Leha like all kinds of strange things. Recently he liked the function F(n, k). Consider all possible k-element subsets of the set [1, 2, ..., n]. For subset find minimal element in it. F(n, k) — mathematical expectation of the minimal element among all k-element subsets.
But only function does not interest him. He wants to do interesting things with it. Mom brought him two arrays A and B, each consists of m integers. For all i, j such that 1 ≤ i, j ≤ m the condition Ai ≥ Bj holds. Help Leha rearrange the numbers in the array A so that the sum is maximally possible, where A' is already rearranged array.
First line of input data contains single integer m (1 ≤ m ≤ 2·105) — length of arrays A and B.
Next line contains m integers a1, a2, ..., am (1 ≤ ai ≤ 109) — array A.
Next line contains m integers b1, b2, ..., bm (1 ≤ bi ≤ 109) — array B.
Output m integers a'1, a'2, ..., a'm — array A' which is permutation of the array A.
5
7 3 5 3 4
2 1 3 2 3
4 7 3 5 3
7
4 6 5 8 8 2 6
2 1 2 2 1 1 2
2 6 4 5 8 8 6
这道题一开始我也没看懂,然后就一直对着数据分析.
搞清楚了是啥意思.
多排几下序就A了.
1 #include <bits/stdc++.h> 2 #define N 200005 3 using namespace std; 4 struct Node{ 5 int index; 6 int num; 7 int val; 8 }; 9 Node node[N]; 10 bool cmp1(Node a,Node b){ 11 if(a.index==b.index) 12 return a.num<b.num; 13 return a.index>b.index; 14 } 15 bool cmp2(Node a,Node b){ 16 return a.num<b.num; 17 } 18 int k[N]; 19 int main(){ 20 int n; 21 scanf("%d",&n); 22 for(int i=0;i<n;i++){ 23 scanf("%d",&k[i]); 24 } 25 sort(k,k+n); 26 for(int i=0;i<n;i++){ 27 scanf("%d",&node[i].index); 28 node[i].num=i; 29 } 30 sort(node,node+n,cmp1); 31 for(int i=0;i<n;i++){ 32 node[i].val=k[i]; 33 } 34 sort(node,node+n,cmp2); 35 for(int i=0;i<n;i++){ 36 printf("%d ",node[i].val); 37 } 38 printf(" "); 39 return 0; 40 }