Codeforces Round #791 (Div. 2) A - D 题解

传送门

A. AvtoBus

直接判断就好了，大的话就尽量用4，小的话就尽量用6，然后根据取余的关系找就行了

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;
typedef long long ll;
#define pii pair
const ll maxn = 2e5 + 10;
const ll inf = 1e17 + 10;

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    int t;
    cin >> t;
    while(t--)
    {
        ll n;
        cin >> n;
        ll a = 0, b = 0;
        if(n <= 3 || n % 2 == 1) cout << -1 << endl;
        else
        {
            n >>= 1;
            b = n >> 1;
            a = n / 3;
            if(n % 3 == 1) a++;
            else if(n % 3 == 2) a++;
            cout << a << " " << b << endl;
        }
    }

    return 0;
}

B. Stone Age Problem

只要在更改的时候，判断一下更改时间的先后，再进行更改就行了，这样就是 \(O(1)\) 的修改

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;
typedef long long ll;
#define pii pair
const ll maxn = 2e5 + 10;
const ll inf = 1e17 + 10;
pii num[maxn];

int main()
{
    int n, m;
    scanf("%d%d", &n, &m);
    long long ans = 0;
    for(int i=0; i

C. Rooks Defenders

用树状数组维护一下行和列是否有车存在，为了防止同行或同列有多个车的存在，还要开两个数组去维护同行同列有多少个车的数量，然后在删除和添加时，车的数量 0-1 变换的时候才用树状数组维护

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;
typedef long long ll;
#define pii pair
const ll maxn = 2e5 + 10;
const ll inf = 1e17 + 10;
int r[maxn], v[maxn];
int row[maxn], vol[maxn];
int n, m;

inline int lowbit(int x)
{
    return x & (-x);
}

void add(int x, int val, int* way)
{
    for(int i=x; i<=n; i+=lowbit(i))
        way[i] += val;
}

int query(int x, int* way)
{
    int ans = 0;
    for(int i=x; i; i-=lowbit(i))
        ans += way[i];
    return ans;
}

int main()
{
    scanf("%d%d", &n, &m);
    while(m--)
    {
        int t, a, b, c, d;
        scanf("%d%d%d", &t, &a, &b);
        if(t == 3)
        {
            scanf("%d%d", &c, &d);
            int x = query(c, r) - query(a - 1, r);
            int y = query(d, v) - query(b - 1, v);
            if(x == c - a + 1 || y == d - b + 1)
                printf("Yes\n");
            else
                printf("No\n");
        }
        else if(t == 1)
        {
            if(++row[a] == 1)
                add(a, 1, r);
            if(++vol[b] == 1)
                add(b, 1, v);
        }
        else
        {
            if(--row[a] == 0)
                add(a, -1, r);
            if(--vol[b] == 0)
                add(b, -1, v);
        }
    }

    return 0;
}

D. Toss a Coin to Your Graph...

二分 + 记忆化搜索

答案是单调的，所以直接二分答案，然后检查的时候就记忆化搜索，看看在限制当前最高值的情况下，能不能走 k 步，如果走的发现是个环，则直接返回可以走 k 步就行

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;
typedef long long ll;
#define pii pair
const ll maxn = 2e5 + 10;
const ll inf = 1e17 + 10;
int num[maxn];
int a[maxn], vis[maxn];
ll dp[maxn];
vectorgra[maxn];
ll n, m, k;

ll dps(int now, int x)
{
    if(dp[now]) return dp[now] == -1 ? 0 : dp[now];
    if(num[now] > x) {dp[now] = -1; return 0;}
    vis[now] = 1;
    ll ans = 0;
    for(int i=0; i= k;
}

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    cin >> n >> m >> k;
    for(int i=1; i<=n; i++) {cin >> num[i]; a[i] = num[i];}
    while(m--)
    {
        int a, b;
        cin >> a >> b;
        gra[a].push_back(b);
    }
    sort(a + 1, a + n + 1);
    int l = 1, r = n;
    while(l < r)
    {
        int mid = l + r >> 1;
        if(judge(a[mid]))
            r = mid;
        else
            l = mid + 1;
    }
    if(judge(a[l]))
        cout << a[l] << endl;
    else
        cout << -1 << endl;

    return 0;
}
CodeForces

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