HDOJ 5143 NPY and arithmetic progression

题意：有a1个1，a2个2，a3个3，a4个4，问能否不重不漏的划分成一些长度大于等于3的等差数列（一个数列可以出现多次）

链接：http://acm.hdu.edu.cn/showproblem.php?pid=5143

思路：四个数字用一个四维数组保存，利用递推对数组预处理，并且当a1，a2，a3，a4全都大于等于三时，等差数列一定存在，缩小数据范围。

注意点：无。

以下为AC代码：

Run ID	Submit Time	Judge Status	Pro.ID	Exe.Time	Exe.Memory	Code Len.	Language	Author
12493321	2014-12-14 00:15:56	Accepted	5143	218MS	1192K	3918 B	G++	luminous11

#include <iostream>
#include <cstdio>
#include <string>
#include <cstring>
#include <vector>
#include <deque>
#include <list>
#include <cctype>
#include <algorithm>
#include <climits>
#include <queue>
#include <stack>
#include <cmath>
#include <map>
#include <set>
#include <iomanip>
#include <cstdlib>
#include <ctime>
#define ll long long
#define ull unsigned long long
#define all(x) (x).begin(), (x).end()
#define clr(a, v) memset( a , v , sizeof(a) )
#define pb push_back
#define mp make_pair
#define read(f) freopen(f, "r", stdin)
#define write(f) freopen(f, "w", stdout)
using namespace std;

bool f[11][11][11][11];
int main()
{
ios::sync_with_stdio( false );
int t;
int a1,a2,a3,a4;
f[0][0][0][0] = true;
for ( a1 = 0; a1 <= 10; a1 ++ )
{
for ( a2 = 0; a2 <= 10; a2 ++ )
{
for ( a3 = 0; a3 <= 10; a3 ++ )
{
for ( a4 = 0; a4 <= 10; a4 ++ )
{
if ( a1 != 0 || a2 != 0 || a3 != 0 || a4 != 0 )
{
f[a1][a2][a3][a4] = false;
for ( int i = a1 - 3; i >= 0; i -- )
{
if ( f[i][a2][a3][a4] )
{
f[a1][a2][a3][a4] = true;
continue;
}
}
for ( int i = a2 - 3; i >= 0; i -- )
{
if ( f[a1][i][a3][a4] )
{
f[a1][a2][a3][a4] = true;
continue;
}
}
for ( int i = a3 - 3; i >= 0; i -- )
{
if ( f[a1][a2][i][a4] )
{
f[a1][a2][a3][a4] = true;
continue;
}
}
for ( int i = a4 - 3; i >= 0; i -- )
{
if ( f[a1][a2][a3][i] )
{
f[a1][a2][a3][a4] = true;
continue;
}
}
if ( a1 > 0 && a2 > 0 && a3 > 0 )
{
if ( f[a1-1][a2-1][a3-1][a4] )
{
f[a1][a2][a3][a4] = true;
continue;
}
}
if ( a2 > 0 && a3 > 0 && a4 > 0)
{
if ( f[a1][a2-1][a3-1][a4-1] )
{
f[a1][a2][a3][a4] = true;
continue;
}
}
if ( a1 > 0 && a2 > 0 && a3 > 0 && a4 > 0 )
{
if ( f[a1-1][a2-1][a3-1][a4-1] )
{
f[a1][a2][a3][a4] = true;
continue;
}
}
}
}
}
}
}
cin >> t;
while ( t -- )
{
cin >> a1 >> a2 >> a3 >> a4;
if ( a1 > 10 ) a1 = 10;
if ( a2 > 10 ) a2 = 10;
if ( a3 > 10 ) a3 = 10;
if ( a4 > 10 ) a4 = 10;
if ( f[a1][a2][a3][a4] )
cout << "Yes" << endl;
else
cout << "No" << endl;
}
return 0;
}