Nine degrees OJ 1534 K small number in the array

Topic 1534: Small number of k in an array

time limit:2 seconds

Memory limit:128 MB

Special question: No

submitted:1524

Resolution:307

topic Description:

given two arrays of integers a and B. We add the element 22 in a and B to get the array C.
For example A is [1,2],b] [3,4]. Then the array c that is added by element 22 in A and B is [4,5,5,6]. The
now gives you arrays A and B, which are the numbers in the array C that are added by A and B 22, and the small number of K.

Input:

The input may contain multiple test cases.
For each test case, enter the first behavior of three integers m,n, K (1<=m,n<=100000, 1<= k <= n *m): The n,m represents the length of the array A and b that will be entered.
The next two lines, respectively, are the number of M and N, representing the elements in arrays A and B. The array element range is [0,1e9].

Output:

corresponding to each test case,
Outputs the small number of k in the array C obtained by adding the element 22 in a and B.

Sample input:

2 2 31 23 43 3 41 2 73 4 5

Sample output:

56

Source:

Google face questions

True, Google interview problem really difficult, (⊙o⊙) ...

#include <stdio.h> #include <stdlib.h>long long m, N;long long K;long long a[100000], B[100000];long long i;int Compare (const void * p, const void * q) {return * (Long long *) P-* (Long long *) Q;} Long long cal (Long long mid) {//This    Although and Cal2 have the same output, but do a lot of useless computation, will time out long I, J;    A long long cnt = 0;    j=0;        for (I=0;i<m;++i) {j=0;//is initialized to 0 each time, restart the count while (J<n&&a[i]+b[j]<=mid) ++j;    Cnt+=j; } return CNT;} Long Long Cal2 (long long mid) {/* This function can save unnecessary calculation value, a small to large traversal, B from large to small traversal can, if a[i]+b[j]<=mid, then at this time J for i+1 speaking, the scope is still some big, continue to decrement J , then, from start to finish, J only needs to be from n-1 to 0.    */Long long I, J;    A long long cnt = 0;    j = n-1;        For (i=0, i<m; ++i) {while (j>=0 && a[i]+b[j]>mid)--j;    CNT + = (j + 1); } return CNT;}    Long Long findkth (long long k) {long long min = a[0] + b[0];    Long Long max = a[m-1] + b[n-1];    Long long mid;     Long long ans; while (min <= max) {mid = (Max-min)/2 + min;        Long Long B=cal2 (mid);        Long Long b=cal (mid);//useless computation too much, will time out if (k <= b) {max = mid-1;    } else min = mid + 1; } return min;}            int main (void) {while (scanf ("%lld%lld%lld", &m, &n, &k)! = EOF) {for (i=0; i<m; ++i)        scanf ("%lld", &a[i]);        for (i=0; i<n; ++i) scanf ("%lld", &b[i]);        Qsort (A, M, sizeof (long long), compare);        Qsort (B, N, sizeof (long long), compare);    printf ("%lld\n", findkth (k)); } return 0;} /************************************************************** problem:1534 User:kirchhoff language:c Resu lt:accepted time:960 Ms Memory:3256 kb****************************************************************/

Nine degrees OJ 1534 K small number in the array