AtCoder Beginning Contest 128

打完心情巨差，题目其实很简单，OJ垃圾。以后坚决不在这破OJ上用STL。

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A：

一眼秒杀。

/* basic header */
#include <bits/stdc++.h>
/* define */
#define ll long long
#define dou double
#define pb emplace_back
#define mp make_pair
#define sot(a,b) sort(a+1,a+1+b)
#define rep1(i,a,b) for(int i=a;i<=b;++i)
#define rep0(i,a,b) for(int i=a;i<b;++i)
#define eps 1e-8
#define int_inf 0x3f3f3f3f
#define ll_inf 0x7f7f7f7f7f7f7f7f
#define lson curPos<<1
#define rson curPos<<1|1
/* namespace */
using namespace std;
/* header end */

int a, p;

int main() {
    cin >> a >> p;
    cout << (a * 3 + p) / 2 << endl;
    return 0;
}

B：

一眼秒杀。

/* basic header */
#include <bits/stdc++.h>
/* define */
#define ll long long
#define dou double
#define pb emplace_back
#define mp make_pair
#define sot(a,b) sort(a+1,a+1+b)
#define rep1(i,a,b) for(int i=a;i<=b;++i)
#define rep0(i,a,b) for(int i=a;i<b;++i)
#define eps 1e-8
#define int_inf 0x3f3f3f3f
#define ll_inf 0x7f7f7f7f7f7f7f7f
#define lson curPos<<1
#define rson curPos<<1|1
/* namespace */
using namespace std;
/* header end */

struct Node {
    string s;
    int score, pos;
    Node() {}
    Node(string a, int b, int c) {
        s = a; score = b; pos = c;
    }
    bool operator<(const Node &rhs)const {
        if (s != rhs.s) return s < rhs.s;
        else return score > rhs.score;
    }
};

vector<Node>v;

int main() {
    int n; cin >> n;
    rep1(i, 1, n) {
        string s; int p; cin >> s >> p;
        v.pb(s, p, i);
    }
    sort(v.begin(), v.end());
    for (auto i : v) cout << i.pos << endl;
    return 0;
}

C：

一眼秒杀，枚举所有开关状态即可。但是用set T了超久都没发现导致极度自闭，比赛结束前十分钟重构改成数组就A了。垃圾OJ。

/* basic header */
#include <bits/stdc++.h>
/* define */
#define ll long long
#define dou double
#define pb emplace_back
#define mp make_pair
#define sot(a,b) sort(a+1,a+1+b)
#define rep1(i,a,b) for(int i=a;i<=b;++i)
#define rep0(i,a,b) for(int i=a;i<b;++i)
#define eps 1e-8
#define int_inf 0x3f3f3f3f
#define ll_inf 0x7f7f7f7f7f7f7f7f
#define lson curPos<<1
#define rson curPos<<1|1
/* namespace */
using namespace std;
/* header end */
 
const int maxn = 15;
int n, m, p[maxn], ans = 0;
set<int>a[maxn]; //每个开关管哪些灯
 
int main() {
    cin >> n >> m;
    rep1(i, 1, m) {
        int k; cin >> k;
        while (k--) {
            int x; cin >> x;
            a[x].insert(i);
        }
    }
    rep1(i, 1, m) cin >> p[i];
    int lim = pow(2, n);
    int swit[maxn], time[maxn];
    for (int i = 0; i < lim; i++) {
        int tmp = i, cnt = m, flag = 1;
        for (int j = 0; j < maxn; j++) swit[j] = time[j] = 0;
        while (tmp) swit[cnt--] = tmp % 2, tmp >>= 1;
        for (int j = 1; j <= m; j++)
            if (swit[j])
                for (auto p : a[j]) time[p]++;
        for (int j = 1; j <= m; j++)
            if (time[j] % 2 != p[j]) {
                flag = 0; break;
            }
        if (flag) ans++;
    }
    cout << ans << endl;
    return 0;
}

/* basic header */
#include <bits/stdc++.h>
/* define */
#define ll long long
#define dou double
#define pb emplace_back
#define mp make_pair
#define sot(a,b) sort(a+1,a+1+b)
#define rep1(i,a,b) for(int i=a;i<=b;++i)
#define rep0(i,a,b) for(int i=a;i<b;++i)
#define eps 1e-8
#define int_inf 0x3f3f3f3f
#define ll_inf 0x7f7f7f7f7f7f7f7f
#define lson curPos<<1
#define rson curPos<<1|1
/* namespace */
using namespace std;
/* header end */

const int maxn = 15;
int n, m, p[maxn], ans = 0, a[maxn][maxn];

int main() {
    scanf("%d%d", &n, &m);
    rep1(i, 1, m) {
        scanf("%d", &a[i][0]);
        rep1(j, 1, a[i][0]) scanf("%d", &a[i][j]);
    }
    rep1(i, 1, m) scanf("%d", &p[i]);
    rep1(i, 0, (1 << n) - 1) {
        int flag = 1, cmp = i, swit[20], light[20], cnt = 0;
        rep1(i, 1, 20) swit[i] = light[i] = 0;
        while (cmp) swit[++cnt] = cmp & 1, cmp >>= 1;
        rep1(i, 1, m)
        rep1(j, 1, a[i][0]) light[i] += swit[a[i][j]];
        rep1(i, 1, m) if ((light[i] % 2) != p[i]) {
            flag = 0; break;
        }
        if (flag) ans++;
    }
    printf("%d
", ans);
    return 0;
}

D：

一眼觉得是贪心，但是被上一题TLE搞到没心情想了。队友直接用了个O(n^3)的暴力枚举AC。先枚举答案由多少个数组成，再枚举从左边选的数的个数，最后枚举从右边选的数的个数。

正解也是枚举，但是如果这题数据量变大，能不能贪心呢？不太确定。

Code via. BWKPlus

/* basic header */
#include <bits/stdc++.h>
/* define */
#define ll long long
#define dou double
#define pb emplace_back
#define mp make_pair
#define sot(a,b) sort(a+1,a+1+b)
#define rep1(i,a,b) for(int i=a;i<=b;++i)
#define rep0(i,a,b) for(int i=a;i<b;++i)
#define eps 1e-8
#define int_inf 0x3f3f3f3f
#define ll_inf 0x7f7f7f7f7f7f7f7f
#define lson curPos<<1
#define rson curPos<<1|1
/* namespace */
using namespace std;
/* header end */

const int maxn = 60;
int a[maxn], now[maxn], ans = 0, n, m;

int main() {
    scanf("%d%d", &n, &m);
    rep1(i, 1, n) scanf("%d", &a[i]);
    for (int i = 1; i <= min(n, m); i++) {
        rep1(j, 0, i) {
            rep1(k, 0, n) now[k] = 0;
            int cnt = 0;
            rep1(k, 1, i - j) now[++cnt] = a[k];
            rep1(k, 1, j) now[++cnt] = a[n - k + 1];
            sot(now, i);
            int tot = m - i, sum = 0, x = 1;
            while (tot && now[x] < 0) x++, tot--;
            for (; x <= i; x++) sum += now[x];
            ans = max(ans, sum);
        }
    }
    printf("%d
", ans);
    return 0;
}

E&&F：

待补。