Description

Consider equations having the following form: 

a1x1 
³+ a2x2 
³+ a3x3 
³+ a4x4 
³+ a5x5 
³=0 

The coefficients are given integers from the interval [-50,50]. 

It is consider a solution a system (x1, x2, x3, x4, x5) that verifies the equation, xi∈[-50,50], xi != 0, any i∈{1,2,3,4,5}. 

Determine how many solutions satisfy the given equation. 

Input

The only line of input contains the 5 coefficients a1, a2, a3, a4, a5, separated by blanks.

Output

The output will contain on the first line the number of the solutions for the given equation.

Sample Input

37 29 41 43 47

Sample Output

654

分析：
把前两个hash一下进行存值
然后再后面查询有没有出现过即可

代码：

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <cmath>
 5 using namespace std;
 6 
 7 const int mod = 100007;
 8 
 9 struct Node {
10     int d;
11     Node* next;
12 };
13 
14 Node* head[mod + 10];
15 Node nd[mod + 10];
16 
17 int main() {
18     int a, b, c, d, e;
19     while(EOF != scanf("%d %d %d %d %d",&a, &b, &c, &d, &e) ) {
20         memset(head, 0, sizeof(head));
21         int n_cnt = 0;
22         for(int i = -50; i <= 50; i++) {
23             for(int j = -50; j <= 50; j++) {
24                 if(i == 0 || j == 0) continue;
25                 int num = a * i * i * i + b *j * j * j;
26                 int xx = num > 0 ? num : -num;
27                 int p = xx % mod;
28                 Node*pt = head[p];
29                 while(pt) {
30                     pt = pt -> next;
31                 }
32                 nd[n_cnt].d = num;
33                 nd[n_cnt].next = head[p];
34                 head[p] = &nd[n_cnt++];
35             }
36         }
37         int ans = 0;
38         for(int i = -50; i <= 50; i++) {
39             for(int j = -50; j <= 50; j++) {
40                 for(int k = -50; k <= 50; k++) {
41                     if(i == 0 || j == 0 || k == 0) continue;
42                     int num = c * i * i * i + d * j * j * j + e * k * k * k;
43                     num = - num;
44                     int xx = num > 0 ? num : - num;
45                     int p = xx % mod;
46                     Node * pt = head[p];
47                     while(pt) {
48                         if(pt -> d == num) ans++;
49                         pt = pt -> next;
50                     }
51                 }
52             }
53         }
54         printf("%d\n", ans);
55     }
56     return 0;
57 }

动态内存申请

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 using namespace std;
 5 
 6 const int mod = 10007;
 7 
 8 struct Node {
 9     int to;
10     int next;
11 }e[mod + 10];
12 
13 int head[mod + 10];
14 
15 int tot;
16 void add(int u, int v) {
17     e[tot].to = v;
18     e[tot].next = head[u];
19     head[u] = tot++;
20 }
21 
22 int Find(int p, int num) {
23     int cnt = 0;
24     for(int i = head[p]; i; i = e[i].next) {
25         if(e[i].to == num) cnt++;
26     }
27     return cnt;
28 }
29 
30 int Fabs(int x) {
31     return x > 0 ? x : - x;
32 }
33 
34 int main() {
35     int a, b, c, d, e;
36     while(EOF != scanf("%d %d %d %d %d",&a, &b, &c, &d, &e) ) {
37         memset(head, 0, sizeof(head));
38         tot = 1;
39         for(int i = -50; i <= 50; i++) {
40             for(int j = -50; j <= 50; j++) {
41                 if(i == 0 || j == 0) continue;
42                 int num = a * i * i * i + b * j * j * j;
43                 int p = Fabs(num) % mod;
44                 add(p, num);
45             }
46         }
47         int ans = 0;
48         for(int i = -50; i <= 50; i++) {
49             for(int j = -50; j <= 50; j++) {
50                 for(int k = -50; k <= 50; k++) {
51                     if(i == 0 || j == 0 || k == 0) continue;
52                     int num = c * i * i * i + d * j * j * j + e * k * k * k;
53                     num = - num;
54                     int p = Fabs(num) % mod;
55                     ans += Find(p, num);
56                 }
57             }
58         }
59         printf("%d\n",ans);
60     }
61     return 0;
62 }

数组模拟