2607. Dress'em in Vests!
The Two-dimensional kingdom is going through hard times... This morning the Three-Dimensional kingdom declared war on the Two-dimensional one. This (possibly armed) conflict will determine the ultimate owner of the straight line.
The Two-dimensional kingdom has a regular army of n people. Each soldier registered himself and indicated the desired size of the bulletproof vest: the i
The Two-dimensional kingdom has m vests at its disposal, the j
Input
The first input line contains four integers n, m
The second line contains n integers a1,a2,...,an (1≤ai≤109) in non-decreasing order, separated by single spaces − the desired sizes of vests.
The third line contains m integers b1,b2,...,bm (1≤bj≤109) in non-decreasing order, separated by single spaces − the sizes of the available vests.
Output
In the first line print a single integer k − the maximum number of soldiers equipped with bulletproof vests.
In the next k lines print k pairs, one pair per line, as "ui vi" (without the quotes). Pair (ui, vi) means that soldier number ui must wear vest number vi. Soldiers and vests are numbered starting from one in the order in which they are specified in the input. All numbers of soldiers in the pairs should be pairwise different, all numbers of vests in the pairs also should be pairwise different. You can print the pairs in any order.
If there are multiple optimal answers, you are allowed to print any of them.
Examples
Input
5 3 0 0 1 2 3 3 4 1 3 5
Output
2 1 1 3 2
Input
3 3 2 2 1 5 9 3 5 7
Output
3 1 1 2 2 3 3
Note
In the first sample you need the vests' sizes to match perfectly: the first soldier gets the first vest (size 1), the third soldier gets the second vest (size 3). This sample allows another answer, which gives the second vest to the fourth soldier instead of the third one.
In the second sample the vest size can differ from the desired size by at most 2 sizes, so all soldiers can be equipped.
注:判斷b裡在範圍[ai-x,ai+y]有多少個,並輸出相應的位置
#include<iostream>
#include<algorithm>
using namespace std;
const int maxn=1e5+5;
int a[maxn];
int b[maxn];
int c[maxn];
int d[maxn];
int ans=0;
int main(){
int n,m,x,y;
cin>>n>>m>>x>>y;
for(int i=0;i<n;i++){
cin>>a[i];
}
for(int i=0;i<m;i++){
cin>>b[i];
}
sort(a,a+n);
sort(b,b+m);
int bt=1;
for(int i=0,j=0;j<m&&i<n;){
if(a[i]-x<=b[j]&&b[j]<=a[i]+y){
ans++;
j++;
i++;
c[bt]=i;
d[bt]=j;
bt++;
}else if(b[j]<a[i]){
j++;
}else{
i++;
}
}
cout<<ans<<endl;
for(int i=1;i<=ans;i++){
cout<<c[i]<<" "<<d[i]<<endl;
}
return 0;
}