Codeforces Round #643 (Div. 2)

A. Sequence with Digits

当数位中出现0时break

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 1e6 + 5;

int main()
{
    ll t, n, k;
    scanf("%lld", &t);
    while(t--)
    {
        scanf("%lld%lld", &n, &k);
        for(ll i = 2; i <= k; ++i)
        {
            ll maxx = 0;
            ll minn = 9;
            ll tmp = n;
            while(tmp)
            {
                maxx = max(maxx, tmp % 10);
                minn = min(minn, tmp % 10);
                tmp /= 10;
            }
            n += maxx * minn;
            if(minn == 0)
            {
                break;
            }
        }
        cout<<n<<'\n';
    }
    return 0;
}

B. Young Explorers

题意：

给出每个人的无经验值，一个队伍中的人数应 >= 该队中每个人的无经验值，问最多组多少队

思路：

相同无经验值的人尽量组在一起，剩下的人按无经验值从小到大排个序，cnt记录当前队的人数，当人数 == 当前人的无经验值时，正好组成一个队。

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 2e5 + 5;

int main()
{
    int t, n, v;
    scanf("%d", &t);
    while(t--)
    {
        scanf("%d", &n);
        map<int, int>mp;
        map<int, int>::iterator it;
        for(int i = 1; i <= n; ++i)
        {
            scanf("%d", &v);
            mp[v]++;
        }
        int ans = 0;
        vector<int>vec;
        for(it = mp.begin(); it != mp.end(); ++it)
        {
            ans += it -> second / it -> first;
            int cnt = it -> second % it -> first;
            for(int i = 1; i <= cnt; ++i)
            {
                vec.push_back(it -> first);
            }
        }
        int siz = vec.size();
        int tmp = 0;
        for(int i = 0; i < siz; ++i)
        {
            tmp++;
            if(vec[i] == tmp)
            {
                ans++;
                tmp = 0;
            }
        }
        cout<<ans<<'\n';
        vec.clear();
        mp.clear();
    }
    return 0;
}

C. Count Triangles

题意：

给出a，b，c，d，令a <= x <= b <= y <= c <= z <= d，问由x，y，z组成的三角形有多少个

思路：

遍历最短边，即 i ~ [a，b]，次短边的范围为[i + b，i + c]，求该区间和最长边可取区间 [c，d] 的交集

分类讨论

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 2e5 + 5;

int main()
{
    ll a, b, c, d;
    while(~scanf("%lld%lld%lld%lld", &a, &b, &c, &d))
    {
        ll ans = 0;
        for(ll i = a; i <= b; ++i)
        {
            if(i + b <= c && i + c <= d)
            {
                ans += i * (i + 1) / 2;
            }
            else if(i + b <= c && i + c > d)
            {
                ans += (d - c + 1) * (i + c - d - 1) + (d - c + 2) * (d - c + 1) / 2;
            }
            else if(i + b > c && i + b <= d && i + c <= d)
            {
                ans += (i - c + b) * (c - b + 1) + (1 + c - b) * (c - b) / 2;
            }
            else if(i + b > c && i + b <= d && i + c > d)
            {
                ans += (i + c - d) * (d - c + 1) + (d - i - b + 1) * (i + b - c) + (1 + d - i - b) * (d - i - b) / 2;
            }
            else if(i + b > d && i + c > d)
            {
                ans += (d - c + 1) * (c - b + 1);
            }
        }
        cout << ans << '\n';
    }
    return 0;
}

D. Game With Array

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 2e5 + 5;

int main()
{
    int n, s, k;
    while(~scanf("%d%d", &n, &s))
    {
        if(s < 2 * n)
        {
            cout<<"NO"<<'\n';
            continue;
        }
        cout<<"YES"<<'\n';
        for(int i = 1; i < n; ++i)
        {
            cout<<2<<' ';
        }
        cout<<s - 2 * n + 2<<'\n';
        cout<<1<<'\n';
    }
    return 0;
}

E. Restorer Distance

题意：

有 n 个砖柱，给出每个砖柱的砖数，要求将所有砖柱处理成高度相同的砖柱，求总花费最小值

（1）放上一块砖花费A

（2）拿下一块砖花费R

（3）从一个砖柱上移动到另一个砖柱花费M

思路：

砖数和最终花费之间成二次函数关系，三分最终砖柱上有多少块砖，如果A + R > M时直接从一个地方移动到另一个地方

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const ll inf = 0x3f3f3f3f3f3f3f3f;
const int N = 1e5 + 5;

ll n, u, d, m, h[N];

ll check(ll num)
{
    ll up = 0, down = 0;
    for(ll i = 1; i <= n; ++i)
    {
        if(h[i] > num)
        {
            down += h[i] - num;
        }
        else
        {
            up += num - h[i];
        }
    }
    ll res = up * u + down * d;
    if(u + d > m)
    {
        ll tmp = min(up, down);
        res += m * tmp - (u + d) * tmp;
    }
    return res;
}

int main()
{
    while(~scanf("%d%d%d%d", &n, &u, &d, &m))
    {
        for(ll i = 1; i <= n; ++i)
        {
            scanf("%lld", &h[i]);
        }
        ll l = 0, r = 1e9;
        while(l + 10 < r)
        {
            ll m1 = l + (r - l) / 3;
            ll m2 = r - (r - l) / 3;
            ll tmp1 = check(m1);
            ll tmp2 = check(m2);
            if(tmp1 > tmp2)
            {
                l = m1;
            }
            else
            {
                r = m2;
            }
        }
        ll res = inf;
        for(ll i = l; i <= r; ++i)
        {
            res = min(res, check(i));
        }
        cout<<res<<'\n';
    }
    return 0;
}