AtCoder Beginner Contest 159

https://atcoder.jp/contests/abc159
EF 是打基础的好题

D - Banned K

#include <bits/stdc++.h>
#define ll long long

using namespace std;
const int N = 2e5 + 5;
int a[N], n;

int main () {
    cin >> n;
    map<int, int> mp;
    for (int i = 1; i <= n; i++)    cin >> a[i], mp[a[i]] ++;
    ll ans = 0;
    for (auto i : mp)   ans += max (0ll, 1ll * i.second * (i.second - 1) / 2);
    for (int i = 1; i <= n; i++)    cout << ans - mp[a[i]] + 1 << endl;
}

E - Dividing Chocolate

二进制枚举 + 贪心 + 前缀和处理
这题思路主要是来源于 \(n\) 那不同寻常小的范围

#include <bits/stdc++.h>

using namespace std;
const int N = 15, M = 1005;
int a[N][M], n, m, k, ans;
int sum[N][M]; //sum[i][j]: 第i行的前缀和(分行考虑)

int count (int x) {
    int cnt = 0;
    while (x)   cnt += (x & 1), x /= 2;
    return cnt;
}

int main () {
    cin >> n >> m >> k;
    ans = n * m;
    for (int i = 1; i <= n; i++) {
        string s;
        cin >> s;
        int cnt = 0;
        for (int j = 1; j <= m; j++) {
            a[i][j] = s[j-1] - '0';
            sum[i][j] = sum[i][j-1] + a[i][j];
            //cout << sum[i][j] << ' ';
        }
        //cout << endl;
    }

    for (int st = 0; st < (1ll << (n - 1)); st++) {
        //如果有单列都超了的就直接剪枝
        int lst = 0, maxn = 0, tsum = 0;
        bool suc = true;
        for (int j = 1; j <= m; j++) {
            tsum = 0;
            for (int i = 1; i <= n; i++) {
                tsum += a[i][j];
                if (tsum > k) {
                    suc = false;
                    break;
                }
                if (st >> (i - 1) & 1)  tsum = 0;
            }
        }
        if (!suc)   continue;

        //cout << st << ": ";
        int cnt = count (st);
        //cout << cnt << endl;
        lst = 0; //列的上一个切割点
        for (int j = 1; j <= m; j++) { //列贪心切割s
            maxn = tsum = 0; //目前最大的一块
            for (int i = 1; i <= n; i++) {
                tsum += sum[i][j] - sum[i][lst];
                maxn = max (maxn, tsum);
                if (st >> (i - 1) & 1)    tsum = 0;
            }
            //cout << maxn << ' ';
            if (maxn > k) {
                cnt ++;
                lst = j - 1;
            }
        }
        ans = min (ans, cnt);
        //cout << endl;
    }
    cout << ans << endl;
}

//二进制枚举行的切割状态，然后贪心切割列

F - Knapsack for All Segments

魔改背包
关键在于状态的设计。
\(f_i\)：和为 \(i\) 的左端点选择方案数

#include <bits/stdc++.h>
#define ll long long

using namespace std;
const int N = 3005, mod = 998244353;
ll n, s, a[N], f[N], ans; //f[i]:和为i的左端点选择方案数

int main () {
    cin >> n >> s;
    for (int i = 1; i <= n; i++)    cin >> a[i];
    for (int i = 1; i <= n; i++) {
        f[s] = 0; //因为上次统计了fs所以要清空防止重复计算
        for (int j = s; j >= a[i]; j--) { //01背包
            (f[j] += f[j-a[i]]) %= mod;
        }
        (f[a[i]] += i) %= mod; //1-i共i种左端点选择
        (ans += (n - i + 1) * f[s]) %= mod; //乘上右端点贡献
    }
    cout << ans;
}