Codeforces 991 车牌号组合数学计算

A

/*Huyyt*/
#include<bits/stdc++.h>
#define mem(a,b) memset(a,b,sizeof(a))
#define pb push_back
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int mod = 1e9 + ;
const int gakki =  +  +  +  + 1e9;
const int MAXN = 3e5 + , MAXM = 6e6 + ;
/*int to[MAXM << 1], nxt[MAXM << 1], Head[MAXN], tot = 1;
inline void addedge(int u, int v, ll val)
{
        to[++tot] = v;
        value[tot] = val;
        nxt[tot] = Head[u];
        Head[u] = tot;
}*/
int main()
{
        ios_base::sync_with_stdio();
        cin.tie();
        int a, b, c, n;
        cin >> a >> b >> c >> n;
        if (max(c, max(a, b)) > n)
        {
                cout << - << endl;
                return ;
        }
        if (c > min(a, b))
        {
                cout << - << endl;
                return ;
        }
        if (n - a - b + c < )
        {
                cout << - << endl;
                return ;
        }
        cout << n - a - b + c << endl;
        return ;
}

B

*10后就不会有精度问题了

/*Huyyt*/
#include<bits/stdc++.h>
#define mem(a,b) memset(a,b,sizeof(a))
#define pb push_back
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int mod = 1e9 + ;
const int gakki =  +  +  +  + 1e9;
const int MAXN = 3e5 + , MAXM = 6e6 + ;
/*int to[MAXM << 1], nxt[MAXM << 1], Head[MAXN], tot = 1;
inline void addedge(int u, int v, ll val)
{
        to[++tot] = v;
        value[tot] = val;
        nxt[tot] = Head[u];
        Head[u] = tot;
}*/
int num[];
priority_queue<int, vector<int>, greater<int> > que;
int main()
{
        ios_base::sync_with_stdio();
        cin.tie();
        int n;
        cin >> n;
        int sum = ;
        for (int i = ; i <= n; i++)
        {
                cin >> num[i];
                num[i] = num[i] * ;
                que.push(num[i]);
                sum += num[i];
        }
        int level =  * n;
        int ans = ;
        while (sum < level)
        {
                sum +=  - que.top();
                que.pop();
                ans++;
        }
        cout << ans << endl;
        return ;
}

C

一眼二分

/*Huyyt*/
#include<bits/stdc++.h>
#define mem(a,b) memset(a,b,sizeof(a))
#define pb push_back
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int mod = 1e9 + ;
const int gakki =  +  +  +  + 1e9;
const int MAXN = 3e5 + , MAXM = 6e6 + ;
/*int to[MAXM << 1], nxt[MAXM << 1], Head[MAXN], tot = 1;
inline void addedge(int u, int v, ll val)
{
        to[++tot] = v;
        value[tot] = val;
        nxt[tot] = Head[u];
        Head[u] = tot;
}*/
int num[];
priority_queue<int, vector<int>, greater<int> > que;
ll n;
ll check(ll x)
{
        ll sum = ;
        ll now = n;
        while (now > )
        {
                ll duce = min(x, now);
                sum += duce;
                now -= duce;
                now -= now / ;
        }
        return sum;
}
int main()
{
        ios_base::sync_with_stdio();
        cin.tie();
        cin >> n;
        ll aim = n /  + n % ;
        ll l = , r =  * n;
        ll mid;
        while (l < r - )
        {
                mid = (l + r) / ;
                if (check(mid) >= aim)
                {
                        r = mid;
                }
                else
                {
                        l = mid;
                }
        }
        cout << r << endl;
        return ;
}

D

DP或者贪心模拟即可

/*Huyyt*/
#include<bits/stdc++.h>
#define mem(a,b) memset(a,b,sizeof(a))
#define pb push_back
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int mod = 1e9 + ;
const int gakki =  +  +  +  + 1e9;
const int MAXN = 3e5 + , MAXM = 6e6 + ;
/*int to[MAXM << 1], nxt[MAXM << 1], Head[MAXN], tot = 1;
inline void addedge(int u, int v, ll val)
{
        to[++tot] = v;
        value[tot] = val;
        nxt[tot] = Head[u];
        Head[u] = tot;
}*/
char f[][];
char s[][];
string ch[];
int n;
int ans = ;
int main()
{
        ios_base::sync_with_stdio();
        cin.tie();
        for (int i = ; i <= ; i++)
        {
                cin >> ch[i];
                n = ch[i].size();
                for (int j = ; j < n; j++)
                {
                        f[i][j + ] = ch[i][j];
                }
        }
        for (int i = ; i <= ; i++)
        {
                for (int j = ; j <= n; j++)
                {
                        s[i][j] = f[i][j];
                }
        }
        int sumnow;
        for (int i = ; i <= n - ; i++)
        {
                sumnow = ;
                for (int j = ; j <= ; j++)
                {
                        for (int k = i; k <= i + ; k++)
                        {
                                if (s[j][k] == '')
                                {
                                        sumnow++;
                                }
                        }
                }
                if (sumnow >= )
                {
                        ans++;
                        for (int j = ; j <= ; j++)
                        {
                                for (int k = i; k <= i + ; k++)
                                {
                                        if (s[j][k] == '')
                                        {
                                                s[j][k] = 'X';
                                        }
                                }
                        }
                        if (sumnow == )
                        {
                                s[][i + ] = '';
                        }
                }
        }
        cout << ans << endl;
        return ;
}

E

题意:

给你一个最多18位的数 你可以任意排列或者删除某个数 但是出现过的数至少要有一个且不能有前导零

问你有几种方案

解:

统计下每个数的个数 然后DFS一下算下答案即可

/*Huyyt*/
#include<bits/stdc++.h>
#define mem(a,b) memset(a,b,sizeof(a))
#define pb push_back
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int mod = 1e9 + ;
const int gakki =  +  +  +  + 1e9;
const int MAXN = 5e2 + , MAXM = 6e6 + ;
/*int to[MAXM << 1], nxt[MAXM << 1], Head[MAXN], tot = 1;
inline void addedge(int u, int v, ll val)
{
        to[++tot] = v;
        value[tot] = val;
        nxt[tot] = Head[u];
        Head[u] = tot;
}*/
int num[MAXN];
int nownum[MAXN];
ll anser = ;
ll facnum[MAXN];
void init()
{
        facnum[] = ;
        for (int i = ; i <= ; i++)
        {
                facnum[i] = i * facnum[i - ];
        }
}
ll calans ()
{
        ll ansnow = ;
        for (int i = ; i <= ; i++)
        {
                ansnow += nownum[i];
        }
        ansnow = facnum[ansnow];
        for (int i = ; i <= ; i++)
        {
                ansnow /= facnum[nownum[i]];
        }
        return ansnow;
}
void getans (int x)
{
        if (x > )
        {
                anser += calans();
                if (nownum[] > )
                {
                        nownum[]--;
                        anser -= calans();
                        nownum[]++;
                }
                return ;
        }
        if (num[x])
        {
                for (nownum[x] = ; nownum[x] <= num[x]; nownum[x]++)
                {
                        getans(x + );
                }

        }
        else
        {
                getans(x + );
        }
}
int main ()
{
        ios_base::sync_with_stdio();
        cin.tie();
        init();
        ll pop;
        cin >> pop;
        while (pop > )
        {
                num[pop % ]++;
                pop /= ;
        }
        getans();
        cout << anser << endl;
        return ;
}