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hdu1496——哈希妙用(暴力+优化)
阅读量:659 次
发布时间:2019-03-15

本文共 1410 字,大约阅读时间需要 4 分钟。

题目链接:

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 -41 1 1 1

Sample Output

390880

这个题直接暴力求x1,x2,x3,算出x4,判断一下,再加一个特判就可以过,不过用哈希来实现更简单,

#include 
#include
#include
#define ll long longusing namespace std;const int maxn=1e6+7;int hash1[maxn],hash2[maxn];int main(int argc, char** argv) { 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&&d<0&&c<0)){ printf("0\n"); continue; } int ans=0; memset(hash1,0,sizeof(hash1)); memset(hash2,0,sizeof(hash2)); for(int i=1;i<=100;++i){ for(int j=1;j<=100;++j){ int x=a*i*i+b*j*j; if(x>=0) hash1[x]++; else hash2[-x]++; } } for(int i=1;i<=100;++i){ for(int j=1;j<=100;++j){ int x=c*i*i+d*j*j; if(x>0) ans+=hash2[x]; else ans+=hash1[-x]; } } printf("%d\n",ans*16); } return 0;}

 

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