题目描述
Ashish has an array a of consisting of 2n positive integers. He wants to compress a into an array b of size n−1. To do this, he first discards exactly 2 (any two) elements from a. He then performs the following operation until there are no elements left in a:
Remove any two elements from a and append their sum to b.
The compressed array b has to have a special property. The greatest common divisor (gcd) of all its elements should be greater than 1.
Recall that the gcd of an array of positive integers is the biggest integer that is a divisor of all integers in the array.
It can be proven that it is always possible to compress array a into an array b of size n−1 such that gcd(b1,b2…,bn−1)>1.
Help Ashish find a way to do so.
Input
The first line contains a single integer t (1≤t≤10) — the number of test cases. The description of the test cases follows.
The first line of each test case contains a single integer n (2≤n≤1000).
The second line of each test case contains 2n integers a1,a2,…,a2n (1≤ai≤1000) — the elements of the array a.
Output
For each test case, output n−1 lines — the operations performed to compress the array a to the array b. The initial discard of the two elements is not an operation, you don’t need to output anything about it.
The i-th line should contain two integers, the indices (1 —based) of the two elements from the array a that are used in the i-th operation. All 2n−2 indices should be distinct integers from 1 to 2n.
You don’t need to output two initially discarded elements from a.
If there are multiple answers, you can find any.
Example
input
3
3
1 2 3 4 5 6
2
5 7 9 10
5
1 3 3 4 5 90 100 101 2 3
output
3 6
4 5
3 4
1 9
2 3
4 5
6 10
Note
In the first test case, b={3+6,4+5}={9,9} and gcd(9,9)=9.
In the second test case, b={9+10}={19} and gcd(19)=19.
In the third test case, b={1+2,3+3,4+5,90+3}={3,6,9,93} and gcd(3,6,9,93)=3.
题目分析
这个题可以分成两种情况：
- a[]中有奇数个奇数，此时可以删掉一个奇数和一个偶数，这样a[]中就是剩下了偶数个偶数和偶数个奇数，这样组合后gcd( b[] )必定大于2。
- a[]中有偶数个奇数，此时就可以删掉两个偶数或者两个奇数（前提是保证a[]中偶数或者奇数的个数大于2），这样a[]中偶数和奇数的个数也还是偶数，组合后gcd( b[] )必定大于2。
代码如下
#include <iostream>
#include <cstdio>
#include <cmath>
#include <string>
#include <cstring>
#include <stack>
#include <map>
#include <unordered_map>
#include <queue>
#include <vector>
#include <set>
#include <algorithm>
#include <iomanip>
#define LL long long
using namespace std;
const int N=2e3+5;
int a[N],ji[N],ou[N];
int main()
{
int t;
cin>>t;
while(t--)
{
int n;
cin>>n;
int cnt1=0,cnt2=0; //统计a[]中偶数和奇数的个数
for(int i=1;i<=2*n;i++)
{
cin>>a[i];
if(a[i]&1) ji[cnt1++]=i;
else ou[cnt2++]=i;
}
if(cnt1%2) //奇数的个数为奇数的情况（删除一奇一偶）
{
for(int i=0;i<cnt1-1;i+=2)
cout<<ji[i]<<' '<<ji[i+1]<<endl;
for(int i=0;i<cnt2-1;i+=2)
cout<<ou[i]<<' '<<ou[i+1]<<endl;
}
else if(cnt1>=2) //奇数的个数为偶数，且奇数个数大于2的情况（删除两个奇数）
{
for(int i=0;i<cnt1-2;i+=2)
cout<<ji[i]<<' '<<ji[i+1]<<endl;
for(int i=0;i<cnt2;i+=2)
cout<<ou[i]<<' '<<ou[i+1]<<endl;
}
else //奇数的个数为偶数，且偶数个数大于2的情况（删除两个偶数）
{
for(int i=0;i<cnt1;i+=2)
cout<<ji[i]<<' '<<ji[i+1]<<endl;
for(int i=0;i<cnt2-2;i+=2)
cout<<ou[i]<<' '<<ou[i+1]<<endl;
}
}
return 0;
}