hyz's blog
0%

发表于

题解按AC数降序排列。

J题：Various Tree

简单bfs。

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79

#include<bits/stdc++.h>
#define x first
#define y second
#define ok cout << "ok" << endl;
using namespace std;
typedef long long ll;
typedef vector<int> vi;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
const double PI = acos(-1.0);
const int INF=0x3f3f3f3f;
const ll LINF=0x3f3f3f3f3f3f3f3f;
const int N=1e6+39;
const int shift=1e3+9;
const double Eps=1e-7;

int cnt[N];

void bfs(int u, int v) {
    bool vis[N];
    int dis[N];
    queue<int> que;
    memset(vis, 0, sizeof vis);
    memset(dis, INF, sizeof dis);
    dis[u] = 0;
    vis[u] = 1;
    que.push(u);
    while(que.size()) {
        int uu = que.front();
        que.pop();
        if(uu == v) {
            printf("%d\n", dis[v]);
            break;
        }
        if(uu + 1 < N && !vis[uu + 1]) {
            dis[uu + 1] = dis[uu] + 1;
            vis[uu + 1] = 1;
            que.push(uu + 1);
        }
        if(uu - 1 >= 0 && !vis[uu - 1]) {
            dis[uu - 1] = dis[uu] + 1;
            vis[uu - 1] = 1;
            que.push(uu - 1);
        }
        if(uu + cnt[uu] < N && !vis[uu + cnt[uu]]) {
            dis[uu + cnt[uu]] = dis[uu] + 1;
            vis[uu + cnt[uu]] = 1;
            que.push(uu + cnt[uu]);
        }
        if(uu - cnt[uu] >= 0 && !vis[uu - cnt[uu]]) {
            dis[uu - cnt[uu]] = dis[uu] + 1;
            vis[uu - cnt[uu]] = 1;
            que.push(uu - cnt[uu]);
        }
    }
}

void pre() {
    memset(cnt, 0, sizeof cnt);
    for(int i = 1; i < N; i++) {
        int t = i;
        while(t) {
            if(t & 1)
                cnt[i]++;
            t >>= 1;
        }
    }
}

int main(void) {
    if(fopen("in", "r")!=NULL) {freopen("in", "r", stdin); freopen("out", "w", stdout);}
    pre();
    int a, b;
    while(cin >> a >> b)
        bfs(a, b);

    return 0;
}

K题：Walking in the Forest

简单二分。

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51

#include<bits/stdc++.h>
#define x first
#define y second
#define ok cout << "ok" << endl;
using namespace std;
typedef long long ll;
typedef vector<int> vi;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
const double PI = acos(-1.0);
const int INF=0x3f3f3f3f;
const ll LINF=0x3f3f3f3f3f3f3f3f;
const int N=1e5+9;
const int shift=1e3+9;
const double Eps=1e-7;

int k, n, a[N];

bool check(ll mid) {
    ll sum = LINF, cnt = 0;
    for(int i = 0; i < k; i++) {
        if(a[i] > mid)
            return false;
        if(sum + a[i] > mid)
            sum = a[i], cnt++;
        else
            sum += a[i];
    }
    return cnt <= n;
}

int main(void) {
    if(fopen("in", "r")!=NULL) {freopen("in", "r", stdin); freopen("out", "w", stdout);}
    while(~scanf("%d%d", &k, &n)) {
        k--;
        for(int i = 0; i < k; i++)
            scanf("%d", a + i);
        ll l = 0, r = 10000000000;
        while(l < r) {
            ll mid = (l + r) >> 1;
            if(check(mid))
                r = mid;
            else 
                l = mid + 1;
        }
        printf("%lld\n", r);
    }

    return 0;
}

B题：Beautiful Trees Cutting

简单数学题。

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52

#include<bits/stdc++.h>
#define x first
#define y second
#define ok cout << "ok" << endl;
using namespace std;
typedef long long ll;
typedef vector<int> vi;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
const double PI = acos(-1.0);
const int INF=0x3f3f3f3f;
const ll LINF=0x3f3f3f3f3f3f3f3f;
const int N=1e5+9;
const int shift=1e3+9;
const double Eps=1e-7;

const int mod = 1e9 + 7;

char s[N];
ll ans, m, n;
vi v;

ll ksm(ll a, ll b) {
    ll ans = 1;
    while(b) {
        if(b & 1)
            (ans *= a) %= mod;
        (a *= a) %= mod;
        b >>= 1;
    }
    return ans;
}

int main(void) {
    if(fopen("in", "r")!=NULL) {freopen("in", "r", stdin); freopen("out", "w", stdout);}
    while(cin >> m >> s) {
        v.clear();
        n = strlen(s);
        for(int i = 0; i < n; i++)
            if((s[i] - '0') % 5 == 0)
                v.push_back(i);
        ans = 0;
        for(int i = 0; i < v.size(); i++)
            (ans += ksm(2, v[i])) %= mod;
        (ans *= ksm(2, n * m) - 1) %= mod;
        (ans *= ksm(ksm(2, n) - 1, mod - 2)) %= mod;
        printf("%lld\n", ans % mod);
    }

    return 0;
}

F题：Sorting Trees

简单思维题。之前在Codejam上做过在k = 2情况下的一道题，这道题算是对那道题的推广。

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60

#include<bits/stdc++.h>
#define x first
#define y second
#define ok cout << "ok" << endl;
using namespace std;
typedef long long ll;
typedef vector<int> vi;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
const double PI = acos(-1.0);
const int INF=0x3f3f3f3f;
const ll LINF=0x3f3f3f3f3f3f3f3f;
const int N=1e5+9;
const int shift=1e3+9;
const double Eps=1e-7;

vi v[N];
int n, k, a[N], b[N];

int main(void) {
    if(fopen("in", "r")!=NULL) {freopen("in", "r", stdin); freopen("out", "w", stdout);}
    while(cin >> n >> k) {
        for(int i = 0; i < n; i++)
            scanf("%d", a + i);
        if(k == 0) {
            memcpy(b, a, sizeof a);
            sort(b, b + n);
            int i;
            for(i = 0; i < n; i++)
                if(a[i] != b[i]) {
                    printf("%d\n", i + 1);
                    break;
                }
            if(i == n)
                printf("-1\n");
            continue;
        }
        for(int i = 0; i < n; i++)
            v[i % k].push_back(a[i]);
        for(int i = 0; i < k; i++) {
            if(v[i].empty()) continue;
            sort(v[i].begin(), v[i].end());
        }
        for(int i = 0; i < k; i++)
            for(int j = 0; j < v[i].size(); j++)
                b[i + j * k] = v[i][j];
        sort(a, a + n);
        int i;
        for(i = 0; i < n; i++)
            if(a[i] != b[i]) {
                printf("%d\n", i + 1);
                break;
            }
        if(i == n)
            printf("-1\n");
    }

    return 0;
}

C题：Professional Manager

留坑待补。