这两题的思路都是将等式化成左右两部分，用一个hash数组，先把左边的结果存起来，然后计算右边的结果直接寻址就行，不过，HDU那道题要注意剪枝，如果系数全为正或者全为负则直接输出0，POJ那道题hash函数定义成char型的，否则会超内存。。

    

HDU_1496 Equations：

#include 
     
       #include 
      
        #include 
       
         using namespace std; const int N = 2000009; int hash[N]; int main() {
    //freopen("data.in", "r", stdin);     int a, b, c, d, i, j; while(scanf("%d%d%d%d", &a, &b, &c, &d) != EOF)     {
    if((a > 0 && b > 0 && c > 0 && d > 0) || (a < 0 && b < 0 && c < 0 && d < 0))         { printf("0\n"); continue;}         memset(hash, 0, sizeof(hash)); for(i = 1; i <= 100; i++) for(j = 1; j <= 100; j++)                 hash[1000000 + a*i*i + b*j*j]++; int sum = 0; for(i = 1 ; i <= 100; i++) for(j = 1; j <= 100; j++) if(hash[1000000 - c*i*i - d*j*j])                     sum += hash[1000000 - c*i*i - d*j*j];         printf("%d\n", sum*16);     } return 0; }
       
      
     

 POJ_1840 Eqs：

#include 
     
       #include 
      
        #include 
       
         using namespace std; const int N = 25000007; char hash[N];    //！！！ int main() {
    //freopen("data.in", "r", stdin);     int a, b, c, d, e, i, j, k; while(scanf("%d%d%d%d%d", &a, &b, &c, &d, &e) != EOF)     {
            memset(hash, 0, sizeof(hash)); for(i = -50; i <= 50; i++)         {
    if(i == 0)    continue; for(j = -50; j <= 50; j++)             {
    if(j == 0)    continue;                 hash[12500000 + a*i*i*i + b*j*j*j]++;             }         } int sum = 0; for(i = -50; i <= 50; i++)         {
    if(i == 0)    continue; for(j = -50; j <= 50; j++)             {
    if(j == 0)    continue; for(k = -50; k <= 50; k++)                 {
    if(k == 0 || 12500000 < c*i*i*i + d*j*j*j + e*k*k*k || -12500000 > c*i*i*i + d*j*j*j + e*k*k*k)    continue;                     sum += hash[12500000-c*i*i*i-d*j*j*j-e*k*k*k];                 }             }         }         printf("%d\n", sum);     } return 0; }