CodeForces 374A. Divisible by Seven

最新推荐文章于 2017-08-23 15:01:54 发布

码代码的猿猿的AC之路

最新推荐文章于 2017-08-23 15:01:54 发布

阅读量1.5k

点赞数 1

CC 4.0 BY-SA版权

分类专栏：数论文章标签： codeforces

本文链接：https://blog.youkuaiyun.com/ck_boss/article/details/17551343

数论专栏收录该内容

37 篇文章

订阅专栏

本文介绍了一种算法，用于重新排列一个包含特定数字（1、6、8、9）的整数a的位数，使其成为7的倍数。输入为一个不含前导零且包含1、6、8、9的正整数a，输出则是重新排列后的7的倍数，若无法排列则输出0。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

构造。。。。

You have number a, whose decimal representation quite luckily contains digits 1, 6, 8, 9. Rearrange the digits in its decimal representation so that the resulting number will be divisible by 7.

Number a doesn't contain any leading zeroes and contains digits 1, 6, 8, 9 (it also can contain another digits). The resulting number also mustn't contain any leading zeroes.

Input

The first line contains positive integer a in the decimal record. It is guaranteed that the record of number a contains digits: 1, 6, 8, 9. Number a doesn't contain any leading zeroes. The decimal representation of number a contains at least 4 and at most 106 characters.

Output

Print a number in the decimal notation without leading zeroes — the result of the permutation.

If it is impossible to rearrange the digits of the number a in the required manner, print 0.

Sample test(s)

input
1689

output
1869

input
18906

output
18690

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>

using namespace std;

bool sh[10010000];

int main()
{
	string str;
	int cnt = 0, mod = 0;
	cin >> str;
	int n = str.size();
	memset(sh, true, sizeof(sh));
	bool n1 = false, n6 = false, n8 = false, n9 = false;
	for (int i = 0; i < n; i++)
	{
		int b = str[i] - '0';
		if (b == 0)
		{
			sh[i] = false;
			cnt++; continue;
		}
		if (b == 1 && !n1)
		{
			n1 = true;
			sh[i] = false;
			continue;
		}
		if (b == 6 && !n6)
		{
			n6 = true;
			sh[i] = false;
			continue;
		}
		if (b == 8 && !n8)
		{
			n8 = true;
			sh[i] = false;
			continue;
		}
		if (b == 9 && !n9)
		{
			n9 = true;
			sh[i] = false;
			continue;
		}
		mod = (mod * 10 + b) % 7;
	}
	mod = (mod * 10000) % 7;
	for (int i = 0; i < n; i++)
	{
		if (sh[i]) putchar(str[i]);
	}
	if (mod == 0)
	{
		printf("1869");
	}
	else if (7 - mod == 1)
	{
		printf("1968");
	}
	else if (7 - mod == 2)
	{
		printf("1689");
	}
	else if (7 - mod == 3)
	{
		printf("6198");
	}
	else if (7 - mod == 4)
	{
		printf("1698");
	}
	else if (7 - mod == 5)
	{
		printf("1986");
	}
	else if (7 - mod == 6)
	{
		printf("1896");
	}
	while (cnt--)
	{
		putchar('0'); 
	}
	///system("pause");
	return 0;
}