父亲给三个儿子分田（水题，但比较麻烦）

                                     做这道题花了很长时间，代码能力太菜，存数组时就出现问题了，下面三个FOR循环都怎样都控制不好。

Three sons inherited from their father a rectangular corn fiend divided into n × m squares. For each square we know how many tons of corn grows on it. The father, an old farmer did
not love all three sons equally, which is why he bequeathed to divide his field into three parts containing A, B and C tons
of corn.

The field should be divided by two parallel lines. The lines should be parallel to one side of the field and to each other. The lines should go strictly between the squares of the field. Each resulting part of the field should consist of at least one square.

Your task is to find the number of ways to divide the field as is described above, that is, to mark two lines, dividing the field in three parts so that on one of the resulting parts grew A tons
of corn, B on another one and C on the remaining one.

Input

The first line contains space-separated integers n and m — the sizes of the original (1 ≤ n, m ≤ 50, max(n, m) ≥ 3).
Then the field's description follows: n lines, each containing m space-separated integers cij,
(0 ≤ cij ≤ 100) — the number of tons of corn each square contains. The last line contains space-separated integers A, B, C (0 ≤ A, B, C ≤ 106).

Output

Print the answer to the problem: the number of ways to divide the father's field so that one of the resulting parts contained A tons of corn, another one contained B tons,
and the remaining one contained C tons. If no such way exists, print 0.

Sample Input

Input

3 3
1 1 1
1 1 1
1 1 1
3 3 3

Output

2

Input

2 5
1 1 1 1 1
2 2 2 2 23 6 6

Output

3

Input

3 31 2 33 1 22 3 1
5 6 7

Output

0

#include<stdio.h>
#include<string.h>
using namespace std;
int a[100][100],b[1000020],c[1000020];
int v[5],a1,a2,a3;
int main()
{
//freopen("input.txt","r",stdin); freopen("output.txt","w",stdout);
int n,m;
while(scanf("%d%d",&n,&m)!=EOF)
{
memset(b,0,sizeof(b));
memset(c,0,sizeof(c));
int k=0,s=0;
for(int i=0; i<n; i++)
for(int j=0; j<m; j++)
scanf("%d",&a[i][j]);
scanf("%d %d %d",&a1,&a2,&a3);
for(int i=0; i<n; i++)
{
for(int j=0;j<m;j++)
{b[k]+=a[i][j];//把每行的和存入数组。
}
k++;

}
for(int i=0;i<m;i++)
{for(int j=0; j<n; j++)
{
c[s]+=a[j][i];//把每列的和存入数组
}
s++;
}
int z=0,sum=0;
for(int i=0; i<n-2; i++)//固定一部分，遍历
{
sum+=b[i];
int num=0;
for(int j=i+1; j<n-1; j++)//固定第二部分，遍历
{
num+=b[j];
int l=0;
for(int x=j+1;x<n;x++)//固定第三部分，遍历
{l+=b[x];}
v[0]=a1;
v[1]=a2;
v[2]=a3;//保证分给三个儿子的田地不变。
int o=0;
for(int d=0; d<3; d+
4000
+)//查询是否分到a1,a2,a3,完成o++，加到3，记录一次
{
if(sum==v[d]&&v[d]!=-1)
{   o++;
v[d]=-1;

}
if(num==v[d]&&v[d]!=-1)
{o++;
v[d]=-1;
}
if(l==v[d]&&v[d]!=-1)
{
o++;
v[d]=-1;
}
if(o==3)
{
z++;
}
}
}
}

//下面为列的所有查询

int sum1=0;
for(int i=0; i<m-2; i++)
{
sum1+=c[i];
int num1=0;
for(int j=i+1; j<m-1; j++)
{
num1+=c[j];
int l1=0;
for(int x=j+1;x<m;x++)
{
l1+=c[x];
}
v[0]=a1;
v[1]=a2;
v[2]=a3;
int o=0;
for(int d=0; d<3; d++)
{
if(sum1==v[d]&&v[d]!=-1)
{
v[d]=-1;
o++;
}
if(num1==v[d]&&v[d]!=-1)
{
v[d]=-1;
o++;
}
if(l1==v[d]&&v[d]!=-1)
{
v[d]=-1;
o++;
}

if(o==3)
{
z++;
}
}
}
}
printf("%d\n",z);
}
}