2-13. Median of two ordered sequences (25) (zjupat)

Link: http://pat.zju.edu.cn/contests/ds/2-13

It is known that there are two unequal non-descending sequence S1 and S2. The design function calculates the median of the union of S1 and S2. Sequential sequence A0, A1... The median of An-1 refers to the value of a (N-1)/2, that is, the number of [(n + 1)/2] (A0 is 1st ).

Input format description:

The input is divided into three rows. The first row shows the common length of the sequence N (0 <n <= 1st). Then, each row enters the information of a sequence, that is, N non-descending integers. Numbers are separated by spaces.

Output format description:

Output the median of the Union sequence of two input sequences in one row.

Sample input and output:

Serial number	Input	Output
1	51 3 5 7 92 3 4 5 6	4
2	6-100 -10 1 1 1 1-50 0 2 3 4 5	1
3	31 2 34 5 6	3
4	34 5 61 2 3	3
5	121	1

PS:

This question is full of tears. I blame myself for my own IQ. When I saw the Union in the question, I took a great deal and then ...... In the last case, WA won't be able to pass the discussion! I used various methods ............

The Code is as follows:

# Include <cstdio> # include <cstring> const int maxn = 100017; int A [maxn], B [maxn], C [maxn]; int main () {int N; while (~ Scanf ("% d", & N) {for (INT I = 0; I <n; I ++) {scanf ("% d ", & A [I]) ;}for (INT I = 0; I <n; I ++) {scanf ("% d", & B [I]);} int L = 0; int I = 0, j = 0; while (I <n & J <n) {if (a [I] = B [J]) {c [L ++] = A [I]; C [L ++] = B [J]; // This line does not start with wa to vomit I ++, j ++;} if (a [I] <B [J]) {C [L ++] = A [I]; I ++ ;} else {C [L ++] = B [J]; j ++ ;}} while (I <n) {C [L ++] = A [I]; I ++;} while (j <n) {C [L ++] = B [J]; j ++;} printf ("% d \ n ", c [L-1)/2]);} return 0 ;}

2-13. Median of two ordered sequences (25) (zjupat)