HDU 4282 A very hard mathematic proble（暴力最適化）
2409 ワード
原題:http://acm.hdu.edu.cn/showproblem.php?pid=4282
A very hard mathematic proble m
Time Limit：2000/1000 MS（Java/Others） メモリLimit：32768/32768 K（Java/Others）Total Submission（s）：3871 Acceepted Submission(s)：1133
Problem Description
Haoron is very goodt at soliving mathematic probles.Today he is working a problem like this:
Find three positive integers X，Y and Z(X1)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 to Haoron.Now he wants to challing more：What’s the total number of different solution？
Surpringly、he is unable to solive this one.It seems that it's really a very hard mathematic proble.
Now，it’s your turn.
Input
The re 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 line for each test case.
Sample Input
题意：求出符合上面式子的个数
A very hard mathematic proble m
Time Limit：2000/1000 MS（Java/Others） メモリLimit：32768/32768 K（Java/Others）Total Submission（s）：3871 Acceepted Submission(s)：1133
Problem Description
Haoron is very goodt at soliving mathematic probles.Today he is working a problem like this:
Find three positive integers X，Y and Z(X
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 to Haoron.Now he wants to challing more：What’s the total number of different solution？
Surpringly、he is unable to solive this one.It seems that it's really a very hard mathematic proble.
Now，it’s your turn.
Input
The re 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 line for each test case.
Sample Input
9 53 6 0
Sample Output
1 1 0
AC代码：
#include
#include
#define LL __int64
/*
author:YangSir
time:2014/5/9
*/
LL qpow(LL x,LL y)
{
LL temp=x,i;
for(i=2;i<=y;i++)
temp*=x;
return temp;
}
int main()
{
LL n,m,x,y,z,k,s,num;
while(~scanf("%I64d",&k)&&k)
{
num=0;
s=sqrt(k);
if(s*s==k)
{
num+=(s-1)/2;// z 2 ， ， s x、y （x=k/2)// xk)//
break;
if(qpow(x,z)+qpow(y,z)+x*y*z==k)
{
num++;
break;//
}
}
}
}
printf("%I64d
",num);
}
return 0;
}