2018-2019 Всероссийская командная олимпиада школьников по программированию, интернет-тур + отборы регионов (ВКОШП 18, интернет-тур) Solution

A：

水。

#include<bits/stdc++.h>

using namespace std;

typedef long long ll;

const ll INFLL = 0x3f3f3f3f3f3f3f3f;

ll n, m, h, w;

ll solve()
{
    ll cnt = 0;
    ll tn = n, tm = m;
    while(tn > h)
    {
        tn = (tn + 1)  >> 1;
        cnt++;
    }
    while(tm > w)
    {
        tm = (tm + 1) >> 1;
        cnt++;
    }
    return cnt;
}

int main()
{
    while(~scanf("%lld %lld %lld %lld", &n, &m, &h, &w))
    {
        ll ans = INFLL;
        ans = min(ans, solve());
        swap(h, w);
        ans = min(ans, solve());
        printf("%lld
", ans);
    }
    return 0;
}

B：留坑。

C:

题意：有n道题目，总时间为T,对于每一道题目有完成的时间和可以得到的分数，做题策略为按照一定顺序做题，如果时间足够，那么就可以得到这道题的分数，安排一个顺序，使得得到的分数最少

思路：如果总时间<=T 那么直接输出所有分数

反之 ，考虑一定存在至少一道题目恰好不能被完成

枚举这道题目，考虑所有时间小于这道题目的所有题目都要被完成

对于每道题目，还需要枚举剩下的时间，使得它完不成

那么排序之后枚举，左边的题目都要做完，右边的做01背包

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

#define ll long long
#define N 2010
#define pll pair <ll, ll> 
#define INF 0x3f3f3f3f3f3f3f3f
int n, m;
pll a[N];
ll dp[N][N];
ll tot_time, tot_sco;

int main()
{
    while (scanf("%d%d", &n, &m) != EOF)
    {
        tot_time = 0, tot_sco = 0;
        for (int i = 1; i <= n; ++i) 
        {
            scanf("%lld%lld", &a[i].first, &a[i].second);
            tot_time += a[i].first;
            tot_sco += a[i].second;
        }
        if (tot_time <= m)
        {
            printf("%lld
", tot_sco);
            continue;
        }
        sort(a + 1, a + 1 + n);
        memset(dp, 0x3f, sizeof dp);    
        dp[n + 1][0] = 0;
        for (int i = n; i > 1; --i)
        { 
            for (int j = m; j >= a[i].first; --j)
                dp[i][j] = min(dp[i][j], dp[i + 1][j - a[i].first] + a[i].second);
            for (int j = 0; j <= m; ++j) dp[i][j] = min(dp[i][j], dp[i + 1][j]);
        }
        ll step = 0, tot = 0, res = INF;
        for (int i = 1; i <= n; ++i)
        {
            ll tmp = INF;
            for (int j = m - step; j + a[i].first + step > m && j >= 0; --j) tmp = min(tmp, tot + dp[i + 1][j]);
            res = min(tmp, res); 
            step += a[i].first;
            tot += a[i].second;
        }
        printf("%lld
", res);
    }
    return 0;
}

D：

题意：给出一棵树中每个点的度数，构造这一棵树

思路：将点按度数从大到小排序，每个点会有一个父亲，最多连出n - 1个点。

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

#define N 200010
#define pii pair <int, int>
int n, d[N], pos[N], fa[N]; 

bool cmp (int a, int b) { return d[a] > d[b]; }

int main()
{
    while (scanf("%d", &n) != EOF)
    {
        for (int i = 1; i <= n; ++i) scanf("%d", d + i), pos[i] = i; 
        sort(pos + 1, pos + 1 + n, cmp);
        for (int i = 2; i <= n; ++i) --d[pos[i]];
        int id = 1;
        for (int i = 1; i <= n; ++i)
        {
            while (d[pos[i]]--) 
                fa[pos[++id]] = pos[i];   
        }
        for (int i = 1; i <= n; ++i) if (i != pos[1])
            printf("%d %d
", fa[i], i);    
    }
    return 0;
}

E：

水。

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

#define ll long long
#define N 1000010
char s[N];

int main()
{
    while (scanf("%s", s + 1) != EOF)
    {
        int cnt = 0;
        for (int i = 1, len = strlen(s + 1); i <= len; ++i) cnt += (s[i] == 'A');
        ll res = 0;
        for (ll i = 1; ; ++i)
        {
            if (i * (i + 1) / 2 > cnt) break;
            res = i;
        }
        printf("%lld
", res);
    }
    return 0;
}

F：留坑。

G：

题意：给出一个n 要构造一个长度为n的序列，满足题目要求

思路：

如果是奇数

那么存在 起始点3   有 $2 cdot 5 cdot 6 = 3 cdot 4 cdot 5$

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

int n;

int main()
{
    while (scanf("%d", &n) != EOF)
    {
        if (n == 1) puts("NO");
        else if (n & 1) 
        {
            puts("YES");
            puts("3"); 
            printf("-++");
            for (int i = 4; i <= n; ++i) printf("%c", "+-"[i & 1]);
            puts("");
        }
        else
        {
            puts("YES");
            puts("1");
            for (int i = 1; i <= n; ++i) printf("%c", "-+"[i & 1]);
            puts("");
        }
    }
    return 0;
}

H：

题意：给出0-9每个数字的个数，构造一个最大的数，使得任意连续三位连起来都是3的倍数。数字可以不用完

思路：先将所有数字模3

就是三种数字 0， 1， 2

再考虑 排列

显然 三个一组三个一组，三个中012至少出现一次，即012的全排列，一共6种，

再考虑 全放0, 1, 2 又是三种

再考虑 只有一位或者两位，可以任意摆放  

一共十种情况，去掉前导0之后取最大即可。

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

#define N 100010
int cnt[15], tmp_cnt[15], cntt[3], tmp_cntt[3], tar[3], Max[3], tmp_Max[3]; 
char ans[10][N];     
int pos;   

void g()
{
    for (int i = 0; i < 15; ++i) tmp_cnt[i] = cnt[i];
    for (int i = 0; i < 3; ++i) 
    {
        tmp_cntt[i] = cntt[i];
        tmp_Max[i] = Max[i];
    } 
} 

void f()
{
    for (int i = 0; i < 15; ++i) cnt[i] = tmp_cnt[i];
    for (int i = 0; i < 3; ++i)
    {
        cntt[i] = tmp_cntt[i];
        Max[i] = tmp_Max[i]; 
    }
} 

void solve(int ide, int *tar)  
{ 
    pos = 0;
    while (1)
    {
        for (int i = 0; i < 3; ++i)
        {
            int id = tar[i];
            if (!cntt[id]) 
                return; 
            if (cnt[Max[id]] == 0) 
            {
                Max[id] = -1;  
                for (int i = 9; i >= 0; --i) if (i % 3 == id && cnt[i]) 
                {
                    Max[id] = i; 
                    break; 
                }
                if (Max[id] == -1) return; 
            }
            ans[ide][++pos] = Max[id] + '0';   
            --cntt[id]; 
            --cnt[Max[id]];  
        }
    }
}

void work(int id)
{
    pos = 0; 
    while (cntt[id])
    {
        if (!cnt[Max[id]]) 
        {
            Max[id] = -1;
            for (int i = 9; i >= 0; --i) if (i % 3 == id && cnt[i])
            {
                Max[id] = i;
                break;    
            }
            if (Max[id] == -1) return;    
        }
        ans[id + 6][++pos] = Max[id] + '0';
        --cnt[Max[id]];
        --cntt[id];
    }
}

bool cmp(char *a, char *b) 
{
    int lena = strlen(a + 1), lenb = strlen(b + 1);
    if (lena != lenb) return lena < lenb;
    for (int i = 1; i <= lena; ++i) if (a[i] != b[i])
        return a[i] < b[i];
    return 1;
}

void cle(int id)
{
    int len = strlen(ans[id] + 1);
    pos = len + 1;
    for (int i = 1, len = strlen(ans[id] + 1); i <= len; ++i) if (ans[id][i] != '0')
    {
        pos = i;
        break;
    }
    if (pos == len + 1) 
    {
        ans[id][1] = '0';
        ans[id][2] = 0;
    }
    else
    {
        for (int i = 1; i + pos - 1 <= len; ++i)
            ans[id][i] = ans[id][i + pos - 1];
        ans[id][len - pos + 2] = 0;
    }
}

int main()
{
    while (scanf("%d", cnt) != EOF)
    {
        memset(cntt, 0, sizeof cntt); 
        memset(Max, -1, sizeof Max);
        memset(ans, 0, sizeof ans); 
        for (int i = 1; i <= 9; ++i) scanf("%d", cnt + i);
        for (int i = 0; i < 10; ++i) if (cnt[i])
        {
            cntt[i % 3] += cnt[i];
            Max[i % 3] = i; 
        } 
        int res = 0;  
        for (int i = 9; i >= 0 && res < 10; --i) if (cnt[i])
        {
            res = res * 10 + i;
            if (res < 10 && cnt[i] > 1)
                res = res * 10 + i; 
        } 
        g(); 
        pos = 0; 
        while (res)
        {
            ans[9][++pos] = res % 10 + '0';  
            res /= 10;
        } 
        ans[9][pos + 1] = 0; 
        reverse(ans[9] + 1, ans[9] + pos + 1);  
        tar[0] = 0, tar[1] = 1, tar[2] = 2;  
        int ide = 0;
        do
        {
            f();
            solve(ide, tar); ans[ide][pos + 1] = 0; 
            ++ide; 
        } while (next_permutation(tar, tar + 3)); 
        for (int i = 0; i < 3; ++i)
        {
            f(); pos = 0; work(i); 
            ans[i + 6][pos + 1] = 0;  
        } 
        for (int i = 0; i < 10; ++i) cle(i);
        pos = 0; 
        for (int i = 1; i < 10; ++i) if (cmp(ans[pos], ans[i]))
            pos = i;   
        printf("%s
", ans[pos] + 1); 
    } 
    return 0;
}

I：

题意：阅读理解题。有一个类似游泳圈的东西，有四道大道路，内圈，外圈，上圈，下圈，每个国家有四个城市，四个城市分别属于四道大道路上，定义经过一个国家为至少经过一条该国家的道路，有从任意一个国家的内城市出发，经过所有城市的最短路径

思路：考虑两种走法，

一种是两个国家之间都走内圈，还有一种是间隔走，走一下内圈，走一下上圈，取Min

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

#define ll long long
const double PI = acos(-1.0);
ll r, R, n;

int main()
{
    while (scanf("%lld%lld%lld", &r, &R, &n) != EOF)
    {
        if (n == 1) 
        {
            printf("%.15f
", PI * r / 2);
            continue;
        }
        double len1 = (R - r) * 2.0 * PI / n; 
        double len2 = R * 2.0 * PI / n;
        double ans1 = (2 * n - 1) * PI * r / 2 + (n - 1) * len1;
        double ans2 = n * PI * r / 2 + (n / 2) * len2 + ((n - 1) / 2) * len1;
        printf("%.15f
", min(ans1, ans2));
    }
    return 0;
}

J：留坑。

k：留坑。

L：水。

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

#define ll long long
#define N 1010
int n, k, x, cnt; 
char s[15];
ll Bit[15];

ll f(char *s)
{
    ll res = 0;
    for (int i = 1, len = strlen(s + 1); i <= len; ++i)
        res = res * 10 + s[i] - '0';
    return res;
}

int main()
{
    Bit[1] = 1;
    for (int i = 2; i <= 10; ++i) Bit[i] = Bit[i - 1] * 10;
    while (scanf("%d", &n) != EOF)
    {
        cnt = 0; ll Min = 0;
        for (int i = 1; i <= n; ++i) 
        {
            scanf("%s", s + 1);
            cnt = max(cnt, (int)strlen(s + 1));
            Min = max(Min, f(s)); 
        }
        Min = max((ll)n, Min); 
        Min = max(Min, Bit[cnt]); 
        printf("%lld
", Min);
        for (int i = 1; i <= cnt; ++i) putchar('9'); putchar('
');
    }
    return 0;
}