5.1培训考试二

今天小姐姐给了我们说良心很良心，说毒瘤很毒瘤的一套题（小姐姐说是可以AK的）

老师一定刚去看了复联4

据说这是道水题，一个set模拟就好了

来我们看看小姐姐的标程（仅限c++11）

#include <cstdio>
#include <set>
#include <algorithm>
using namespace std;

class People {
    int id, force;
    People() {}
    People(int i, int f): id(i), force(f) {}//构造函数，意思是把id赋值为i,把force赋值为f
};

bool operator <(People x, People y) {
    return x.force < y.force;
}

int n;
set <People> s;


int main() {
    scanf("%d", &n);
    s.insert(People(1, 1000000000));
    for (int i = 1, x, y; i <= n; ++i) {
        scanf("%d%d", &x, &y);
        auto it1 = s.upper_bound(People(x, y));//找到第一个大于当前能力的人，这里返回的是指针
        if (it1 == s.begin()) printf("%d %d
", x, it1->id);//如果是第一个人，就让他和超人比
        else {
            auto it3 = it1;
            auto it2 = --it3;//上面的数的前一个数的指针
            int a1 = abs(it1->force - y);//与第一个大于其的数的能力值比较
            int a2 = abs(it2->force - y);//和it2指向的数的能力值比较
            if (a2 <= a1) it1 = it2;//选取最小的（这里依旧是指针）
            printf("%d %d
", x, it1->id);//把那个数输出来
        }
        s.insert(People(x, y));//比较完后加入set
    }
    return 0;
}

bat-man:

60pts:

O(n^2)暴力枚举

80pts:

st表讲解（里面的程序没有加小姐姐的神奇输入输出，但主要部分就是里面的程序辣）

加了小姐姐的输入输出：

#include <cstdio>
#include <ctime>
#include <cstdlib>
#include <algorithm>
#include <cmath>
using namespace std;

const int N = 1e7 + 5;

long long st[1<<23][23];
int n, m, a[N], ans[N];
int gen, cute1, cute2;
int number() {
    gen = (1LL * gen * cute1) ^ cute2;
    return (gen & (n - 1)) + 1;
}
int seach(int l,int r)
{  int t=log((double)(r-l+1))/log(2.0);
   return max(st[l][t],st[r-(1<<t)+1][t]);
}
int main() {
    scanf("%d%d", &n, &m);
    scanf("%d%d%d", &gen, &cute1, &cute2);
     for (int i = 1; i <= n; ++i)//小姐姐的输入输出
        a[i] = number(); 
    for(int i=1;i<=n;i++)
      st[i][0]=a[i];
      double sx=log((double) n)/log(2.0);
    for(int j=1;j<=sx;j++)//剩下的就是st表了qwq
     { for(int i=1;(i+(1<<j))-1<=n;i++)
      {st[i][j]=max(st[i][j-1],st[i+(1<<(j-1))][j-1]);
      }
     }
    for (int i = 1; i <= m; ++i) {
        int l = number(), r = number();
        if (l > r) swap(l, r);
         ans[i]=seach(l,r);
    }
    int sum = 0;
    
    for (int i = 1; i <= m; ++i)
        sum = (1LL * sum * cute1 + ans[i]) % cute2;
    printf("%d
", sum);
}

100pts:

我们使用并查集。

 小姐姐的代码（看不懂QwQ嘤~）：

#include <cstdio>
#include <ctime>
#include <cstdlib>
#include <algorithm>
using namespace std;

const int N = 1e7 + 5;

int n, m;
int gen, cute1, cute2;
int number() {
    gen = (1LL * gen * cute1) ^ cute2;
    return (gen & (n - 1)) + 1;
}


int hd[N], nxt[N], id[N], to[N], cnt;
int ans[N], a[N], p[N], q[N];

int add(int x, int y, int i) {
    ++cnt;
    nxt[cnt] = hd[x];
    to[cnt] = y;
    id[cnt] = i;
    hd[x] = cnt;
}


int getfa(int x, int y) {
    int fa = x;
    for (int i = x; i; i = p[i])
        if (p[i] < y || p[i] == i) {
            fa = i;
            break;
        }
    for (int j, i = x; i != fa; i = j) {
        j = p[i], p[i] = fa;
    }
    return fa;
}

int main() {
    scanf("%d%d", &n, &m);
    scanf("%d%d%d", &gen, &cute1, &cute2);

    for (int i = 1; i <= n; ++i)
        a[i] = number();
    for (int i = 1; i <= m; ++i) {
        int l = number(), r = number();
        if (l > r) swap(l, r);
        add(l, r, i);
    }
    double t1;
    fprintf(stderr, "%lf
", t1 = (double)clock()/CLOCKS_PER_SEC);

    int ind = 0;
    for (int i = 1; i <= n; ++i) {
        while (ind && a[q[ind]] <= a[i]) --ind;
        if (ind) p[i] = q[ind];
        else p[i] = i;
        q[++ind] = i;
    }

    for (int i = n; i; --i) {
        for (int j = hd[i]; j; j = nxt[j])
            ans[id[j]] = a[getfa(to[j], i)];
    }


    fprintf(stderr, "%lf
", (double)clock()/CLOCKS_PER_SEC - t1);

    int sum = 0;
    for (int i = 1; i <= m; ++i)
        sum = (1LL * sum * cute1 + ans[i]) % cute2;
    printf("%d
", sum);
}

 

40pts：

暴力模拟（胡说，我暴力只有10分）

100pts:

二分答案+check

我们枚举一个答案t,将原来的所有的a[i]都减去t.

那我们看有多少个区间的和等于0，如果刚好有k个，那t就是答案。

如何看区间的和？

既然我们二分，就是为了快，所以肯定不能O（n）暴力算是吧。

我们使用前缀和与归并排序来进行这波操作。

扯不下去了233

具体的看小姐姐的代码吧

看不懂233qwq

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <cctype>
#include <climits>
#include <cassert>
#include <ctime>
#include <iostream>
#include <fstream>
#include <algorithm>
#include <functional>
#include <string>

#define x first
#define y second
#define MP std::make_pair
#define DEBUG(...) fprintf(stderr, __VA_ARGS__)
#ifdef __linux__
#define getchar getchar_unlocked
#define putchar putchar_unlocked
#endif
#pragma GCC optimize("O3")

typedef long long LL;
typedef std::pair<int, int> Pii;

const int oo = 0x3f3f3f3f;

template<typename T> inline bool chkmax(T &a, T b) { return a < b ? a = b, true : false; }
template<typename T> inline bool chkmin(T &a, T b) { return b < a ? a = b, true : false; }
std::string procStatus()
{
    std::ifstream t("/proc/self/status");
    return std::string(std::istreambuf_iterator<char>(t), std::istreambuf_iterator<char>());
}
template<typename T> T read(T &x)
{
    int f = 1;
    char ch = getchar();
    for (; !isdigit(ch); ch = getchar())
        if (ch == '-')
            f = -1;
    for (x = 0; isdigit(ch); ch = getchar())
        x = 10 * x + ch - '0';
    return x *= f;
}
template<typename T> void write(T x)
{
    if (x == 0) {
        putchar('0');
        return;
    }
    if (x < 0) {
        putchar('-');
        x = -x;
    }
    static char s[20];
    int top = 0;
    for (; x; x /= 10)
        s[++top] = x % 10 + '0';
    while (top)
        putchar(s[top--]);
}

const int MAXN = 1e5 + 5;

int N;
LL K;
int A[MAXN];

LL mergeSort(long double a[], register int n)
{
    if (n <= 1)
        return 0;
    register int mid = n >> 1;
    register LL ret = mergeSort(a, mid) + mergeSort(a + mid, n - mid);

    static long double t[MAXN];
    register int p = 0, q = mid, tot = 0;
    while (p < mid || q < n) {
        if (p < mid && (q == n || a[p] <= a[q])) {
            t[tot++] = a[p++];
            ret += q - mid;
        } else
            t[tot++] = a[q++];
    }
    memcpy(a, t, sizeof(*a) * n);
    return ret;
}

bool check(register long double mid)
{
    static long double sum[MAXN];

    sum[0] = 0;
    for (int i = 1; i <= N; ++i) {
        sum[i] = sum[i - 1] + A[i] - mid;
    }

    return mergeSort(sum, N + 1) >= K;
}

void input()
{
    read(N); read(K);
    for (int i = 1; i <= N; ++i) {
        read(A[i]);
    }
}

void solve()
{
    long double l = *std::min_element(A + 1, A + N + 1), r = *std::max_element(A + 1, A + N + 1);
    while (clock() < 0.9 * CLOCKS_PER_SEC) {
        long double mid = (l + r) / 2;
        (check(mid) ? r : l) = mid;
    }
    printf("%.4f
", (double)r);
}

int main()
{
    input();
    solve();

    return 0;
}