cf ed#70 D

The subsequence is a sequence that can be derived from another sequence by deleting some elements without changing the order of the remaining elements.

You are given an integer n.

You have to find a sequence s consisting of digits {1,3,7} such that it has exactly n subsequences equal to 1337.

For example, sequence 337133377 has 6 subsequences equal to 1337:

3371–33–3–77– (you can remove the second and fifth characters);
3371–3–33–77– (you can remove the third and fifth characters);
3371–3–3–377– (you can remove the fourth and fifth characters);
3371–33–3–7–7 (you can remove the second and sixth characters);
3371–3–33–7–7 (you can remove the third and sixth characters);
3371–3–3–37–7 (you can remove the fourth and sixth characters).
Note that the length of the sequence s must not exceed 105.

You have to answer t independent queries.

Input
The first line contains one integer t (1≤t≤10) — the number of queries.

Next t lines contains a description of queries: the i-th line contains one integer ni (1≤ni≤109).

Output
For the i-th query print one string si (1≤|si|≤105) consisting of digits {1,3,7}. String si must have exactly ni subsequences 1337. If there are multiple such strings, print any of them.

Example
Input
2
6
1
Output
113337
1337

将答案分解成133 7777（若干个7）33333(若干个3） 7 33333（若干个3）7
第一个1和最后一个7加上中间所有的3（假设有i1个3）组成的1337个数为i1*(i1-1)/2;
第一个1和中间的7加上两者之间的3（假设有i2个3）组成的1337个数为i2*(i2-1)/2;
剩下再用 n-i1*(i1-1)/2-i2*(i2-1)/2 个7放在最开始的133后面解决剩下的次数。

#include<stdio.h>
#include<math.h>
int n,t;
int main()
{
	scanf("%d",&t);
	while(t--)
	{
		scanf("%d",&n);
		int i1;
		int maxn=1+sqrt(2*n);
		int tmp;
		for(i1=maxn;i1>1;i1--)
		{
			tmp=i1*(i1-1)/2;
			if(n>=tmp)
			break;
		}
		if(i1>1)
		n-=i1*(i1-1)/2;
		
		int i2;
		maxn=1+sqrt(2*n);
		for(i2=maxn;i2>1;i2--)
		{
			tmp=i2*(i2-1)/2;
			if(n>=tmp)
			break;
		}
		if(i2>1)
		n-=i2*(i2-1)/2;
		
		printf("133");
		for(int i=1;i<=n;i++)
		printf("7");
		
		if(i2>1)
		{
			for(int i=1;i<=i2-2;i++)
			printf("3");
			printf("7");
		}
		
		for(int i=1;i<=i1-i2&&i<=i1-2;i++)
		printf("3");
		printf("7\n");
	}
}