Everything s Okay
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Everything s Okay
时间限制:1000 ms | 内存限制:65535 KB
难度:3
- 描述
输入N个数,M次查询。
每次查询给出一个数x。
要求:每次查询输出前x个数中第i小的数。(i为第i次查询)
你可以假设M <= N,Xi <= Xi+1 <= Xi+2 <= ……. <= Xm (Xm <= N).
- 输入
- Line0:T
Line1: N,M
Line2…LineN+1:num1,......,numN
LineN+2…LineN+2+M:x1,……,xM
N < 30000, num < 2000000000 - 输出
- 每次查询输出前i小的数,单独一行。
详细格式请参考样例。 - 样例输入
17 43 1 -4 2 8 -1000 21 2 6 6
- 样例输出
3312
#include<stdio.h>#include<stdlib.h>int com(const void *a,const void *b){return *(int *)a-*(int *)b;}int main(void){int t,n,m,k,i;int s[30001];scanf("%d",&t);while(t--){scanf("%d%d",&n,&m);for(i=0;i<n;i++)scanf("%d",&s[i]);for(i=0;i<m;i++){scanf("%d",&k);qsort(s,k,sizeof(s[0]),com);printf("%d\n",s[i]);}}return 0;}
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