题目链接：https://www.51nod.com/onlineJudge/questionCode.html#!problemId=1218

题解：先要确定这些点是不是属于最长递增序列然后再确定这些数在最长递增序列中出现的次数，如果大于1次显然是可能出现只出现1次肯定是必然出现。那么就是怎么判断是不是属于最长递增序列，这个只要顺着求一下最长递增标一下该点属于长度几然后再逆着求一下最长递减标一下该点属于长度几如果两个下标之和等于最长长度+1那么该点就属于最长递增序列，然后就是求1～len（len表示最长的长度）中各个长度出现的次数就行。

#include <iostream>
#include <cstring>
#include <cstdio>
#include <vector>
#define inf 0X3f3f3f3f
using namespace std;
const int M = 5e4 + 10;
int a[M] , b[M] , dpa[M] , dpb[M] , vis[M];
bool vs[M];
int binsearch(int l , int r , int num) {int mid = (l + r) >> 1;int ans = 0;while(l <= r) {mid = (l + r) >> 1;if(b[mid] > num) r = mid - 1;else {ans = mid;l = mid + 1;}}return ans;
}
int binsearch2(int l , int r , int num) {int mid = (l + r) >> 1;int ans = 0;while(l <= r) {mid = (l + r) >> 1;if(b[mid] < num) r = mid - 1;else {ans = mid;l = mid + 1;}}return ans;
}
int main() {int n;scanf("%d" , &n);for(int i = 1 ; i <= n ; i++) scanf("%d" , &a[i]);int len = 0;b[0] = -1;for(int i = 1 ; i <= n ; i++) {if(a[i] > b[len]) {len++;b[len] = a[i];dpa[i] = len;continue;}else {int pos = binsearch(1 , len , a[i]);b[pos + 1] = min(b[pos + 1] , a[i]);dpa[i] = pos + 1;}}int len2 = 0;memset(b , inf , sizeof(b));for(int i = n ; i >= 1 ; i--) {if(a[i] < b[len2]) {len2++;b[len2] = a[i];dpb[i] = len2;}else {int pos = binsearch2(1 , len2 , a[i]);b[pos + 1] = max(b[pos + 1] , a[i]);dpb[i] = pos + 1;}}memset(vs , false , sizeof(vs));for(int i = 1 ; i <= n ; i++) {if(dpa[i] + dpb[i] == len + 1) {vis[dpa[i]]++;vs[i] = true;}}printf("A:");for(int i = 1 ; i <= n ; i++) {if(vis[dpa[i]] > 1 && vs[i]) printf("%d " , i);}printf("\n");printf("B:");for(int i = 1 ; i <= n ; i++) {if(vis[dpa[i]] == 1 && vs[i]) printf("%d " , i);}printf("\n");return 0;
}