「Fortuna OJ」Feb 16th – Group B 解题报告

A – Binary & B – Chess

傻逼题,不分析。

C – Sum

这道题非常的有趣(精心调参之后随机化程序可以拿 90 分),正解非常的巧妙。我们可以把问题化为求\(min\{ S_{i,j} \mod P, K \leq S_{i,j} \mod P\}\)。我们先用\(O(n)\)的时间来进行前缀和的预处理,之后我们倒序处理前缀和,从后往前加入 set。在加入 set 的过程中二分查找\(s[i]+k\)和\(s[i]+k-p\)这两个目标最优解,之后记录答案即可,非常的巧妙。

// C.cpp
#include <bits/stdc++.h>
#define ll long long
using namespace std;
const int MAX_N = 100010;
int n, k, p, ans = 0x7f7f7f7f, s[MAX_N];
set<int> st;
int main()
{
    scanf("%d%d%d", &n, &k, &p);
    for (int i = 1; i <= n; i++)
        scanf("%d", &s[i]), (s[i] += s[i - 1]) %= p;
    set<int>::iterator it;
    st.insert(s[n]);
    for (int i = n - 1; i >= 1; i--)
    {
        it = st.lower_bound(s[i] + k);
        if (it != st.end())
            ans = min(*it - s[i], ans);
        it = st.lower_bound(s[i] + k - p);
        if (it != st.end() && *it < s[i])
            ans = min(*it - s[i] + p, ans);
        st.insert(s[i]);
    }
    printf("%d", ans);
    return 0;
}

D – Jail

哦,傻逼题。——crazydave

这道题算是套路吧,用状态压缩枚举每一维符号的状态,求出最大最小值,最大值减去最小值(最小值代表着状态相反的曼哈顿贡献)即可。

// D.cpp
#include <bits/stdc++.h>
using namespace std;
const int MAX_N = 1000010;
int n, d, info[MAX_N][6], st[10], ans;
void calc(int stat)
{
    int mn = (1 << 29), mx = -mn;
    for (int i = 1; i <= d; i++, stat >>= 1)
        st[i] = stat & 1;
    for (int i = 1; i <= n; i++)
    {
        int acc = 0;
        for (int j = 1; j <= d; j++)
            acc += (st[j] == 1) ? info[i][j] : -info[i][j];
        if (acc != 0)
            mn = min(mn, acc), mx = max(mx, acc);
    }
    ans = max(mx - mn, ans);
}
int main()
{
    scanf("%d%d", &n, &d);
    for (int i = 1; i <= n; i++)
        for (int j = 1; j <= d; j++)
            scanf("%d", &info[i][j]);
    for (int i = 0; i < (1 << d); i++)
        calc(i);
    printf("%d", ans);
    return 0;
}

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