Brute Force - B. Candy Boxes ( Codeforces Round #278 (Div. 2)
B. Candy Boxes
Problem's Link: http://codeforces.com/contest/488/problem/B
Mean:
T题目意思很简单,不解释。
analyse:
这道题还是很有意思的,需要考虑到各种情况才能AC。
解这个题目之前,首先要推出两条式子
x4=3x1
4x1=x2+x3
然后就是分类讨论,枚举各种情况就可。
Time complexity: O(1)
Source code:
// Memory Time
// 1347K 0MS
// by : Snarl_jsb
// 2014-11-26-21.20
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<vector>
#include<queue>
#include<stack>
#include<map>
#include<string>
#include<climits>
#include<cmath>
#define N 1000010
#define LL long long
using namespace std; int ans[10], input[10];
bool flag = true;
bool Solve(int n)
{
if (n == 0)
{
puts("YES");
printf("%d\n%d\n%d\n%d\n", 1, 1, 3, 3);
}
if (n == 1)
{
puts("YES");
ans[1] = input[1], ans[4] = ans[1] * 3, ans[2] = ans[1], ans[3] = ans[4];
for (int i = 2; i <= 4; i++) printf("%d\n", ans[i]);
}
if (n == 2)
{
if (input[2] % 3)
{
if (input[1] * 3 < input[2]) return flag = false;
puts("YES");
ans[1] = input[1], ans[4] = ans[1] * 3;
int tmp = 4 * ans[1] - input[2];
printf("%d\n%d\n", ans[4], tmp);
}
else
{
if (input[2] / 3 > input[1]) return flag = false;
puts("YES");
ans[1] = input[2] / 3;
int tmp = 4 * ans[1] - input[1];
printf("%d\n%d\n", ans[1], tmp);
}
}
if (n == 3)
{
if (input[3] % 3)
{
if (input[1] * 3 < input[3]) return flag = false;
for (int i = 1; i <= 3; i++) ans[i] = input[i];
ans[4] = ans[1] * 3;
if (4 * ans[1] != ans[2] + ans[3]) return flag = false;
puts("YES");
printf("%d\n", ans[4]);
}
else
{
if (input[3] / 3 > input[1]) return flag = false;
if (input[3] / 3 == input[1])
{
puts("YES");
printf("%d\n", input[1] * 4 - input[2]);
return true;
}
ans[1] = input[3] / 3, ans[4] = input[3], ans[2] = input[1], ans[3] = input[2];
if (4 * ans[1] != ans[2] + ans[3]) return flag = false;
puts("YES");
printf("%d\n", ans[1]);
}
}
if (n == 4)
{
for (int i = 1; i <= 4; i++) ans[i] = input[i];
if (4 * ans[1] == ans[2] + ans[3]) puts("YES");
else flag = false;
}
} int main()
{
ios_base::sync_with_stdio(false);
cin.tie(0);
// freopen("C:\\Users\\ASUS\\Desktop\\cin.cpp","r",stdin);
// freopen("C:\\Users\\ASUS\\Desktop\\cout.cpp","w",stdout);
int n, i, j;
scanf("%d", &n);
for (i = 1; i <= n; i++) scanf("%d", &input[i]);
sort(input + 1, input + n + 1);
Solve(n);
if (!flag)
{
puts("NO");
return 0;
}
return 0;
}
/* */
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