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\n", 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\n", 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\n", 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\n", 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\n", (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\n", 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\n", (double)r);
}
int main()
{
input();
solve();
return 0;
}
来源：https://www.cnblogs.com/lcez56jsy/p/10813792.html