【HDU6197 2017 ACM ICPC Asia Regional Shenyang Online D】【LIS 最长不下降序列】array array array 数列删除恰好K个数，使得恰好

array array array

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 233    Accepted Submission(s): 141


Problem Description

One day, Kaitou Kiddo had stolen a priceless diamond ring. But detective Conan blocked Kiddo's path to escape from the museum. But Kiddo didn't want to give it back. So, Kiddo asked Conan a question. If Conan could give a right answer, Kiddo would return the
ring to the museum. 

Kiddo: "I have an array A and
a number k,
if you can choose exactly k elements
from A and
erase them, then the remaining array is in non-increasing order or non-decreasing order, we say A is
a magic array. Now I want you to tell me whether A is
a magic array. " Conan: "emmmmm..." Now, Conan seems to be in trouble, can you help him?

 

Input

The first line contains an integer T indicating the total number of test cases. Each test case starts with two integers n and k in
one line, then one line with n integers: A1,A2…An.
1≤T≤20
1≤n≤105
0≤k≤n
1≤Ai≤105

 

Output

For each test case, please output "A is a magic array." if it is a magic array. Otherwise, output "A is not a magic array." (without quotes).

 

Sample Input

3
4 1
1 4 3 7
5 2
4 1 3 1 2
6 1
1 4 3 5 4 6

 

Sample Output

A is a magic array.
A is a magic array.
A is not a magic array.

 

Source

2017 ACM/ICPC Asia Regional Shenyang Online

 

#include<stdio.h>
#include<iostream>
#include<string.h>
#include<string>
#include<ctype.h>
#include<math.h>
#include<set>
#include<map>
#include<vector>
#include<queue>
#include<bitset>
#include<algorithm>
#include<time.h>
using namespace std;
void fre() { freopen("c://test//input.in", "r", stdin); freopen("c://test//output.out", "w", stdout); }
#define MS(x, y) memset(x, y, sizeof(x))
#define ls o<<1
#define rs o<<1|1
typedef long long LL;
typedef unsigned long long UL;
typedef unsigned int UI;
template <class T1, class T2>inline void gmax(T1 &a, T2 b) { if (b > a)a = b; }
template <class T1, class T2>inline void gmin(T1 &a, T2 b) { if (b < a)a = b; }
const int N = 1010, M = 0, Z = 1e9 + 7, inf = 0x3f3f3f3f;
template <class T1, class T2>inline void gadd(T1 &a, T2 b) { a = (a + b) % Z; }
int casenum, casei;
int K;
int fib
;
bitset<100010>f[24];
namespace DU
{
#define rep(i,a,n) for (int i=a;i<n;i++)
#define per(i,a,n) for (int i=n-1;i>=a;i--)
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define SZ(x) ((int)(x).size())
typedef vector<int> VI;
typedef long long ll;
typedef pair<int, int> PII;
const ll mod = 998244353;
ll powmod(ll a, ll b)
{
ll res = 1; a %= mod;
//assert(b >= 0);
for (; b; b >>= 1) { if (b & 1)res = res*a%mod; a = a*a%mod; }
return res;
}
// head

int _, n;
namespace linear_seq {
const int N = 10010;
ll res
, base
, _c
, _md
;

vector<int> Md;
void mul(ll *a, ll *b, int k) {
rep(i, 0, k + k) _c[i] = 0;
rep(i, 0, k) if (a[i]) rep(j, 0, k) _c[i + j] = (_c[i + j] + a[i] * b[j]) % mod;
for (int i = k + k - 1; i >= k; i--) if (_c[i])
rep(j, 0, SZ(Md)) _c[i - k + Md[j]] = (_c[i - k + Md[j]] - _c[i] * _md[Md[j]]) % mod;
rep(i, 0, k) a[i] = _c[i];
}
int solve(ll n, VI a, VI b) { // a 系数 b 初值 b[n+1]=a[0]*b
+...
// printf("%d\n",SZ(b));
ll ans = 0, pnt = 0;
int k = SZ(a);
//assert(SZ(a) == SZ(b));
rep(i, 0, k) _md[k - 1 - i] = -a[i]; _md[k] = 1;
Md.clear();
rep(i, 0, k) if (_md[i] != 0) Md.push_back(i);
rep(i, 0, k) res[i] = base[i] = 0;
res[0] = 1;
while ((1ll << pnt) <= n) pnt++;
for (int p = pnt; p >= 0; p--) {
mul(res, res, k);
if ((n >> p) & 1) {
for (int i = k - 1; i >= 0; i--) res[i + 1] = res[i]; res[0] = 0;
rep(j, 0, SZ(Md)) res[Md[j]] = (res[Md[j]] - res[k] * _md[Md[j]]) % mod;
}
}
rep(i, 0, k) ans = (ans + res[i] * b[i]) % mod;
if (ans<0) ans += mod;
return ans;
}
VI BM(VI s) {
VI C(1, 1), B(1, 1);
int L = 0, m = 1, b = 1;
rep(n, 0, SZ(s)) {
ll d = 0;
rep(i, 0, L + 1) d = (d + (ll)C[i] * s[n - i]) % mod;
if (d == 0) ++m;
else if (2 * L <= n) {
VI T = C;
ll c = mod - d*powmod(b, mod - 2) % mod;
while (SZ(C)<SZ(B) + m) C.pb(0);
rep(i, 0, SZ(B)) C[i + m] = (C[i + m] + c*B[i]) % mod;
L = n + 1 - L; B = T; b = d; m = 1;
}
else {
ll c = mod - d*powmod(b, mod - 2) % mod;
while (SZ(C)<SZ(B) + m) C.pb(0);
rep(i, 0, SZ(B)) C[i + m] = (C[i + m] + c*B[i]) % mod;
++m;
}
}
return C;
}
int gao(VI a, ll n) {
VI c = BM(a);
c.erase(c.begin());
rep(i, 0, SZ(c)) c[i] = (mod - c[i]) % mod;
return solve(n, c, VI(a.begin(), a.begin() + SZ(c)));
}
};

int solve(int n)
{
return linear_seq::gao(VI{ 1,4,12,33,88,232,609,1596,4180,10945 }, n);
}
}
void table()
{
fib[0] = 0; fib[1] = 1;
for (int i = 2; i <= 30; ++i)
{
fib[i] = fib[i - 1] + fib[i - 2];
}
f[0][0] = 1;
for (int i = 0; i <= 20; ++i)
{
for (int k = 0; k <= 30; ++k)
{
f[i + 1] |= (f[i] << fib[k]);
}
for (int j = 1; j < 100000; ++j)if (!f[i][j])
{
printf("%d %d\n", i, j);
break;
}
}
}
int main()
{
table();
while(~scanf("%d", &K))
{
printf("%d\n", DU::solve(K));
}
return 0;
}
/*
【trick&&吐槽】

【题意】
输入一个K（1e9）范围的数，让你求出，无法用K个斐波那契数表示的最小整数。

【分析】

【时间复杂度&&优化】

【数据】
1 1 2 3 5 8 13 21 34 55 89
_ _ _ _ _
1 4 12 33 88 232 609

*/