Codeforces Round #523 (Div. 2) Solution

A. Coins

Water.

#include <bits/stdc++.h>
using namespace std;
int n, s;

int main()
{
    while (scanf("%d%d", &n, &s) != EOF)
    {
        int res = 0;
        for (int i = n; i >= 1; --i) while (s >= i) 
        {
            ++res;
            s -= i;
        }
        printf("%d\n", res);
    }
    return 0;
}

 

 

B. Views Matter

Solved.

题意：

有n个栈，不受重力影响，在保持俯视图以及侧视图不变的情况下，最多可以移掉多少个方块

思路：

考虑原来那一列有的话那么这一列至少有一个，然后贪心往高了放

#include <bits/stdc++.h>
using namespace std;

#define ll long long
#define N 100010
int n, m;
ll a[N], sum;

int main()
{
    while (scanf("%d%d", &n, &m) != EOF)
    {
        sum = 0;
        for (int i = 1; i <= n; ++i) scanf("%lld", a + i), sum += a[i];
        sort(a + 1, a + 1 + n);
        ll res = 0, high = 0;
        for (int i = 1; i <= n; ++i) if (a[i] > high) 
            ++high;
        printf("%lld\n", sum - n - a[n] + high);
    }
    return 0;
}

 

C. Multiplicity

Upsolved.

题意：

定义一个序列为好的序列即$b_1, b_2, ...., b_k 中i \in [1, k] 使得 b_i % i == 0$

求有多少个好的子序列

思路：

考虑$Dp$

$令dp[i][j] 表示第i个数，长度为j的序列有多少种方式$

$dp[i][j] = dp[i - 1][j - 1] + dp[i - 1][j]   (arr[j] % j == 0)$

否则 $dp[i][j] = dp[i - 1][j]$

然后不能暴力递推，只需要更新$arr[j] % j == 0 的j即可$

#include <bits/stdc++.h>
using namespace std;

#define ll long long
#define N 1000100
const ll MOD = (ll)1e9 + 7;
int n; ll a[N], dp[N];

int main()
{
    while (scanf("%d", &n) != EOF)
    {
        for (int i = 1; i <= n; ++i) scanf("%lld", a + i);
        memset(dp, 0, sizeof dp);
        dp[0] = 1;    
        for (int i = 1; i <= n; ++i)
        {
            vector <int> cur;
            for (int j = 1; j * j <= a[i]; ++j)
            {
                if (a[i] % j == 0)
                {
                    cur.push_back(j);
                    if (j != a[i] / j)
                        cur.push_back(a[i] / j);
                }
            }
            sort(cur.begin(), cur.end());
            reverse(cur.begin(), cur.end());
            for (auto it : cur)
                dp[it] = (dp[it] + dp[it - 1]) % MOD;
        }
        ll res = 0;
        for (int i = 1; i <= 1000000; ++i) res = (res + dp[i]) % MOD;
        printf("%lld\n", res);    
    }
    return 0;
}

 

D. TV Shows

Upsolved.

题意：

有n个电视节目，每个节目播放的时间是$[l, r]，租用一台电视机的费用为x + y \cdot time$

一台电视机同时只能看一个电视节目，求看完所有电视节目最少花费

思路：

贪心。

因为租用电视机的初始费用是相同的，那么我们将电视节目将左端点排序后

每次选择已经租用的电视机中上次放映时间离自己最近的，还要比较租用新电视机的费用，

如果是刚开始或者没有一台电视机闲着，则需要租用新的电视机

#include <bits/stdc++.h>
using namespace std;

#define ll long long
#define N 100010
struct node
{
    ll l, r;
    void scan() { scanf("%lld%lld", &l, &r); }
    bool operator < (const node &r) const 
    {
        return l < r.l || (l == r.l && this->r < r.r);
    }
}arr[N];
int n; ll x, y;
const ll MOD = (ll)1e9 + 7;
multiset <ll> se;

int main()
{
    while (scanf("%d%lld%lld", &n, &x, &y) != EOF) 
    {
        for (int i = 1; i <= n; ++i) arr[i].scan();
        sort(arr + 1, arr + 1 + n);
        ll res = 0;
        for (int i = 1; i <= n; ++i)
        {
            if (se.lower_bound(arr[i].l) == se.begin())
                res = (res + x + y * (arr[i].r - arr[i].l) % MOD) % MOD;
            else
            {
                int pos = *(--se.lower_bound(arr[i].l)); 
                if (x < y * (arr[i].l - pos)) 
                    res = (res + x + y * (arr[i].r - arr[i].l) % MOD) % MOD;
                else
                {
                    se.erase(--se.lower_bound(arr[i].l));
                    res = (res + y * (arr[i].r - pos) % MOD) % MOD; 
                }    
            }
            se.insert(arr[i].r); 
        }
        printf("%lld\n", res);
    }
    return 0;
}

 

E. Politics

Unsolved.

 

F. Lost Root

Unsolved.