标签：lin ecif put his ant note define 规则 line
Array of integers is unimodal, if:
The first block (increasing) and the last block (decreasing) may be absent. It is allowed that both of this blocks are absent.
For example, the following three arrays are unimodal: [5,?7,?11,?11,?2,?1], [4,?4,?2], [7], but the following three are not unimodal: [5,?5,?6,?6,?1], [1,?2,?1,?2], [4,?5,?5,?6].
Write a program that checks if an array is unimodal.
The first line contains integer n (1?≤?n?≤?100) — the number of elements in the array.
The second line contains n integers a1,?a2,?...,?an (1?≤?ai?≤?1?000) — the elements of the array.
Print "YES" if the given array is unimodal. Otherwise, print "NO".
You can output each letter in any case (upper or lower).
6
1 5 5 5 4 2
YES
5
10 20 30 20 10
YES
4
1 2 1 2
NO
7
3 3 3 3 3 3 3
YES
In the first example the array is unimodal, because it is strictly increasing in the beginning (from position 1 to position 2, inclusively), that it is constant (from position 2 to position 4, inclusively) and then it is strictly decreasing (from position 4 to position 6, inclusively).
1 /* 2 题目大意：有一个序列 3 如果满足左部严格上升（可以为空），中间恒定，右部严格下降（可以为空） 4 那么就输出YES，否则输出NO 5 解题思路：按照规则走一遍，如果没走完，那么就NO，否则YES。 6 */ 7 #include <iostream> 8 using namespace std; 9 const int MAXN=105; 10 const int INF=0x3f3f3f3f; 11 int a[MAXN]; 12 int main(){ 13 int n; 14 while(cin>>n){ 15 for(int i=1;i<=n;i++) 16 cin>>a[i]; 17 a[n+1]=INF; 18 int p=2; 19 while(a[p]>a[p-1]) p++; 20 while(a [p]==a[p-1]) p++; 21 while(a[p]<a[p-1]) p++; 22 if(p<=n) cout<<"NO"<<endl; 23 else cout<<"YES"<<endl; 24 } 25 return 0; 26 }
1 #include <iostream> 2 #define N 105 3 using namespace std; 4 int n,maxn=1,a[N]; 5 bool check(){ 6 for(int i=1;i<=n;i++){ 7 if(a[i]==maxn){ 8 for(int j=i-1;j>=1;j--) 9 if(a[j]>=a[j+1]) return false; 10 while(a[i]==maxn&&i<=n) 11 i++; 12 for(int j=i;j<=n;j++) 13 if(a[j-1]<=a[j]||a[j]>=maxn) 14 return false; 15 break; 16 } 17 } 18 return true; 19 } 20 int main(){ 21 cin>>n; 22 for(int i=1;i<=n;i++) 23 cin>>a[i],maxn=max(a[i],maxn); 24 if(check()) 25 cout<<"YES"<<endl; 26 else cout<<"NO"<<endl; 27 return 0; 28 }
标签：lin ecif put his ant note define 规则 line
原文地址：http://www.cnblogs.com/z-712/p/7307935.html
