A very hard mathematic problem HDU
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A very hard mathematic problem
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 7411 Accepted Submission(s): 2232
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
Source
2012 ACM/ICPC Asia Regional Tianjin Online
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#include <iostream>#include <cstdio>#include <cstring>#include <cmath>#define maxx 1e9using namespace std;typedef long long ll;ll powmod(ll a, ll b){ ll ans = 1; while(b) { if(b%2==1) ans = (ans * a); b=b/2; a = (a*a); } return ans;}int main(){ int i,j; int k; int z; while(cin>>k, k) { int ans = 0; ll tx,ty; ll lt,rt; ll tk = (int)sqrt(k*1.0); if(tk*tk == k) ans += (tk-1)/2; for(z=3; z<31; z++) { for(ll y=2; ;y++) { ty = powmod(y, z); if(ty > k) break; lt =1; rt=y-1; while(lt<=rt) { ll x= (lt+rt)/2; ll ta=pow(x, z) + pow(y, z) + x*y*z; if(ta < k) { lt++; } else if(ta > k) { rt--; } else { ans++; break; } } } } printf("%d\n", ans); } return 0;}
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