POJ1840Eqs（哈希表）

Eqs 
Time Limit: 5000MS Memory Limit: 65536K 
Total Submissions: 16763 Accepted: 8223 
Description

Consider equations having the following form: 
a1x13+ a2x23+ a3x33+ a4x43+ a5x53=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 

Source

思路：分成两段循环降低时间复杂度，避免超时。

#include <stdio.h>#include <string.h>int hash[25000001];int main(){    int a1,a2,a3,a4,a5,x1,x2,x3,x4,x5,sum;    while(~scanf("%d%d%d%d%d",&a1,&a2,&a3,&a4,&a5))    {        memset(hash,0,sizeof(hash));        for(x1 = -50; x1<=50; x1++)        {            if(!x1)                continue;            for(x2 = -50; x2<=50; x2++)            {                if(!x2)                {                    //printf("x2=%d\n",x2);                    continue;                }                sum = -1*(a1*x1*x1*x1+a2*x2*x2*x2);                if(sum<0)                    sum+=25000000;//数字下标没有负数，换一种出储存方式                hash[sum]++;            }        }        int cnt = 0;        for(x3 = -50; x3<=50; x3++)        {            if(!x3)                continue;            for(x4 = -50; x4<=50; x4++)            {                if(!x4)                    continue;                for(x5 = -50; x5<=50; x5++)                {                    if(!x5)                        continue;                    sum = a3*x3*x3*x3+a4*x4*x4*x4+a5*x5*x5*x5;                    if(sum<0)                        sum+=25000000;                    cnt+=hash[sum];                }            }        }        printf("%d\n",cnt);    }    return 0;}