codeChef----Partition the numbers_codechef optimal partition-优快云博客

本文探讨了给定两个整数x和N时如何将1到N（除x外）的整数集划分为两组，使两组的元素之和相等。介绍了问题的输入输出格式及限制条件，并通过示例解释了解决方案。

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You are given two integers x and N. Consider all integers between 1 and N inclusive, except x. We want to partition these integers into two disjoint sets (each integer has to appear in exactly one set) such that the sums of numbers in these sets are equal.

Find one valid partition or determine that it doesn't exist.

Input

The first line of the input contains a single integer T denoting the number of test cases. The description of T test cases follows.
The first and only line of each test case contains two space-separated integers xand N.

Output

For each test case:

If there's no way to partition the numbers into two sets with equal sums, print a single line containing the string "impossible" (without quotes).
Otherwise, print a single line containing a string with length N.
The x-th character of this string should be '2'.
For each valid i ≠ x, the i-th character of this string should be '0' if number i should be in the first set or '1' if it should be in the second set.

Constraints

1 ≤ T ≤ 10,000
2 ≤ N ≤ 1,000,000
1 ≤ x ≤ N
1 ≤ sum of N in all test cases ≤ 1,000,000

Subtasks

Subtask #1 (20 points): sum of N in all test cases ≤ 50

Subtask #2 (80 points): original constraints

Example

Input:

3
2 4
5 5
1 2

Output:

0201
01102
impossible

Author:6★kingofnumbers

Date Added:4-01-2018

Time Limit:1 secs

Source Limit:50000 Bytes

Languages:ADA, ASM, BASH, BF, C, CAML, CLOJ, CLPS, CPP 4.3.2, CPP 6.3, CPP14, CS2, D, ERL, FORT, FS, GO, HASK, ICK, ICON, JAVA, JS, kotlin, LISP clisp, LISP sbcl, LUA, NEM, NICE, NODEJS, PAS fpc, PAS gpc, PERL, PERL6, PHP, PIKE, PRLG, PYPY, PYTH, PYTH 3.5, RUBY, rust, SCALA, SCM chicken, SCM guile, SCM qobi, ST, swift, TCL, TEXT, WSPC

思路：题目其实不是很难，但是我却花了很长时间，最后我想出来的方法是计算出两边相等的和（至于不满足条件的代码中有所体现），然后从最大的值开始循环，依次减去，若在相减过程中遇到去掉的数，则回退一步，具体实现代码如下：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long ll;
ll x,n;
ll a[1000010];
int main()
{
	int t;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%lld%lld",&x,&n);
		ll m=(n*(n+1)/2);
		ll z=x;
		if((m-x)%2!=0)
		{
			printf("impossible\n");
			continue;
		}
		else
		{
			memset(a,0,sizeof(a));
			a[x]=2;
			ll k=(m-x)/2;
			ll sum=0;
			for(ll i=n;i>=1;i--)
			{
				if(k==0) break;
				else if(i!=x && k-i>=0)
				{
					a[i]=1;
					k-=i;
					if(k==x)
					{
						k+=i;
						a[i]=0;
					}
				}
			}
			if(k==0)
			{
				for(ll i=1;i<=n;i++) printf("%lld",a[i]);
				printf("\n");
			}
			else printf("impossible\n");
		}
	}
	return 0;
}