【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

34 11 4 3 75 24 1 3 1 26 11 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|1typedef 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[N];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; }// headint _, n;namespace linear_seq {const int N = 10010;ll res[N], base[N], _c[N], _md[N];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[n]+...  //        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】1123581321345589        _       _       _       _       _14123388232609*/