poj1840 Eqs（哈希表）

来源：互联网发布：韩国人交友软件编辑：程序博客网时间：2024/06/06 09:27

原题链接：Eqs

题意：求满足方程的解的个数

枚举打表吧

#include <cstring>#include <iostream>using namespace std;const int MAX_S = 50 * 50 * 50 * 50 * 2;//前两项和的最大值 short ht[MAX_S * 2 + 1];//*2 此处用short类型，内存勉强过了 int main(){int a1, a2, a3, a4, a5, sum;while(cin >> a1 >> a2 >> a3 >> a4 >> a5 && (a1 || a2 || a3 || a4 || a5)){memset(ht, 0, sizeof(ht));for(int x1 = -50;x1 < 51;x1 ++){if(x1 == 0) continue;for(int x2 = -50;x2 < 51;x2 ++){if(x2 == 0) continue;sum = -1 * (a1 * x1* x1* x1 + a2 * x2* x2* x2) + MAX_S;//将负的变为正的 ++ ht[sum];//将前两项的和的个数保存在数组中 }}int ans = 0;for(int x3 = -50;x3 < 51;x3 ++){if(x3 == 0) continue;for(int x4 = -50;x4 < 51;x4 ++){if(x4 == 0) continue;for(int x5 = -50;x5 < 51;x5 ++){if(x5 == 0) continue;sum = a3 * x3* x3* x3 + a4 * x4* x4* x4 + a5 * x5* x5* x5 + MAX_S;if(sum >= 0 && sum < 2 * MAX_S && ht[sum]){//sum<0或者sum小于前两项和的二倍说明不会相等 ans += ht[sum];}}}}cout << ans << endl;}return 0;}

构造散列函数，用散列表空间上节省好多

#include <cstdio>#include <cstring>using namespace std;const int MOD = 100003;const int MAX_N = 1000000 + 5;int hash[MOD + 1];//hash[key]保存当前关键字key的对应data的下标 int data[MAX_N], next[MAX_N];//data[i]枚举的第i个前三项和的值，next[i]保存同一关键字下data的下标int sz;void insert(int x){data[sz] = x;int t = x > 0 ? x : -x;int key = (t / MOD + t % MOD) % MOD;//哈希函数 next[sz] = hash[key];hash[key] = sz;sz ++;}int find(int x){int t = x > 0 ? x : -x;int key = (t / MOD + t % MOD) % MOD;int dex = hash[key], count = 0;while(dex > 0){ //同一关键下搜索 if(data[dex] == x)count ++;dex = next[dex];}return count;}int main(){int a1, a2, a3, a4, a5;while(~scanf("%d%d%d%d%d", &a1, &a2, &a3, &a4, &a5)){int i, j, k, ans = 0;sz = 1;memset(hash, 0, sizeof(hash));//初始化为每个关键子的下标都为0 for(i = -50;i <= 50;i ++){if(!i)continue;for(j = -50;j <= 50;j ++){if(!j)continue;for(k = -50;k <= 50;k ++){if(!k)continue;insert(a1 * i * i * i + a2 * j * j * j + a3 * k * k * k);}}}for(i = -50;i <= 50;i ++){if(!i)continue;for(j = -50;j <= 50;j ++){if(!j)continue;ans += find(-(a4 * i * i * i + a5 * j * j * j));}}printf("%d\n", ans);}return 0;}

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