POJ 1840：Eqs
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
芒果君：这道题是裸暴力，但是今天我们有一个新的思路，就是用哈希来优化暴力。我们把五项i=1->5 表示为Xi,那么很明显X1+X2+X3==-X4-X5，这样我们把等式左边存到哈希表，然后让等式右边去找左边，大大减小了枚举的时间复杂度。
1 #include<cstdio> 2 #include<cstring> 3 #include<algorithm> 4 #include<map> 5 #define mod 100003 6 #define ll long long 7 using namespace std; 8 int hl[mod],a1,a2,a3,a4,a5,cnt,ans; 9 int cal(int x){return x*x*x;} 10 struct H{ 11 int val,ne; 12 }Hash[1100000]; 13 void insert(int x) 14 { 15 int key=abs(x)%mod; 16 Hash[++cnt].val=x; 17 Hash[cnt].ne=hl[key]; 18 hl[key]=cnt; 19 } 20 int search(int x) 21 { 22 int sum=0; 23 int key=abs(x)%mod; 24 for(int i=hl[key];i;i=Hash[i].ne) if(Hash[i].val==x) sum++; 25 return sum; 26 } 27 int main() 28 { 29 scanf("%d%d%d%d%d",&a1,&a2,&a3,&a4,&a5); 30 for(int i=-50;i<=50;++i)if(i) 31 for(int j=-50;j<=50;++j)if(j) 32 for(int k=-50;k<=50;++k)if(k) 33 insert(a1*cal(i)+a2*cal(j)+a3*cal(k)); 34 for(int i=-50;i<=50;++i)if(i) 35 for(int j=-50;j<=50;++j)if(j) 36 ans+=search(-a4*cal(i)-a5*cal(j)); 37 printf("%d",ans); 38 return 0; 39 }
