1243. Divorce of the Seven Dwarfs

Time limit: 1.0 second
Memory limit: 64 MB
After the Snow White with her bridegroom had left the house of the seven dwarfs, their peaceful and prosperous life has come to an end. Each dwarf blames others to be the reason of the Snow White's leave. To stop everlasting quarrels, the dwarfs decided to part. According to an ancient law, their common possessions should be divided in the most fair way, which means that all the dwarfs should get equal parts. Everything that the dwarfs cannot divide in a fair way they give to the Snow White. For example, after dividing 26 old boots, each dwarf got 3 old boots, and the Snow White got the remaining 5 old boots. Some of the numbers are very large, for example, the dwarfs have 123456123456 poppy seeds, so it is not easy to calculate that the Snow White gets only one seed. To speed up the divorce, help the dwarfs to determine quickly the Snow White's part.
 
 
 

Input

The only line contains an integer N that represents the number of similar items that the dwarfs want to divide (1 ≤ N ≤ 1050).

Output

You should output the number of items that pass into the possession of the Snow White.

Sample

input output
123456123456
1
Problem Author: Stanislav Vasilyev
Problem Source: Ural State University Personal Programming Contest, March 1, 2003
Difficulty: 57
 
题意:输出x%7
分析:高精度除法
 #include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
using namespace std;
const int N = ;
int Arr[N], n;
char Str[N];
#define For(i, s, t) for(int i = (s); i <= (t); i++) int main()
{
scanf("%s", Str + );
n = strlen(Str + );
For(i, , n) Arr[i] = Str[i] - '';
int Last = ;
For(i, , n)
{
Last = Last * + Arr[i];
Last %= ;
}
cout << Last << endl;
return ;
}

ural 1243. Divorce of the Seven Dwarfs的更多相关文章

  1. URAL - 1243 - Divorce of the Seven Dwarfs (大数取模)

    1243. Divorce of the Seven Dwarfs Time limit: 1.0 second Memory limit: 64 MB After the Snow White wi ...

  2. 后缀数组 POJ 3974 Palindrome && URAL 1297 Palindrome

    题目链接 题意:求给定的字符串的最长回文子串 分析:做法是构造一个新的字符串是原字符串+反转后的原字符串(这样方便求两边回文的后缀的最长前缀),即newS = S + '$' + revS,枚举回文串 ...

  3. ural 2071. Juice Cocktails

    2071. Juice Cocktails Time limit: 1.0 secondMemory limit: 64 MB Once n Denchiks come to the bar and ...

  4. ural 2073. Log Files

    2073. Log Files Time limit: 1.0 secondMemory limit: 64 MB Nikolay has decided to become the best pro ...

  5. ural 2070. Interesting Numbers

    2070. Interesting Numbers Time limit: 2.0 secondMemory limit: 64 MB Nikolay and Asya investigate int ...

  6. ural 2069. Hard Rock

    2069. Hard Rock Time limit: 1.0 secondMemory limit: 64 MB Ilya is a frontman of the most famous rock ...

  7. ural 2068. Game of Nuts

    2068. Game of Nuts Time limit: 1.0 secondMemory limit: 64 MB The war for Westeros is still in proces ...

  8. ural 2067. Friends and Berries

    2067. Friends and Berries Time limit: 2.0 secondMemory limit: 64 MB There is a group of n children. ...

  9. ural 2066. Simple Expression

    2066. Simple Expression Time limit: 1.0 secondMemory limit: 64 MB You probably know that Alex is a v ...

随机推荐

  1. 关于showModalDialog()对话框点击按钮弹出新页面的问题

    页面a.aspx上,单击按钮a,走脚本,弹出showModalDialog("b.aspx",....) 在b.aspx上有个服务器控件按钮b,单击按钮,更新数据后,会弹出一个新的 ...

  2. sql分页查询语句

    有关分页 SQL 的资料很多,有的使用存储过程,有的使用游标.本人不喜欢使用游标,我觉得它耗资.效率低:使用存储过程是个不错的选择,因为存储过程是经过预编译的,执行效率高,也更灵活.先看看单条 SQL ...

  3. [POJ1007]DNA Sorting

    [POJ1007]DNA Sorting 试题描述 One measure of ``unsortedness'' in a sequence is the number of pairs of en ...

  4. 新浪云php与java连接MySQL数据库

    PHP新浪云连接MySQL <?php $con=mysql_connect(SAE_MYSQL_HOST_M.':'.SAE_MYSQL_PORT,SAE_MYSQL_USER,SAE_MYS ...

  5. 最长公共子序列 NYOJ37

    http://acm.nyist.net/JudgeOnline/problem.php?pid=37 先逆转原来的字符串,再用原来的字符串跟逆转后的字符串进行比较,求得的最长公共子序列就是回文串,也 ...

  6. [转]sql语句中出现笛卡尔乘积 SQL查询入门篇

    本篇文章中,主要说明SQL中的各种连接以及使用范围,以及更进一步的解释关系代数法和关系演算法对在同一条查询的不同思路. 多表连接简介 在关系数据库中,一个查询往往会涉及多个表,因为很少有数据库只有一个 ...

  7. ASP.NET小知识

    所有System.Web.UI.*命名空间下的内容可以称为Web From,而System.Web.*命名空间下的其他内容可以称为ASP.NET. @section用法:配合母版页中的@RenderS ...

  8. iOS 用protocol 和 用继承小体会

    最近写程序时,2个类都有相同的函数,又因为在用oc,所以就用了protocol来实现.后来发现其实这2个类除了相同的函数,还需要一些相同的变量,当初用继承的话会更简单.

  9. Java for LeetCode 076 Minimum Window Substring

    Given a string S and a string T, find the minimum window in S which will contain all the characters ...

  10. jquery给height拼接动态变量

    var sizeLength = "${list.size()}"; if(sizeLength==''){ sizeLength=0; } sizeLength=400*size ...