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poj1840Eqs【散列表】
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.
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 }
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 }
poj1840Eqs【散列表】
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