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 + 7;
const int gakki = 5 + 2 + 1 + 19880611 + 1e9;
const int MAXN = 3e5 + 5, MAXM = 6e6 + 5;
/*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(0);
        cin.tie(0);
        int a, b, c, n;
        cin >> a >> b >> c >> n;
        if (max(c, max(a, b)) > n)
        {
                cout << -1 << endl;
                return 0;
        }
        if (c > min(a, b))
        {
                cout << -1 << endl;
                return 0;
        }
        if (n - a - b + c < 1)
        {
                cout << -1 << endl;
                return 0;
        }
        cout << n - a - b + c << endl;
        return 0;
}

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 + 7;
const int gakki = 5 + 2 + 1 + 19880611 + 1e9;
const int MAXN = 3e5 + 5, MAXM = 6e6 + 5;
/*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[105];
priority_queue<int, vector<int>, greater<int> > que;
int main()
{
        ios_base::sync_with_stdio(0);
        cin.tie(0);
        int n;
        cin >> n;
        int sum = 0;
        for (int i = 1; i <= n; i++)
        {
                cin >> num[i];
                num[i] = num[i] * 10;
                que.push(num[i]);
                sum += num[i];
        }
        int level = 45 * n;
        int ans = 0;
        while (sum < level)
        {
                sum += 50 - que.top();
                que.pop();
                ans++;
        }
        cout << ans << endl;
        return 0;
}

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 + 7;
const int gakki = 5 + 2 + 1 + 19880611 + 1e9;
const int MAXN = 3e5 + 5, MAXM = 6e6 + 5;
/*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[105];
priority_queue<int, vector<int>, greater<int> > que;
ll n;
ll check(ll x)
{
        ll sum = 0;
        ll now = n;
        while (now > 0)
        {
                ll duce = min(x, now);
                sum += duce;
                now -= duce;
                now -= now / 10;
        }
        return sum;
}
int main()
{
        ios_base::sync_with_stdio(0);
        cin.tie(0);
        cin >> n;
        ll aim = n / 2 + n % 2;
        ll l = 0, r = 2 * n;
        ll mid;
        while (l < r - 1)
        {
                mid = (l + r) / 2;
                if (check(mid) >= aim)
                {
                        r = mid;
                }
                else
                {
                        l = mid;
                }
        }
        cout << r << endl;
        return 0;
}

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 + 7;
const int gakki = 5 + 2 + 1 + 19880611 + 1e9;
const int MAXN = 3e5 + 5, MAXM = 6e6 + 5;
/*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[3][105];
char s[3][105];
string ch[3];
int n;
int ans = 0;
int main()
{
        ios_base::sync_with_stdio(0);
        cin.tie(0);
        for (int i = 1; i <= 2; i++)
        {
                cin >> ch[i];
                n = ch[i].size();
                for (int j = 0; j < n; j++)
                {
                        f[i][j + 1] = ch[i][j];
                }
        }
        for (int i = 1; i <= 2; i++)
        {
                for (int j = 1; j <= n; j++)
                {
                        s[i][j] = f[i][j];
                }
        }
        int sumnow;
        for (int i = 1; i <= n - 1; i++)
        {
                sumnow = 0;
                for (int j = 1; j <= 2; j++)
                {
                        for (int k = i; k <= i + 1; k++)
                        {
                                if (s[j][k] == '0')
                                {
                                        sumnow++;
                                }
                        }
                }
                if (sumnow >= 3)
                {
                        ans++;
                        for (int j = 1; j <= 2; j++)
                        {
                                for (int k = i; k <= i + 1; k++)
                                {
                                        if (s[j][k] == '0')
                                        {
                                                s[j][k] = 'X';
                                        }
                                }
                        }
                        if (sumnow == 4)
                        {
                                s[2][i + 1] = '0';
                        }
                }
        }
        cout << ans << endl;
        return 0;
}

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 + 7;
const int gakki = 5 + 2 + 1 + 19880611 + 1e9;
const int MAXN = 5e2 + 5, MAXM = 6e6 + 5;
/*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 = 0;
ll facnum[MAXN];
void init()
{
        facnum[0] = 1;
        for (int i = 1; i <= 20; i++)
        {
                facnum[i] = i * facnum[i - 1];
        }
}
ll calans ()
{
        ll ansnow = 0;
        for (int i = 0; i <= 9; i++)
        {
                ansnow += nownum[i];
        }
        ansnow = facnum[ansnow];
        for (int i = 0; i <= 9; i++)
        {
                ansnow /= facnum[nownum[i]];
        }
        return ansnow;
}
void getans (int x)
{
        if (x > 9)
        {
                anser += calans();
                if (nownum[0] > 0)
                {
                        nownum[0]--;
                        anser -= calans();
                        nownum[0]++;
                }
                return ;
        }
        if (num[x])
        {
                for (nownum[x] = 1; nownum[x] <= num[x]; nownum[x]++)
                {
                        getans(x + 1);
                }

        }
        else
        {
                getans(x + 1);
        }
}
int main ()
{
        ios_base::sync_with_stdio(0);
        cin.tie(0);
        init();
        ll pop;
        cin >> pop;
        while (pop > 0)
        {
                num[pop % 10]++;
                pop /= 10;
        }
        getans(0);
        cout << anser << endl;
        return 0;
}