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Educational Codeforces Round 33

#	Who	=	Penalty	A	B	C	D	E
479	arkethos	4	247	+ 00:08	+ 00:19	+1 00:59	+2 01:41
479	ne-leonardinkret	4	247	+ 00:04	-1	+1 00:27	+2 01:47	+ 00:49
481	MeePwn^#	4	248	+ 00:14	+ 00:21	+ 00:54	+3 01:39
482	shreypandey	4	251	+ 00:06	+ 00:10	+ 00:17		+5 01:58
483	iskander232	4	252	+1 00:06	+2 00:23	+ 00:17	-3	+2 01:46

Chess For Three

默认第一场比赛由AB进行

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<vector>
#include<algorithm>
#include<iostream>
#include<map>
#include<queue>
#include<string>
using std::vector;
using std::queue;
using std::map;
using std::sort;
using std::string;
using std::lower_bound;
using std::upper_bound;
#define read(x) scanf("%d", &x)
#define reads(x) scanf("%s", x)
#define write(x) printf("%d ", x)
#define writes(x) printf("%s ", x)
#define writeln(x) printf("%d
", x)
#define writesln(x) printf("%s
", x)
#define fillchar(x, a) memset(x, a, sizeof(x))
typedef long long llint;
/*data structure*/

/*data structure*/
int cmp(const void * x, const void * y) {
#define datatype int
    datatype dx = *((datatype *)(x)), dy = *((datatype *)(y));
    //x < y
    return dx > dy ? 1 : -1;
#undef datatype
}
/*global variable*/
int a[200];
/*global variable*/
int main() {
#ifndef ONLINE_JUDGE
    freopen("input.txt", "r", stdin);
#endif
    int n;
    read(n);
    for (int i = 0; i < n; i++) {
        read(a[i]);
        a[i]--;
    }
    int p1 = 0, p2 = 1, spec = 2;
    bool flag1 = true;
    for (int i = 0; i < n; i++) {
        if (a[i] == spec) {
            flag1 = false;
            break;
        }
        if (p1 == a[i]) {
            int t = p2;
            p2 = spec;
            spec = t;
        } else {
            int t = p1;
            p1 = spec;
            spec = t;
        }
    }
    if (flag1) printf("YES
");
    else printf("NO
");
    return 0;
}

View Code

Beautiful Divisors

直接把符合条件的数列出来找

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<vector>
#include<algorithm>
#include<iostream>
#include<map>
#include<queue>
#include<string>
using std::vector;
using std::queue;
using std::map;
using std::sort;
using std::string;
using std::lower_bound;
using std::upper_bound;
#define read(x) scanf("%d", &x)
#define reads(x) scanf("%s", x)
#define write(x) printf("%d ", x)
#define writes(x) printf("%s ", x)
#define writeln(x) printf("%d
", x)
#define writesln(x) printf("%s
", x)
#define fillchar(x, a) memset(x, a, sizeof(x))
typedef long long llint;
/*data structure*/

/*data structure*/
int cmp(const void * x, const void * y) {
#define datatype int
    datatype dx = *((datatype *)(x)), dy = *((datatype *)(y));
    //x < y
    return dx > dy ? 1 : -1;
#undef datatype
}
/*global variable*/
int a[100];
/*global variable*/
int main() {
#ifndef ONLINE_JUDGE
    freopen("input.txt", "r", stdin);
#endif
    int n;
    for (int i = 1;; i++) {
        a[i] = ((1 << i) - 1) * (1 << (i - 1));
        if (a[i] > 100000) {
            n = i;
            break;
        }
    }
    int x;
    read(x);
    for (int i = n; i >= 1; i--) {
        if (x % a[i] == 0) {
            writeln(a[i]);
            break;
        }
    }
    return 0;
}

View Code

Rumor

贪心，每次贿赂还不知道的人里边底线最低的

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<vector>
#include<algorithm>
#include<iostream>
#include<map>
#include<queue>
#include<string>
#include<functional>
using std::priority_queue;
using std::vector;
using std::queue;
using std::map;
using std::sort;
using std::string;
using std::lower_bound;
using std::upper_bound;
#define read(x) scanf("%d", &x)
#define reads(x) scanf("%s", x)
#define write(x) printf("%d ", x)
#define writes(x) printf("%s ", x)
#define writeln(x) printf("%d
", x)
#define writesln(x) printf("%s
", x)
#define fillchar(x, a) memset(x, a, sizeof(x))
typedef long long llint;
/*data structure*/
struct cha {
    int id;
    llint c;
    friend bool operator< (cha n1, cha n2) {
        return n1.c > n2.c;
    }
};
/*data structure*/
int cmp(const void * x, const void * y) {
#define datatype int
    datatype dx = *((datatype *)(x)), dy = *((datatype *)(y));
    //x < y
    return dx > dy ? 1 : -1;
#undef datatype
}
/*global variable*/
priority_queue<cha> q;
vector<int> G[100005];
bool v[100005];
/*global variable*/
void dfs(int x) {
    for (int i = 0; i < G[x].size(); i++) {
        int u = G[x][i];
        if (v[u]) continue;
        v[u] = true;
        dfs(u);
    }
}
int main() {
#ifndef ONLINE_JUDGE
    freopen("input.txt", "r", stdin);
#endif
    int n, m;
    read(n);
    read(m);
    while (!q.empty()) q.pop();
    for (int i = 1; i <= n; i++) {
        cha ccc;
        ccc.id = i;
        scanf("%lld", &ccc.c);
        q.push(ccc);
        G[i].clear();
    }
    for (int i = 0; i < m; i++) {
        int x, y;
        read(x);
        read(y);
        G[x].push_back(y);
        G[y].push_back(x);
    }
    fillchar(v, false);
    long long ans = 0;
    while (!q.empty()) {
        cha x = q.top();
        q.pop();
        if (v[x.id]) continue;
        ans += x.c;
        v[x.id] = true;
        dfs(x.id);
    }
    printf("%I64d
", ans);
    return 0;
}

View Code

Credit Card

首先，早上不存钱，把所有操作模拟一遍，记录每天晚上结束时的钱数m，如果某一天超过d，输出-1。

因为钱不会凭空消失，所以在某一天存入钱后，当天及以后每天的m值增加相同大小。

为了尽量少去，显然只有当a为0的日子才去，存的钱应该在不会导致之后超出d的情况下尽量多，为d减去之后的m值中最大的，再减去之前已经存过的钱数。

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<vector>
#include<algorithm>
#include<iostream>
#include<map>
#include<queue>
#include<string>
#include<functional>
using std::priority_queue;
using std::vector;
using std::queue;
using std::map;
using std::sort;
using std::string;
using std::lower_bound;
using std::upper_bound;
#define read(x) scanf("%d", &x)
#define reads(x) scanf("%s", x)
#define write(x) printf("%d ", x)
#define writes(x) printf("%s ", x)
#define writeln(x) printf("%d
", x)
#define writesln(x) printf("%s
", x)
#define fillchar(x, a) memset(x, a, sizeof(x))
typedef long long llint;
/*data structure*/

/*data structure*/
int cmp(const void * x, const void * y) {
#define datatype int
    datatype dx = *((datatype *)(x)), dy = *((datatype *)(y));
    //x < y
    return dx > dy ? 1 : -1;
#undef datatype
}
/*global variable*/
int a[100005];
llint m[100005], max[100005];
/*global variable*/
int main() {
#ifndef ONLINE_JUDGE
    freopen("input.txt", "r", stdin);
#endif
    int n;
    llint d;
    scanf("%d%I64d", &n, &d);
    for (int i = 0; i < n; i++) read(a[i]);
    m[0] = a[0];
    for (int i = 1; i < n; i++) {
        m[i] = m[i - 1] + a[i];
    }
    for (int i = 0; i < n; i++) {
        if (m[i] > d) {
            printf("-1
");
            return 0;
        }
    }
    max[n - 1] = m[n - 1];
    for (int i = n - 2; i >= 0; i--) {
        if (m[i] > max[i + 1]) max[i] = m[i];
        else max[i] = max[i + 1];
    }
    int ans = 0;
    llint add = 0, l, r;
    bool flag = true;
    for (int i = 0; i < n; i++) {
        if (a[i] != 0) continue;
        if (m[i] + add >= 0) continue;
        l = (llint) - 1 * m[i] - add;
        r = d - (max[i] + add);
        if (l > r) {
            flag = false;
            break;
        }
        ans++;
        add += r;
    }
    if (flag) printf("%d
", ans);
    else printf("-1
");
    return 0;
}

View Code

Counting Arrays

把x分解质因数，对于有cnt个的质因子，答案乘以C(y+cnt-1,cnt)，关于这类问题，有一张总结的图（懒得编辑了，治疗一下颈椎病吧）：

这题被这个数据搞了一下：

1
524288 1000000

这个524288=1<<19，y是最大值1000000，加起来就是1000018，而我数组开的是1000005，于是就跪了。脸打的啪啪啪，牛批啊老哥。

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<vector>
#include<algorithm>
#include<iostream>
#include<map>
#include<queue>
#include<string>
#include<functional>
#include<iostream>
//#include<bits/stdc++.h>
using namespace std;
#define fillchar(x, a) memset(x, a, sizeof(x))
typedef long long lint;
/*data structure*/
template <int size> class NoCombination {
public:
    lint fac[size], inv[size], f[size];
    lint mod;
    void init(lint m) {
        mod = m;
        fac[0] = fac[1] = inv[0] = inv[1] = f[0] = f[1] = 1;
        for (int i = 2; i < size; i++) {
            fac[i] = fac[i - 1] * i % mod;
            f[i] = (mod - mod / i) * f[mod % i] % mod;
            inv[i] = inv[i - 1] * f[i] % mod;
        }
    }
    lint C(lint a, lint b) {
        return fac[a] * inv[b] % mod * inv[a - b] % mod;
    }
};
/*data structure*/
int cmp(const void * x, const void * y) {
#define datatype int
    datatype dx = *((datatype *)(x)), dy = *((datatype *)(y));
    //x < y
    return dx > dy ? 1 : -1;
#undef datatype
}
/*global variable*/
NoCombination<1100005> nc;
const lint mod = 1000000007;
bool f[1100005];
int p[1100005], m = 0;
/*global variable*/
/*function*/
inline lint pow(lint a, lint b, lint p) {
    lint rtn = 1;
    while (b) {
        if (b & 1) rtn = rtn * a % p;
        a = a * a % p;
        b >>= 1;
    }
    return rtn;
}
/*function*/
int main() {
#ifndef ONLINE_JUDGE
    freopen("input.txt", "r", stdin);
#endif
    std::ios::sync_with_stdio(0), cin.tie(0);
    nc.init(mod);
    fillchar(f, true);
    f[0] = f[1] = false;
    for (int i = 2; i < 1000000; i++) {
        if (f[i]) {
            p[m++] = i;
            for (int j = i + i; j < 1000000; j+=i) f[j] = false;
        }
    }
    int q, x, y;
    lint ans;
    cin >> q;
    while (q--) {
        cin >> x >> y;
        ans = 1;
        int maxp = (int)sqrt(x);
        for (int i = 0; i < m; i++) {
            if (x == 1 || p[i] > maxp) break;
            if (x % p[i] != 0) continue;
            int cnt = 0;
            while (x % p[i] == 0) x /= p[i], cnt++;
            ans = ans * nc.C(y + cnt - 1, cnt) % mod;
        }
        if (x > 1) ans = ans * y % mod;
        ans = ans * pow(2, y - 1, mod) % mod;
        cout << ans << endl;
    }
    return 0;
}
//33353503

View Code

Subtree Minimum Query

这道题跟以前做过的一道题解法差不多，记录从每个点往下走1步，2步，4步，8步......能得到的最小的a，然后对于查询的k步用二进制表示出来各个位一凑就行。比如走7步相当于从第0个点走到第4个点的最小值、从第4个点走到第6个点的最小值、第6个点走到第7个点的最小值这3个最小值中的最小值，这样复杂度就变成了对数级别的。有几个可能会出错的坑，卡了我很久才弄明白，整理成两组测试数据：

2 1

1 2

1 100

4 1

10000 10000 10000 1

1 2

2 3

1 4

1 2

#pragma comment(linker, "/STACK:102400000,102400000")
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<vector>
#include<algorithm>
#include<iostream>
#include<map>
#include<queue>
#include<string>
#include<functional>
//#include<bits/stdc++.h>
using namespace std;
typedef long long lint;

vector<int> G[100005];
vector<int> P[100005][25];
int Q[100005][25];
int a[100005], fa[100005], dep[100005];
vector<int> vi[2];

int cmp(const void * x, const void * y) {
#define datatype int
    datatype dx = *((datatype *)(x)), dy = *((datatype *)(y));
    //x < y
    return dx > dy ? 1 : -1;
#undef datatype
}

void dfs(int x) {
    for(int i = 0; i < G[x].size(); i++) {
        int u = G[x][i];

        if(u == fa[x]) continue;

        fa[u] = x;
        dfs(u);

        if(dep[u] + 1 > dep[x]) dep[x] = dep[u] + 1;
    }
}

int main() {
#ifndef ONLINE_JUDGE
    freopen("input.txt", "r", stdin);
#endif
    std::ios::sync_with_stdio(0), cin.tie(0);
    int n, r, x, y;
    cin >> n >> r;

    for(int i = 1; i <= n; i++) cin >> a[i];

    for(int i = 1; i < n; i++) {
        cin >> x >> y;
        G[x].push_back(y);
        G[y].push_back(x);
    }

    memset(fa, 0, sizeof(fa));
    memset(dep, 0, sizeof(dep));
    fa[r] = -1;
    dfs(r);

    for(int i = 1; i <= n; i++) {
        Q[i][0] = a[i];

        for(int j = 0; j < G[i].size(); j++) {
            int u = G[i][j];

            if(u == fa[i]) continue;

            P[i][0].push_back(u);
            Q[i][0] = min(Q[i][0], a[u]);
        }
    }

    for(int j = 1; j < 25; j++) {
        for(int i = 1; i <= n; i++) {
            Q[i][j] = Q[i][j - 1];

            for(int k = 0; k < P[i][j - 1].size(); k++) {
                int u = P[i][j - 1][k];
                Q[i][j] = min(Q[i][j], Q[u][j - 1]);

                for(int p = 0; p < P[u][j - 1].size(); p++) {
                    int v = P[u][j - 1][p];
                    P[i][j].push_back(v);
                }
            }
        }
    }

    int t, p, q, cur, step, ans, last = 0;
    cin >> t;

    while(t--) {
        cin >> p >> q;

        //p = (p + last) % n + 1, q = (q + last) % n;
        if(q > dep[p]) q = dep[p];

        int pp = p, qq = q;
        vi[0].clear(), vi[1].clear();
        ans = a[p];
        cur = 0;
        vi[cur].push_back(p);
        step = 0;

        while(q) {
            vi[1 - cur].clear();

            if(q & 1) {
                for(int i = 0; i < vi[cur].size(); i++) {
                    int u = vi[cur][i];
                    ans = min(ans, Q[u][step]);

                    for(int j = 0; j < P[u][step].size(); j++) {
                        vi[1 - cur].push_back(P[u][step][j]);
                    }
                }

                cur = 1 - cur;
            }

            step++;
            q >>= 1;
        }

        cout << ans << endl;
        last = ans;
    }

    return 0;
}

View Code

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原文地址：https://www.cnblogs.com/dramstadt/p/7891020.html