hdu 1496 hash表求表达式的解的个数 2中hash方法

EquationsTime Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3356    Accepted Submission(s): 1347


Problem DescriptionInputOutputSample InputSample OutputAuthorSource/** hash+数学，很好的题 * 对整数求hash，采用除余法，及线性探测解决冲突 * 注意：devc++中不能定义全局变量count，它和库函数中的函数名同名了 */#include <cstdio>#include <cstring>#include <iostream>using namespace std;const int M = 175447;int counts[M];int result[M];int temp[101];int hashInt(int s){ int k = s % M; if (k < 0) k += M; while (counts[k] && result[k]!=s) k = (k + 1) % M;//如果当前对应不为空 且对应的不是s 则继续寻找 return k;}int solve(int a, int b, int c, int d) { if (a>0&&b>0&&c>0&&d>0 || a<0&&b<0&&c<0&&d<0) return 0; memset(counts, 0, sizeof(counts)); int s, p, i, j; for (i=1; i<101; ++i) { for (j=1; j<101; ++j) { s = a * temp[i] + b * temp[j]; p = hashInt(s); result[p] = s;// ++counts[p]; } } int ans = 0; for (i=1; i<101; ++i) { for (j=1; j<101; ++j) { s = -(c * temp[i] + d * temp[j]); p = hashInt(s); ans += counts[p]; } } return ans;}int main() { int a, b, c, d; for (int i=1; i<101; ++i) temp[i] = i * i; while (scanf("%d%d%d%d", &a, &b, &c, &d) != EOF) { int ans = solve(a, b, c, d); printf ("%d\n", ans<<4); } return 0;}