Poj 1840

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

解析：我使用的是hash，来标记，将数据分成两部分,用cnt来记录满足等式的个数，例如：sum1=a1*x1*x1*x1+a2*x2*x2*x2与sum2=a3*x3*x3*x3+a4*x4*x4*x4+a5*x5*x5*x5；比如sum1=125,如果sum2=125,那么cnt+=hash[sum2],因为125与-125的个数是相等的，就是加个负号而已。这里注意标记的时候要下标可能为负号（将值扩大+maxn）详细见代码。

#include<cstdio>#include<cstring>#include<string>#include<cmath>#include<algorithm>using namespace std;const int maxn=12500000;short has[maxn*2+1];int a1,a2,a3,a4,a5;int main(){    while(scanf("%d%d%d%d%d",&a1,&a2,&a3,&a4,&a5)!=EOF)    {        int cnt=0;        //memset(has,0,sizeof(has));        for(int x1=-50; x1<=50; x1++)            if(x1!=0)            {                for(int x2=-50; x2<=50; x2++)                    if(x2!=0)                        has[(a1*x1*x1*x1+a2*x2*x2*x2)+maxn]++;            }        for(int x3=-50; x3<=50; x3++)        {            if(x3!=0)            {                for(int x4=-50; x4<=50; x4++)                    if(x4!=0)                    {                        for(int x5=-50; x5<=50; x5++)                        {                            if(x5!=0)                                {int sum=a3*x3*x3*x3+a4*x4*x4*x4+a5*x5*x5*x5;                            if(sum>=-12500000&&sum<=12500000)                                cnt+=has[sum+maxn];}                        }                    }            }        }        printf("%d\n",cnt);    }    return 0;}