Codeforces Round #565 div3

https://codeforces.com/contest/1176

https://codeforces.com/contest/1176/standings

 

A: Divide it! 

题意：每个数可以n/2, 2n/3, 3n/5，求变为1的最小次数，不行的话输出-1

贪心

 

B:Merge it!

题意：一个数组，可以选泽把两个数求和后合并为一个数，求最多的能被3整数的数的个数

贪心：

mod 3 == 0不处理， 取min(mod3==1, mod3==2)，再把剩下的 mod3==1的三个合为一个, mod3==2的三个合为一个

 

C:Lose it!

实际上就是求原数组可以分为多少个长度为6的数组 [4,8,15,16,23,42]

怎么说呢。。看代码吧

#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cctype>
#include <string>
#include <cstring>
#include <ctime>
#include <cassert>
#include <string.h>
#include <unordered_set>
#include <unordered_map>
using namespace std;

#define rep(i, a, b) for(int i = (a); i <= (b); i++)
#define reps(i, a, b) for(int i = (a); i < (b); i++)
#define pb push_back
#define ps push
#define mp make_pair
#define CLR(x,t) memset(x,t,sizeof x)
#define LEN(X) strlen(X)
#define F first
#define S second
#define Debug(x) cout<<#x<<"="<<x<<endl;


const double euler_r = 0.57721566490153286060651209;
const double pi = 3.141592653589793238462643383279;
const double E = 2.7182818284590452353602874713526;
const int inf = ~0U >> 1;
const int MOD = int (1e9) + 7;
const double EPS = 1e-6;

typedef long long LL;

int n, x;
int cnt[10];

int cal (int x) {
    if (x == 4) return 1;
    if (x == 8) return 2;
    if (x == 15) return 3;
    if (x == 16) return 4;
    if (x == 23) return 5;
    if (x == 42) return 6;
}

int main() {
    cin >> n;
    cnt[0] = 10000000;
    rep (i, 1, n) {
        cin >> x;
        x = cal (x);
        if (cnt[x - 1] > 0) {
            cnt[x - 1]--;
            cnt[x]++;
        }

    }
    cout << n - 6 * cnt[6] << endl;

}

D:Recover it!

题意：有两个数组A和B,初始B=A,遍历A数组，如果A[i]是素数，就把第A[i]个素数加到B数组，否则就把A[i]的最大约数（不是自己）加到B数组.

题目给了B数组，求满足条件的A数组

贪心:

预处理出素数，和素数的位置

倒序枚举每个数，如果是素数，这个素数认为是被A[i]是素数的情况导出来的，否则认为这个数是A数组的数，那么它可以导出一个约数。

#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cctype>
#include <string>
#include <cstring>
#include <ctime>
#include <cassert>
#include <string.h>
#include <unordered_set>
#include <unordered_map>
using namespace std;

#define rep(i, a, b) for(int i = (a); i <= (b); i++)
#define reps(i, a, b) for(int i = (a); i < (b); i++)
#define pb push_back
#define ps push
#define mp make_pair
#define CLR(x,t) memset(x,t,sizeof x)
#define LEN(X) strlen(X)
#define F first
#define S second
#define Debug(x) cout<<#x<<"="<<x<<endl;


const double euler_r = 0.57721566490153286060651209;
const double pi = 3.141592653589793238462643383279;
const double E = 2.7182818284590452353602874713526;
const int inf = ~0U >> 1;
const int MOD = int (1e9) + 7;
const double EPS = 1e-6;

typedef long long LL;

const int maxn = 3000000;
int n;
int a[maxn];
int id[maxn];
bool used[maxn];

//max divisor
int fi (int x) {
    for (int i = 2; i * i <= x; i++) if (x % i == 0) return x / i;
}



int main() {
    ios::sync_with_stdio (false);
    cin.tie (0);
    vector<int> p;
    vector<int> res;
    for (int i = 2; i < maxn; i++) {
        if (used[i]) continue;
        p.push_back (i);
        id[i] = p.size();
        for (int j = i + i; j < maxn; j += i) {
            used[j] = true;
        }
    }
    cin >> n;
    n <<= 1;
    for (int i = 1; i <= n; i++) {
        int x;
        cin >> x;
        a[x]++;
    }
    for (int i = maxn - 1; i >= 2; i--) {
        while (a[i]) {
            if (!used[i]) {
                //i is prime
                res.push_back (id[i]);
                a[id[i]]--;
            } else {
                res.push_back (i);
                a[fi (i)]--;
            }
            a[i]--;
        }

    }
    for (int x : res) {
        cout << x << " ";
    }
    cout << endl;
    return 0;
}

E:Cover it!

一个无相连通图，求选择一个顶点集( 个数<= n/2)，使得剩下的每个顶点都和所选的顶点集有边相连.

黑白染色，选个数<=n/2的那个集合

#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cctype>
#include <string>
#include <cstring>
#include <ctime>
#include <cassert>
#include <string.h>
#include <unordered_set>
#include <unordered_map>
using namespace std;

#define rep(i, a, b) for(int i = (a); i <= (b); i++)
#define reps(i, a, b) for(int i = (a); i < (b); i++)
#define pb push_back
#define ps push
#define mp make_pair
#define CLR(x,t) memset(x,t,sizeof x)
#define LEN(X) strlen(X)
#define F first
#define S second
#define Debug(x) cout<<#x<<"="<<x<<endl;


const double euler_r = 0.57721566490153286060651209;
const double pi = 3.141592653589793238462643383279;
const double E = 2.7182818284590452353602874713526;
const int inf = ~0U >> 1;
const int MOD = int (1e9) + 7;
const double EPS = 1e-6;

typedef long long LL;

const int maxn = 200020;
int n, m;
int a[maxn];
vector<int> v[maxn];

//black-white ranse
void dfs (int x, int c) {
    if (a[x] > 0) {
        return ;
    }
    a[x] = c;
    for (int y : v[x]) {
        dfs (y, 3 - c);
    }

}

void solve() {
    cin >> n >> m;
    rep (i, 1, n) v[i].clear(), a[i] = 0;
    rep (i, 1, m) {
        int x, y;
        cin >> x >> y;
        v[x].push_back (y);
        v[y].push_back (x);
    }
    dfs (1, 1);
    vector<int> A, B;
    rep (i, 1, n) {
        if (a[i] == 1) A.push_back (i);
        else B.push_back (i);
    }
    if (B.size() < A.size()) A = B;
    cout << A.size() << endl;
    for (int x : A) {
        cout << x << " ";
    }
    cout << endl;
}

int main() {
    int T;
    cin >> T;
    while (T--) {
        solve();
    }
}

F:Destroy it!

打boss，一共有n轮，每轮你有一些卡片，每个卡牌有两个属性cost、damage.每一轮你可以选择一些卡片,他们c的和不能超过3，会造成d的伤害（选的卡牌的d的和）.每第10张卡片会造成双倍伤害，求最大值.

可以看下样例给的解释，非常清楚。

 

做法:应该是dp，但是我不会。。。