安徽大学第九届程序设计竞赛决赛题解

A.调酒师

很水的一道题，算出每个参数的最小值就行，手速选手被碾压的开始QAQ。

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

int main() {
    //freopen("in.txt", "r", stdin);
    int T;
    int n, a, b, c, d, e, f, g;
    cin >> T;
    while (T--) {
        cin >> n;
        cin >> a >> b >> c >> d >> e >> f >> g;
        int A = a * b;
        int B = c * d;
        int a1, a2, a3;
        a1 = A / f; a2 = B; a3 = e / g;
        int ans = min(a1, min(a2, a3));
        cout << ans / n << endl;
    }

    return 0;
}

B.下一个幸运数

本场比赛难度第三的题目，一开始当成数位DP做，一个半小时debug无果，最后发现幸运数总共就1000多个，存起来暴力扫描相加就行 。

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
vector<ll> vec;
vector<ll>::iterator it;
set<ll> ss;
char s[15];

void dfs(int cur, bool zero, ll sum) {
    if (cur == 11) return;
    if (!ss.count(sum)) {
        ss.insert(sum);
        vec.push_back(sum);
    }
    if (zero) {
        dfs(cur+1, 1, 0);
        dfs(cur+1, 0, 4);
        dfs(cur+1, 0, 7);
    }
    else {
        dfs(cur+1, 0, sum*10+4);
        dfs(cur+1, 0, sum*10+7);
    }
}

int main() {
    dfs(0, 1, 0);
    sort(vec.begin(), vec.end());
    it = vec.begin();
    int T;
    cin >> T;
    while (T--) {
        ll l, r;
        cin >> l >> r;
        int cnt = 0, len = vec.size();
        for (int i = 0; i < len; ++i, ++cnt) if (l <= vec[i]) break;
        int xx = min(vec[cnt], r) - l + 1;
        ll ans = xx * vec[cnt];
        //cout << xx << endl;
        ll temp = vec[cnt];
        int cnt2 = cnt + 1;
        for (int i = cnt + 1; i < len; ++i, ++cnt2) {
            if (vec[i] > r) break;
            int yy = vec[i] - temp;
            temp = vec[i];
            ans += yy * vec[i];
        }
        if (vec[cnt2-1] < r) {
            int zz = r-vec[cnt2-1];
            ans += zz * vec[cnt2];
        }
        cout << ans << endl;
    }

    return 0;
}

C.最短路径和

做出这道题目需要灵活掌握Floyd的原理，比赛时只会个形式便没写出来。实际上只需要记录删点的次序，倒叙用来当中转点更新路径，统计ans的时候只统计目前存在的点。

#include<bits/stdc++.h>
using namespace std;
#define M(a, b) memset(a, b, sizeof(a))
#define INF 0x3f3f3f3f
typedef long long ll;
const int N = 400 + 5;
int G[N][N], d[N][N], n, q[N], ans[N];

void Floyd() {
    M(ans, 0);
    for (int kk = n; kk >= 1; --kk) {
        int k = q[kk];
        for (int i = 1; i <= n; ++i)
            for (int j = 1; j <= n; ++j) {
                d[i][j] = min(d[i][j], d[i][k]+d[k][j]);
            }
        for (int i = kk; i <= n; ++i)
            for (int j = kk; j <= n; ++j)
                ans[kk] += d[q[i]][q[j]];
    }
    cout << ans[1];
    for (int i = 2; i <= n; ++i) cout << " " << ans[i];
    cout << endl;
}

int main() {
    ios::sync_with_stdio(false);
    while (cin >> n) {
        M(G, 0); M(d, INF);
        for (int i = 1; i <= n; ++i)
            for (int j = 1; j <= n; ++j)
                cin >> d[i][j];
        for (int i = 1; i <= n; ++i) cin >> q[i];
        Floyd();
    }

    return 0;
}

D. 西瓜的编译原理作业

贼裸的字典树，不需要任何技巧，不会字典树拿指针模拟都可以做。

#include<bits/stdc++.h>
using namespace std;
#define INF 0x3f3f3f3f
#define M(a, b) memset(a, b, sizeof(a))
#define idx(a) a - 'A'
const int maxn = 5000 + 10;
int ch[maxn*70][27], val[maxn*70], tot;
char s[100];

void Insert(char *s) {
    int now = 0, len = strlen(s), id;
    for (int i = 0; i < len; ++i) {
        id = idx(s[i]);
        if (!ch[now][id]) ch[now][id] = ++tot;
        now = ch[now][id];
    }
    val[now] = 1;
}

int ans = 0;
void dfs(int now) {
    ans++;
    for (int i = 0; i < 26; ++i)
        if (ch[now][i]) dfs(ch[now][i]);
}

int main() {
    //freopen("D://in.txt", "r", stdin);
    while (cin >> s) {
        Insert(s);
    }
    dfs(0);
    cout << ans << endl;

    return 0;
}

E. 晋级下一轮

排序

#include<bits/stdc++.h>
using namespace std;
int a[200];

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

int main() {
    //freopen("D://in.txt", "r", stdin);
    int T;
    int n, k;
    cin >> T;
    while (T--) {
        cin >> n >> k;
        for (int i = 0; i < n; ++i) cin >> a[i];
        sort(a, a+n, cmp);
        int cnt = 1;
        int temp = a[0], kk;
        int i;
        bool flag = 0;
        kk = a[k-1];
        //cout << kk << endl;
        int ans = 0;
        for (int i = 0; i < n; ++i) if (a[i] >= kk && a[i]) ++ans;
        cout << ans << endl;
    }

    return 0;
}

F. 西瓜和病毒

把特征串建立成AC自动机，再从root开始跑，如果存在一个环（环的路径上不包括任何特征串，也就是不经过val值为1的点），就存在一个安全串，这题的一个trick就是如果某个点u的失配指针f[u]的val为1，要把val[u]也赋值为1，因为root~f[u]为root~u的后缀。

#include<bits/stdc++.h>
using namespace std;
#define idx(a) a - '0'
#define M(a, b) memset(a, b, sizeof(a))
#define INF 0x3f3f3f3f
const int N = 30000 + 5;
char s[N];

struct ac_mata {
    int ch[N][2], f[N], in[N], val[N], tot;
    bool vis[N];

    void init() {
        tot = 0;
        M(f, 0); M(in, 0); M(vis, 0); M(val, 0); M(ch, 0);
    }

    void insert(char *s) {
        int now = 0, id, len = strlen(s);
        for (int i = 0; i < len; ++i) {
            id = idx(s[i]);
            if (!ch[now][id]) ch[now][id] = ++tot;
            now = ch[now][id];
        }
        val[now] = 1;
    }

    void getFail() {
        queue<int> q;
        int u;
        for (int i = 0; i < 2; ++i) {
            u = ch[0][i];
            if (u) {f[u] = 0; q.push(u);}
        }
        while (!q.empty()) {
            u = q.front(); q.pop();
            for (int i = 0; i < 2; ++i) {
                int v = ch[u][i];
                if (!v) {ch[u][i] = ch[f[u]][i]; continue;}
                q.push(v);
                int k = f[u];
                if (k && !ch[k][i]) k = f[k];
                f[v] = ch[k][i];
                val[v] |= val[f[v]];
            }
        }
    }

    bool findcircle(int x) {
        in[x] = 1;
        for (int i = 0; i < 2; ++i) {
            int u = ch[x][i];
            if (in[u]) return 1;
            if (vis[u] || val[u]) continue;
            vis[u] = 1;
            if (findcircle(u)) return 1;
        }
        in[x] = 0;
        return 0;
    }

}ac;

int main() {
    ios::sync_with_stdio(false);
    int n;
    while (cin >> n) {
        ac.init();
        while (n--) {
            cin >> s;
            ac.insert(s);
        }
        ac.getFail();
        if (ac.findcircle(0)) cout << "TAK" << endl;
        else cout << "NIE" << endl;
    }

    return 0;
}

G. 黑白棋

大水题，题面把行写成了列，debug很久，头皮发麻。

#include<bits/stdc++.h>
using namespace std;
char s[20];
char a[12][12];
bool b[13];
vector<int> vec;

int main() {
    //freopen("D://in.txt", "r", stdin);
    int n;
    cin >> n;
    while (n--) {
        memset(b, 0, sizeof(b));
        vec.clear();
        int ans = 0;
        scanf("%s", s+1);
        bool flag = 0;
        for (int j = 1; j <= 12; j++) {
            if (12 % j) continue;
            for (int i = 1; i <= j; ++i) {
                flag = 1;
                for (int k = i; k <= 12; k += j)
                    if (s[k] == 'O') flag = 0;
                if (flag) break;
            }
            if (flag) vec.push_back(j), ++ans;
        }
        cout << ans;
        for (int i = vec.size()-1; i >= 0; --i) {
            int c = vec[i];
            cout << " " << 12 / c << "x" << c;
        }
        cout << endl;
    }

    return 0;
}

H. 勤劳的出题人

两个指针指向两个数组，一一比较，最后总数减去成功匹配的数量即可。

#include<bits/stdc++.h>
using namespace std;
const int maxn = 3000 + 10;
int a[maxn], b[maxn];
bool vis[maxn], have[1000010];

int main() {
    //freopen("D://in.txt", "r", stdin);
    int n, m;
    while (cin >> n >> m) {
        memset(have, 0, sizeof(have));
        memset(vis, 0, sizeof(vis));
        for (int i = 1; i <= n; ++i) cin >> a[i];
        for (int j = 1; j <= m; ++j) {
                cin >> b[j];
        }
        sort(a+1, a+n+1);
        sort(b+1, b+m+1);
        int cnt = 0;
        int i = 1, j = 1;
        while (i <= n && j <= m){
            if (a[i] <= b[j]) {
                ++i;
                ++j;
                ++cnt;
            }
            else ++j;
        }
        cout << n - cnt << endl;
    }

    return 0;
}