Eqs (哈希)

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 10695		Accepted: 5185

Description

Consider equations having the following form: 
a1x1 ³+ a2x2 ³+ a3x3 ³+ a4x4 ³+ a5x5 ³=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
题意：求出满足上述等式的解的个数，但x1，x2,x3,x4,x5不能为0；

解题思路：第一感觉是五个循环枚举，那样果断会超时，其时我们可以把x1,x1移项，先枚举x1,x2,哈希之后再枚举x3,x4,x5;
但-a1x1-a2x2结果有可能是负值，这时可以-a1x1-a2x2+12500000将结果变为正的；了解了传说中的哈希函数，值得学习；

 1 #include<stdio.h>

 2 #include<string.h>

 3 char hash[25000000];

 4 

 5 int main()

 6 {

 7     int a1,a2,a3,a4,a5;

 8     int x1,x2,x4,x5,x3,ans;

 9     memset(hash,0,sizeof(hash));

10 

11     scanf("%d %d %d %d %d",&a1,&a2,&a3,&a4,&a5);

12 

13     for(x1 = -50; x1 <= 50; x1++)

14     {

15         if(x1 == 0)continue;

16         for(x2 = -50; x2 <= 50; x2++)

17         {

18             if(x2 == 0)continue;

19             ans = -(x1*x1*x1*a1+x2*x2*x2*a2)+12500000;

20             hash[ans]++;

21         }

22     }

23     int cnt = 0;

24     for(x3 = -50; x3 <= 50; x3++)

25     {

26         if(x3 == 0)continue;

27         for(x4 = -50; x4 <= 50; x4++)

28         {

29             if(x4 == 0)continue;

30             for(x5 = -50; x5 <= 50; x5++)

31             {

32                 if(x5 == 0)continue;

33                 ans = x3*x3*x3*a3+x4*x4*x4*a4+x5*x5*x5*a5+12500000;

34                 if(ans >= 0 && ans < 25000000)

35                     cnt += hash[ans];

36             }

37         }

38     }

39     printf("%d\n",cnt);

40     return 0;

41 }

View Code