poj 1840 Eqs

poj 1840 Eqs,第1张

poj 1840 Eqs
#include <iostream>#include <cstdio>#include <cstdlib>#include <cstring>#include <map>using namespace std;int a[6];map<int, int> h;int main(){for (int i =1; i <6; i++)        scanf("%d", &a[i]);    for (int i =-50; i <=50; i++)        for (int j =-50; j <=50; j++)        { if (i ==0|| j ==0)     continue; int temp = i * i * i * a[1] + j * j * j * a[2]; if (h.find(temp) == h.end())     h[temp] =1; else     h[temp]++;        }    int ans =0;    for (int i =-50; i <=50; i++)        for (int j =-50; j <=50; j++) for (int k =-50; k <=50; k++) {     if (i ==0|| j ==0|| k ==0)         continue;     int temp = i * i * i * a[3] + j * j * j * a[4] + k * k * k  * a[5];     if (h.find(0- temp) == h.end())         continue;     ans += h[0- temp]; }    printf("%dn", ans);    return 0;}

欢迎分享,转载请注明来源:内存溢出

原文地址:https://54852.com/zaji/4925928.html

(0)
打赏 微信扫一扫微信扫一扫 支付宝扫一扫支付宝扫一扫
上一篇 2022-11-12
下一篇2022-11-12

发表评论

登录后才能评论

评论列表(0条)

    保存