hdu 4928 Series 2

Series 2

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 499    Accepted Submission(s): 110

Problem Description

Let A be an integral series {A₁, A₂, . . . , A_n}. 

The zero-order series of A is A itself.

The first-order series of A is {B₁, B₂, . . . , B_n-1, where B_i = A_i+1 - A_i.

The ith-order series of A is the first-order series of its (i - 1)th-order series (2<=i<=n - 1).

We say A is monotonic iff A₁<=A₂<=. . . <=A_n or A₁>= A₂ >=. . . >= A_n.

A is kth-order monotonic iff all ith-order series (0<=i<=k) are monotonic, and (k + 1)th-order are not.

Specially, if the zero-order series of A is not monotonic, then A is named ugly series. If all ith-order (0<=i<=n - 1) series of A are monotonic, then A is a nice series.

Given A, determine whether it’s a ugly series or nice series. If both are not, determine k.

 

Input

The input consists of several test cases. The first line of input gives the number of test cases T (T<=50).

For each test case:
The first line contains a single integer n(1<=n<=10⁵), which denotes the length of series A.
The second line consists of n integers, describing A₁, A₂, . . . , A_n. (0<=|A_i|<=2⁶⁰)

 

Output

For each test case, output either ugly series, nice series or a single integer k.

 

Sample Input

431 3 241 4 6 741 3 4 75-1 0 3 11 29

 

Sample Output

ugly series nice series 0nice series

 

Author

BUPT

 

Source

2014 Multi-University Training Contest 6

 

题解看这：http://weibo.com/p/1005051809706204/weibo?from=page_100505_home&wvr=5.1&mod=weibomore#3740344145782887

看完题解后发现这道题只要缩点，就不会超时。复杂度为NlogV

代码：

#include <iostream>#include <algorithm>#include <cstdio>#include <string>#include <cstring>#include <cmath>#include <vector>#include <list>#include <map>#include <set>#include <deque>#include <queue>#include <stack>#include <bitset>#include <functional>#include <sstream>#include <iomanip>#include <cmath>#include <cstdlib>#include <ctime>#pragma comment(linker, "/STACK:102400000,102400000")typedef long long ll;#define INF 1e9#define maxn 100005#define maxm 100005+10const ll mod = 1e10+7;#define eps 1e-7#define PI acos(-1.0)#define rep(i,n) for(int i=0;i<n;i++)#define rep1(i,n) for(int i=1;i<=n;i++)#define scan(n) scanf("%d",&n)#define scan2(n,m) scanf("%d%d",&n,&m)#define scans(s) scanf("%s",s);#define ini(a) memset(a,0,sizeof(a))#define out(n) printf("%d\n",n)using namespace std;#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1ll a[2][maxn];  int main(){#ifndef ONLINE_JUDGEfreopen("in.txt","r",stdin);//  freopen("out.txt","w",stdout);#endifint T;int n;cin>>T;while(T--){scanf("%d",&n);ini(a);rep1(i,n) scanf("%I64d",&a[0][i]);int s = 0,start = 1,end = n;bool finish = 1;rep(i,n){while(start < end && a[s][start] == 0) start++; //缩点start = max(1 ,start-1);while(end > start && a[s][end] == 0) end--;end = min(end + 1,n - i);if(start >= end) break;int inc = 1, dec = 1;for(int j = start;j < end; j++){if(a[s][j] > a[s][j+1]) inc = 0;else if(a[s][j] < a[s][j+1]) dec = 0;}finish = inc | dec;if(!finish){if(i == 0) puts("ugly series");else printf("%d\n",i - 1);break;}for(int j = start;j < end; j++){a[s^1][j] = a[s][j+1] - a[s][j];}s ^= 1;}if(finish)puts("nice series");}return 0;}