题目描述 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.
输入描述 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.
输出描述 Output the answer S in one line for each case.
输入样例 5 2 3 8 6 1
输出样例 3
简单题，从两边向中间枚举，直到找出ai[j]>ai[i]即可
#include"iostream" using namespace std; int ai[60000]; int main() { int n; while(cin>>n) { for(int i=0;i<n;i++) cin>>ai[i]; for(int i=n-1;;i--) { for(int j=0;j+i<n;j++) { if(ai[i+j]>=ai[j]) { cout<<i<<endl; goto end; } } } end:; } return 0; }