Manthan Codefest 19 (div1+div2)

有点难的一场。

题目链接：https://codeforces.com/contest/1208

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A：

水题。

/* 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 t;

int main() {
    scanf("%d",&t);
    while (t--){
        ll a,b,c,n; scanf("%lld%lld%lld",&a,&b,&n);
        c=a^b;
        if (n%3==0) printf("%lld\n",a);
        else if (n%3==1) printf("%lld\n",b);
        else printf("%lld\n",c);
    }
    return 0;
}

B：

这题很容易想到一种看似很对其实有bug的O(n)做法：维护双指针，往左往右扫，确定最小区间。

比较靠谱的做法是n方枚举区间，并维护不同元素个数，统计答案。

/* 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=2e3+10;
int n,a[maxn],mul=0,ans=int_inf;
map<int,int>m;

int main() {
    m.clear();
    scanf("%d",&n);
    for (int i=1;i<=n;i++){
        scanf("%d",&a[i]);
        m[a[i]]++;
    }
    for (auto i:m)
        if (i.second>1) mul++;
    if (!mul) return puts("0"),0;
    for (int i=1;i<=n;i++){
        int ret=-1,tmp=mul;
        for (int j=i;j<=n;j++){
            ret=j;
            if (--m[a[j]]==1) tmp--;
            if (!tmp){
                ans=min(ans,j-i+1);
                break;
            }
        }
        for (int j=i;j<=ret;j++) m[a[j]]++;
    }
    printf("%d\n",ans);
    return 0;
}

这题还有一个很好的O(nlogn)做法：二分尺取。

#include <bits/stdc++.h>
#define int int64_t
using namespace std;
int n;

bool judge(vector<int> &v, int mid) {
    unordered_map<int, int> m;
    unordered_set<int> s; // counter > 1
    for (int i = 0; i < n; ++i) {
        m[v[i]]++; // counter
        if (m[v[i]] > 1) s.emplace(v[i]);
    }
    for (int i = 0; i < mid; ++i) { // pick all from [0,mid)
        m[v[i]]--;
        if (m[v[i]] < 2) s.erase(v[i]);
    }
    if (s.empty()) return true; // all were picked
    int beg = 0, end = mid;
    while (end < n) {
        m[v[beg]]++;
        if (m[v[beg]] > 1) s.emplace(v[beg]);
        m[v[end]]--;
        if (m[v[end]] < 2) s.erase(v[end]);
        if (s.empty()) return true;
        end++;
        beg++;
    }
    return false;
}

signed main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    cin >> n;
    vector<int> v(n);
    for (int &i : v) cin >> i;
    int lo = 0, hi = n;
    while (lo < hi) { // 二分数量
        int mid = (lo + hi) / 2;
        if (judge(v, mid)) hi = mid - 1;
        else lo = mid + 1;
    }
    if (lo > 0 and judge(v, lo - 1)) cout << lo - 1;
    else if (judge(v, lo)) cout << lo;
    else cout << max(0ll, lo + 1);
    return 0;
}

C：

一个很有意思的xor幻方构造题。留意到给定的数n==4k，所以最后矩阵元素个数为16k^2，那么就可以把整个矩阵分为k^2个含有16个元素的子矩阵，第一个子矩阵从0~15依次分布，第二个子矩阵从16~31依次分布……这样做必然满足题意。

/* 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 = 1010;
int n, a[maxn][maxn];

int main() {
    scanf("%d", &n);
    int cnt = 0;
    for (int i = 0; i < 4; i++)
        for (int j = 0; j < 4; j++) a[i][j] = cnt++;
    for (int i = 0; i < n / 4; i++)
        for (int j = 0; j < n / 4; j++)
            if (i || j) {
                for (int p = 0; p < 4; p++)
                    for (int q = 0; q < 4; q++)
                        a[i * 4 + p][j * 4 + q] = cnt++;
            }
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < n; j++) printf("%d ", a[i][j]);
        puts("");
    }
    return 0;
}

D：

不停地找区间最小值(如果有多个最小值，找最靠后的)即可。可以线段树，也可以树状数组+二分。

/* 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 */

template <class T>
class BIT {
public:
    vector<T> _bit;
    int n;

    BIT(int _n): n(_n) {
        _bit.resize(n);
    }

    void add(int x, T v) {
        while (x < n) {
            _bit[x] += v;
            x |= (x + 1);
        }
    }

    T get(int x) {
        T ret{};
        while (x >= 0) {
            ret += _bit[x];
            x = (x & (x + 1)) - 1;
        }
        return ret;
    }
};

int main() {
    int n; scanf("%d", &n);
    vector<ll> s(n);
    for (int i = 0; i < n; i++) scanf("%lld", &s[i]);
    vector<int> a(n);
    BIT<ll> bit(n);
    for (int i = 0; i < n; i++) bit.add(i, i + 1);
    for (int i = n - 1; i >= 0; i--) {
        int low = 0, high = n - 1;
        while (low < high) {
            int mid = low + high >> 1;
            if (bit.get(mid) > s[i])
                high = mid;
            else low = mid + 1;
        }
        a[i] = low + 1;
        bit.add(low, -low - 1);
    }
    for (int i = 0; i < n; i++) {
        if (i) printf(" ");
        printf("%d", a[i]);
    }
    puts("");
    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 = 2e5 + 10;
struct Node {
    ll minn, tag;
} segt[maxn << 2];
int n, a[maxn];
ll p[maxn];

void maintain(int curpos) {
    segt[curpos].minn = min(segt[lson].minn, segt[rson].minn);
}

void pushdown(int curpos) {
    segt[lson].tag += segt[curpos].tag; segt[rson].tag += segt[curpos].tag;
    segt[lson].minn += segt[curpos].tag; segt[rson].minn += segt[curpos].tag;
    segt[curpos].tag = 0;
}

void build(int curpos, int curl, int curr) {
    if (curl == curr) {
        segt[curpos].minn = p[curl];
        return;
    }
    int mid = curl + curr >> 1;
    build(lson, curl, mid); build(rson, mid + 1, curr);
    maintain(curpos);
}

void update(int curpos, int curl, int curr, int ql, int qr, ll val) {
    if (ql <= curl && curr <= qr) {
        segt[curpos].tag += val;
        segt[curpos].minn += val;
        return;
    }
    int mid = curl + curr >> 1;
    if (ql <= mid) update(lson, curl, mid, ql, qr, val);
    if (mid < qr) update(rson, mid + 1, curr, ql, qr, val);
    maintain(curpos);
}

ll queryMin(int curpos, int curl, int curr, int ql, int qr) {
    if (ql <= curl && curr <= qr)
        return segt[curpos].minn;
    pushdown(curpos);
    int mid = curl + curr >> 1;
    ll ans = ll_inf;
    if (ql <= mid) ans = min(ans, queryMin(lson, curl, mid, ql, qr));
    if (mid < qr) ans = min(ans, queryMin(rson, mid + 1, curr, ql, qr));
    return ans;
}

ll queryLastZero(int curpos, int curl, int curr) {
    if (curl == curr)
        return curl;
    int mid = curl + curr >> 1;
    if (queryMin(1, 1, n, mid + 1, curr) == 0)
        return queryLastZero(rson, mid + 1, curr);
    else return queryLastZero(lson, curl, mid);
    maintain(curpos);
}

int main() {
    scanf("%d", &n);
    rep1(i, 1, n) scanf("%lld", &p[i]);
    build(1, 1, n);
    rep1(i, 1, n) {
        int pos = queryLastZero(1, 1, n);
        a[pos] = i;
        update(1, 1, n, pos + 1, n, -i);
        update(1, 1, n, pos, pos, ll_inf + i);
    }
    rep1(i, 1, n) printf("%d ", a[i]);
    puts("");
    return 0;
}