2013
12-23

# Sum Zero

There are 5 Integer Arrays and each of them contains no more than 300 integers whose value are between -100,000,000 and 100,000,000, You are to find how many such groups (i,j,k,l,m) can make A[0][i]+A[1][j]+A[2][k]+A[3][l]+A[4][m]＝0. Maybe the result is too large, you only need tell me the remainder after divided by 1000000007.

In the first line, there is an Integer T(0<T<20), means the test cases in the input file, then followed by T test cases.
For each test case, there are 5 lines Integers, In each line, the first one is the number of integers in its array.

For each test case, just output the result, followed by a newline character.

1
3 4 -2 3
5 -5 -1 -7 -10 -1
5 -10 2 4 -6 2
2 -4 -1
5 -7 -7 -1 -4 -6

11

/*hdu 1895  A[0][i]+A[1][j]+A[2][k]+A[3][l]+A[4][m]＝0

#include <algorithm>
#include <iostream>
#include <cstdio>
#define MOD 1000000007
#define MAXN 305
using namespace std;
int a[5][MAXN], b[2][MAXN*MAXN];
int cmp( int p1, int p2 )
{
return p1 < p2;
}
int Find( int x, int c[] )
{
int l = 1, r = c[0], mid;
while ( l <= r )
{
mid = ( l + r ) >> 1;
if( x > c[r] || x < c[l] )
{
return 0;
}
if( c[mid] == x )
{
int sum = 1;
for( l = mid - 1; l >= 0 && c[l] == x; l--, sum++ );
for( r = mid + 1; r <= c[0] && c[r] == x; r++, sum++ );
return sum;
}
else if( x > c[mid] )
{
l = mid + 1;
}
else
{
r = mid - 1;
}
}
return 0;
}
void Connect( int k1, int k2, int k3 )
{
int i, j, len = 0;
for( i = 1; i <= a[k1][0]; i++ )
{
for( j = 1; j <= a[k2][0]; j++ )
{
b[k3][++len] = a[k1][i] + a[k2][j];
}
}
b[k3][0] = len;
}
void Solve( int c1[], int c2[], int c3[] )
{
int i, j, num, sum = 0;
sort( c1+1, c1+c1[0]+1, cmp );
sort( c2+1, c2+c2[0]+1, cmp );
sort( c3+1, c3+c3[0]+1, cmp );
for( i = 1; i <= c1[0]; i++ )
{
for( j = 1; j <= c2[0]; j++ )
{
num = -( c1[i] + c2[j] );
if( num < c3[1] || num > c3[c3[0]] )
{
continue;
}
sum += Find( num, c3 );
}
}
printf("%d\n", sum);
}
int main()
{
int cas;
scanf("%d", &cas);
while ( cas-- )
{
int i, j;
for( i = 0; i < 5; i++ )
{
scanf("%d", &a[i][0]);
for( j = 1; j <= a[i][0]; j++ )
{
scanf("%d", &a[i][j]);
}
}
Connect( 0, 1, 0 );
Connect( 2, 3, 1 );
Solve( a[4], b[0], b[1] );
}
return 0;
}

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2. 可以根据二叉排序树的定义进行严格的排序树创建和后序遍历操作。如果形成的排序树相同，其树的前、中、后序遍历是相同的，但在此处不能使用中序遍历，因为，中序遍历的结果就是排序的结果。经在九度测试，运行时间90ms，比楼主的要快。

3. 算法是程序的灵魂，算法分简单和复杂，如果不搞大数据类，程序员了解一下简单点的算法也是可以的，但是会算法的一定要会编程才行，程序员不一定要会算法，利于自己项目需要的可以简单了解。