American group reviews 2017 Spring Recruit Programming Questions _ American Group

The programming question has the question number, should be randomly extracted. The main effect of the topic

Given a sequence, the length of its longest ascending subsequence is obtained. Idea-DP

Very simple DP, first to do the programming questions, the first quickly wrote an O (N2) solution then to do the previous question. Set Dp[i] Represents the length of the longest ascending sequence ending with the number I, and the state transition equation is: Dp[i]=max (Dp[j]) +1,j<i&&num[j]<num[i]

All done, found that there are half an hour, began to think O (NLOGN) solution, although I have forgotten how to write it, I remember that in the process of processing, the length of the ascending subsequence is subscript, and the last number of ascending subsequence is the value (same length, smaller value), and a monotonically increasing array can be constructed. Then you can use a second to find the length of the longest ascending subsequence of the last number that is less than the current number.

After proficiency, you don't need a DP array to code

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

int n, NUM, L, R, Mid, MX;
int Mn[100003];//mn[i] Represents the minimum value

int main () {
    while (1 = scanf ("%d", &n)) in the ascending subsequence of length i {
        mx = 0;
        Memset (MN, 0x3f, sizeof (MN));
        while (n--> 0) {
            scanf ("%d", &num);
            L = 1;
            r = mx;
            while (L <= R) {
                mid = (L + r) >> 1;
                if (Mn[mid] < num) {
                    L = mid + 1;
                }
                else {
                    r = mid-1;
                }
            }
            Mn[r + 1] = min (mn[r + 1], num);
            mx = max (mx, r + 1);
        }
        printf ("%d\n", MX);
    }
    return 0;
}

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