hdu4282 二分

A very hard mathematic problem

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 1685    Accepted Submission(s): 510

Problem Description
　　Haoren is very good at solving mathematic problems. Today he is working a problem like this: 　　Find three positive integers X, Y and Z (X < Y, Z > 1) that holds 　　 X^Z + Y^Z + XYZ = K 　　where K is another given integer. 　　Here the operator “^” means power, e.g., 2^3 = 2 * 2 * 2. 　　Finding a solution is quite easy to Haoren. Now he wants to challenge more: What’s the total number of different solutions? 　　Surprisingly, he is unable to solve this one. It seems that it’s really a very hard mathematic problem. 　　Now, it’s your turn.
 
Input
　　There are multiple test cases. 　　For each case, there is only one integer K (0 < K < 2^31) in a line. 　　K = 0 implies the end of input. 　　
 
Output
　　Output the total number of solutions in a line for each test case.
 
Sample Input
9
53
6
0
 
Sample Output
1
1
0 　　
Hint
9 = 1^2 + 2^2 + 1 * 2 * 2
53 = 2^3 + 3^3 + 2 * 3 * 3
比赛的时候想着数论找规律，终究没找到规律，暴力超时，明明是解方程，二分，却想不起来，越来越脑残了！！
View Code 
 1 #include<cmath>
 2 #include<cstdio>
 3 int n,zz;
 4 int q_pow(int a, int k)
 5 {
 6     if(k==0) return 1;
 7     if(k==1) return a;
 8     if(k%2==0) return q_pow(a*a,k/2);
 9     if(k%2==1) return a*q_pow(a,k-1);
10 }
11 int find(int low,int high,int j,int i)
12 {
13     int mid,res;
14     while(low<=high)
15     {
16         mid=(high+low)>>1;
17         res=zz+pow(mid,i)+i*j*mid-n;
18         if(res==0)  return mid;
19         if(res<0)   low=mid+1;
20         else high=mid-1;
21     }
22    return -1;
23 }
24 int main()
25 {
26     while(scanf("%d",&n)!=EOF&&n)
27     {
28        int res=0;
29        for(int i=2;i<=30;i++)
30        {
31            int _max=(int)(pow(n/2.0,1.0/i)+1)+1;
32            for(int j=1;j<=_max;j++)
33            {
34                 if(q_pow(j,i)+q_pow(j+1,i)+j*(j+1)>n)  break;
35                 zz=(int)q_pow(j,i);
36                 int now=find(j+1,(int)(pow(n+0.0,1.0/i)),j,i);
37                 if (now>0)res++;
38            }
39        }
40        printf("%d\n",res);
41     }
42 }

pow函数还是自己定义吧……否则会WA，再WA……

 
 
Source
2012 ACM/ICPC Asia Regional Tianjin Online
posted @ 2012-09-12 16:01  _sunshine  阅读(252)  评论(0)  编辑  收藏  举报