x + y;
    for( int i = 1 , n = sqrt(x) + sqrt(y) + 2 ; i <= n ; i ++ )
        res = min( res , ( x + i - 1 ) / i + ( y + i - 1 ) / i + i - 1 );
    cout << res << "\n";
}

int32_t main() {
    for (int t = read(); t; t--)
        solve();
    return 0;
}

C. Search in Parallel

访问次数越多的放的位置越靠前，所以直接贪心一下就好了

#include <bits/stdc++.h>

using namespace std;

#define int long long

int read() {
    int x = 0, f = 1, ch = getchar();
    while ((ch < '0' || ch > '9') && ch != '-') ch = getchar();
    if (ch == '-') f = -1, ch = getchar();
    while (ch >= '0' && ch <= '9') x = (x << 3) + (x << 1) + ch - '0', ch = getchar();
    return x * f;
}

void solve() {
    int n = read(), s1 = read(), s2 = read(), t1 = s1, t2 = s2;
    vector<pair<int, int>> p;
    vector<int> a, b;
    for (int i = 1, x; i <= n; i++)
        x = read(), p.emplace_back(x, i);
    sort(p.begin(), p.end(), greater<pair<int, int>>());
    for( auto [  w , i ] : p ){
        if( t1 < t2 ) a.push_back(i) , t1 += s1;
        else b.push_back(i) ,t2 += s2;
    }
    printf("%lld" , a.size() );
    for( auto i : a ) printf(" %lld", i );
    printf("\n%lld" , b.size() );
    for( auto i : b ) printf(" %lld", i );
    printf("\n");
    return;
}

int32_t main() {
    for (int t = read(); t; t--)
        solve();
    return 0;
}

D. Balancing Weapons

枚举一下上届$r$，然后算一下下届$r-k$，然后二分查找一下有多少个$p_i$在$[r-k,r]$之间，在区间之外都都是要操作的。

现在其实就是要看区间外的能否通过操作移动进来，首先如果$f_i \le k+1$就一定可以移动进进一个宽度为$k$的区间，否则就要看是否满足$\left \lfloor \frac{r}{f} \right \rfloor f \ge r-k$，这个式子可以化成$k \ge r - \left \lfloor \frac{r}{f} \right \rfloor f = r \mod f$。其实不难想出已经在区间内的$p_i$也是满足刚才的条件。所以只要判断一下所有的$f$时候满足即可，不用区分是区间内对应的还是区间外对应的。

#include <bits/stdc++.h>

using namespace std;

#define int long long

int read() {
    int x = 0, f = 1, ch = getchar();
    while ((ch < '0' || ch > '9') && ch != '-') ch = getchar();
    if (ch == '-') f = -1, ch = getchar();
    while (ch >= '0' && ch <= '9') x = (x << 3) + (x << 1) + ch - '0', ch = getchar();
    return x * f;
}

int calc(vector<int> &a, int l, int r) {
    return upper_bound(a.begin(), a.end(), r) -
           lower_bound(a.begin(), a.end(), l);
}

void solve() {
    int n = read(), k = read();
    vector<int> f(n), d(n), a(n);
    for (auto &i: f) i = read();
    for (auto &i: d) i = read();
    for (int i = 0; i < n; i++) a[i] = f[i] * d[i];
    sort(f.begin(), f.end());
    f.resize(unique(f.begin(), f.end()) - f.begin());
    reverse(f.begin(), f.end());
    while (f.size() && f.back() <= k + 1) f.pop_back();
    sort(a.begin(), a.end());
    int res = n, r = k + 1;
    for (auto v: a) {
        for (r = max(v, r); r <= v + k; r++) {
            bool flag = true;
            for (auto u: f)
                if (r % u > k) {
                    flag = false;
                    break;
                }
            if (flag) res = min(res, n - calc(a, r - k, r));
        }
    }
    cout << res << "\n";
    return;
}

int32_t main() {
    for (int t = read(); t; t--)
        solve();
    return 0;
}