HZAU 网络赛 Sequence Number
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1205: Sequence Number
Time Limit: 1 Sec Memory Limit: 1280 MBSubmit: 869 Solved: 217
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Description
In Linear algebra, we have learned the definition of inversion number:
Assuming A is a ordered set with n numbers ( n > 1 ) which are different from each other. If exist positive integers i , j, ( 1 ≤ i < j ≤ n and A[i] > A[j]), <A[i], A[j]> is regarded as one of A’s inversions. The number of inversions is regarded as inversion number. Such as, inversions of array <2,3,8,6,1> are <2,1>, <3,1>, <8,1>, <8,6>, <6,1>,and the inversion number is 5.
Similarly, we define a new notion —— sequence number, If exist positive integers i, j, ( 1 ≤ i ≤ j ≤ n and A[i] <= A[j], <A[i], A[j]> is regarded as one of A’s sequence pair. The number of sequence pairs is regarded as sequence number. Define j – i as the length of the sequence pair.
Now, we wonder that the largest length S of all sequence pairs for a given array A.
Input
There are multiply test cases.
In each case, the first line is a number N(1<=N<=50000 ), indicates the size of the array, the 2th ~n+1th line are one number per line, indicates the element Ai (1<=Ai<=10^9) of the array.
Output
Output the answer S in one line for each case.
Sample Input
52 3 8 6 1
Sample Output
3
暴力(72ms)
#include <bits/stdc++.h>using namespace std;#define mst(a,b) memset((a),(b),sizeof(a))#define f(i,a,b) for(int i=(a);i<(b);++i)const int maxn = 50005;const int mod = 10000007;#define ll long long#define rush() int T;scanf("%d",&T);while(T--)int a[maxn];int n;int main(){ while(~scanf("%d",&n)) { bool flag=0; f(i,0,n) { scanf("%d",&a[i]); if(i!=0&&a[i]>a[i-1]) flag=true; } if(flag==0) //优化一 { printf("0\n"); continue; } int ans=0; int k=0; f(i,0,n) { k=max(k,i+1); //优化二 for(int j=k;j<n;j++) { if(a[j]>=a[i]&&j-i>ans) { ans=j-i; k=j; } } } printf("%d\n",ans); } return 0;}
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