题意:求x在(-100<=x<=100)区间上，已知a,b,c,d，满足a*x1^2+b*x2^2+c*x3^2+d*x4^2=0 的情况有多少种

思路：很明显四层for循环肯定超时，可用三层for循环来判断，或是，两两组合，求多少种情况。分正负故最后

要乘上2^4；

题目

Consider equations having the following form: 

a*x1^2+b*x2^2+c*x3^2+d*x4^2=0 
a, b, c, d are integers from the interval [-50,50] and any of them cannot be 0. 

It is consider a solution a system ( x1,x2,x3,x4 ) that verifies the equation, xi is an integer from [-100,100] and xi != 0, any i ∈{1,2,3,4}. 

Determine how many solutions satisfy the given equation. 

Input

The input consists of several test cases. Each test case consists of a single line containing the 4 coefficients a, b, c, d, separated by one or more blanks. 
End of file.

Output

For each test case, output a single line containing the number of the solutions. 

Sample Input

1 2 3 -4
1 1 1 1

Sample Output

39088
0

AC代码

哈希：

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int f1[1000005],f2[1000005];//f1保存得数是正的  f2保存得数是负的
int a,b,c,d;
int main()
{while(~scanf("%d%d%d%d",&a,&b,&c,&d)){if((a>0&&b>0&&c>0&&d>0)||(a<0&&b<0&&c<0&&d<0)) //abcd全部大于0或者小于0，肯定无解。要加上这个，不然超时{printf("0\n");continue;}memset(f1,0,sizeof(f1));memset(f2,0,sizeof(f2));for(int i=1; i<=100; i++)for(int j=1; j<=100; j++){int ans=a*i*i+b*j*j;if(ans>=0) f1[ans]++;else f2[-ans]++;}int sum=0;for(int i=1; i<=100; i++)for(int j=1; j<=100; j++){int ans=c*i*i+d*j*j;if(ans>0) sum+=f2[ans];else sum+=f1[-ans];}printf("%d\n",16*sum);//每个解有正有负，结果有2^4种}return 0;
}

三层for循环：

#include<iostream>
#include<cstdio>
#include<cmath>/*暴力*/
#include<cstring>
using namespace std;
int main()
{int a,b,c,d;while(~scanf("%d%d%d%d",&a,&b,&c,&d)){if(a>0&&b>0&&c>0&&d>0||a<0&&b<0&&c<0&&d<0){printf("0\n");continue;}int sum=0;for(int i=1; i<=100; i++)//因为正负号不同不影响平方的效果，所以只枚举正数for(int j=1; j<=100; j++)for(int l=1; l<=100; l++){int f=a*i*i+b*j*j+c*l*l;if(f%d==0){int h=sqrt(abs((0-f)/d));if(h<=100&&h>0&&h*h==abs((0-f)/d)&&f+d*h*h==0)sum++;}}printf("%d\n",sum*16);}return 0;
}