[UVA][hash] 1152 - 4 Values whose Sum is 0
The SUM problem can be formulated as follows: given four lists A, B, C, D
of integer values, compute how many quadruplet
(a, b, c, d ) 
A x B x C x D
are such that
a + b + c + d = 0
. In the following, we assume that all lists have the same size n
.
Input
The input begins with a single positive integer on a line by itself indicating the number of the cases following, each of them as described below. This line is followed by a blank line, and there is also a blank line between two consecutive inputs.
The first line of the input file contains the size of the lists n
(this value can be as large as 4000). We then have n
lines containing four integer values (with absolute value as large as 228
) that belong respectively to A, B, C
and D
.
Output
For each test case, the output must follow the description below. The outputs of two consecutive cases will be separated by a blank line.
For each input file, your program has to write the number quadruplets whose sum is zero.
Sample Input
1 6 -45 22 42 -16 -41 -27 56 30 -36 53 -37 77 -36 30 -75 -46 26 -38 -10 62 -32 -54 -6 45
Sample Output
5
Sample Explanation:
Indeed, the sum of the five following quadruplets is zero: (-45, -27,
42, 30), (26, 30, -10, -46), (-32, 22, 56, -46),(-32, 30, -75, 77),
(-32, -54, 56, 30).
STL 的 map 過不了, 因此自己寫 hash
O(n*n*k)
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define hashR 2000000
int hash[hashR], size;
struct Node {
int v, t, next;
} Node[16000000];
void addHash(int v) {
int m = abs(v)%hashR;
int head, prev;
prev = 0;
head = hash[m];
while(head != 0) {
if(Node[head].v == v) {
Node[head].t++;
return;
} else if(Node[head].v < v) {
prev = head;
head = Node[head].next;
} else
break;
}
++size;
if(!prev)
hash[m] = size;
else
Node[prev].next = size;
Node[size].v = v, Node[size].t = 1;
Node[size].next = head;
}
int getHash(int v) {
int m = abs(v)%hashR;
int head, prev;
prev = 0;
head = hash[m];
while(head != 0) {
if(Node[head].v == v) {
return Node[head].t;
} else if(Node[head].v < v) {
prev = head;
head = Node[head].next;
} else
return 0;
}
}
int main() {
int t, n, A[4000], B[4000], C[4000], D[4000];
int i, j;
scanf("%d", &t);
while(t--) {
scanf("%d", &n);
memset(hash, 0, sizeof(hash));
size = 0;
for(i = 0; i < n; i++) {
scanf("%d %d %d %d", &A[i], &B[i], &C[i], &D[i]);
}
for(i = 0; i < n; i++) {
for(j = 0; j < n; j++) {
addHash(A[i]+B[j]);
}
}
int ans = 0;
for(i = 0; i < n; i++) {
for(j = 0; j < n; j++) {
ans += getHash(-C[i]-D[j]);
}
}
printf("%d\n", ans);
if(t)
puts("");
}
return 0;
}