Hash和枚舉 解決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.
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
用這個來做hash的入門是再好不過了
在網上也看到很有關hash的代碼,但是
總是看不懂呀,心里老是不明白是怎么
一回事,自從看到這個代碼之后,一下就
大悟了。哈哈。
View Code
View Code #include <iostream> #include <cmath> #define size 200005 using namespace std; int hash[size][10],num[size]; int main() { memset(num, 0, sizeof(num)); int i,j,k,l; int count=0; int temp,mark; int a[5]; scanf("%d%d%d%d%d",&a[0],&a[1],&a[2],&a[3],&a[4]); for(i=-50;i<=50;i++) for(j=-50;j<=50;j++){ if(i!=0 && j!=0){ temp=a[0]*i*i*i+a[1]*j*j*j; mark=abs(temp)%size;//設置key值 hash[mark][num[mark]]=temp;//建立哈希表:mark為key信息,temp為要存儲的信息 num[mark]++;//防止沖突 } } for(i=-50;i<=50;i++) for(j=-50;j<=50;j++) for(k=-50;k<=50;k++){ if(i!=0 && j!=0 && k!=0){ temp=a[2]*i*i*i+a[3]*j*j*j+a[4]*k*k*k; mark=abs(temp)%size; for(l=0;l<num[mark];l++){ //匹配 if(temp==hash[mark][l]) count++; } } } printf("%d\n",count); return 0; }
posted on 2011-09-26 19:21 More study needed. 閱讀(250) 評論(0) 收藏 舉報

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