Virtual Judge Contest 1-E

E - Lucky Tickets. Easy!

Time Limit:2000MS     Memory Limit:16384KB  

Description

Background

The public transport administration of Ekaterinburg is anxious about the fact that passengers don’t like to pay for passage doing their best to avoid the fee. All the measures 

that had been taken (hard currency premiums for all of the chiefs, increase in conductors’ salaries, reduction of number of buses) were in vain. An advisor especially invited 

from the Ural State University says that personally he doesn’t buy tickets because he rarely comes across the lucky ones (a ticket is lucky if the sum of the first three digits 

in its number equals to the sum of the last three ones). So, the way out is found — of course, tickets must be numbered in sequence, but the number of digits on a ticket 

may be changed. Say, if there were only two digits, there would have been ten lucky tickets (with numbers 00, 11, …, 99). Maybe under the circumstances the ratio of

 the lucky tickets to the common ones is greater? And what if we take four digits? A huge work has brought the long-awaited result: in this case there will be 670 lucky tickets. 

But what to do if there are six or more digits?

Problem

So you are to save public transport of our city. Write a program that determines a number of lucky tickets for the given number of digits. By the way, there can’t be more 

than nine digits on one ticket.

Input

contains a positive even integer not greater than 9.

Output

should contain a number of tickets such that the sum of the first half of digits is equal to the sum of the second half of digits.

Sample Input

inputoutput

4

670

Solution:一开始被题目欺骗了，原本说前三位数字之和等于后三位数字之和的票为Lucky Ticket，原来是前一半数字之和和后一半数字之和
相等的票才是。由于输入数字为偶数，且小于9，时间内存存足，直接强破。

#include<iostream>using namespace std;void f2() {int count = 0;for (int i = 0; i <= 9; i++)for (int j = 0; j <= 9; j++)if (i == j)count++;cout << count << endl;}void f4() {int count = 0;for (int i = 0; i <= 9; i++)for (int j = 0; j <= 9; j++)for (int k = 0; k <= 9; k++)for (int l = 0; l <= 9; l++)if (i + j == k + l)count++;cout << count << endl;}void f6() {int count = 0;for (int i = 0; i <= 9; i++)for (int j = 0; j <= 9; j++)for (int k = 0; k <= 9; k++)for (int l = 0; l <= 9; l++)for (int m = 0; m <= 9; m++)for (int n = 0; n <= 9; n++)if (i + j + k == l + m + n)count++;cout << count << endl;}void f8() {int count = 0;for (int i = 0; i <= 9; i++)for (int j = 0; j <= 9; j++)for (int k = 0; k <= 9; k++)for (int l = 0; l <= 9; l++)for (int m = 0; m <= 9; m++)for (int n = 0; n <= 9; n++)for (int o = 0; o <= 9; o++)for (int p = 0; p <= 9; p++)if (i + j + k + l == m + n + o + p)count++;cout << count << endl;}int main() {int n;cin >> n;if (n == 2)f2();else if (n == 4)f4();else if (n == 6)f6();else if (n == 8)f8();return 0;}