Posted on 2010-08-11 13:29
MiYu 閱讀(908)
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ACM ( 搜索 )
MiYu原創, 轉帖請注明 : 轉載自 ______________白白の屋
題目地址:
http://acm.hdu.edu.cn/showproblem.php?pid=2199
題目描述:
Can you solve this equation?
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 322 Accepted Submission(s): 148
Problem Description
Now,given the equation 8*x^4 + 7*x^3 + 2*x^2 + 3*x + 6 == Y,can you find its solution between 0 and 100;
Now please try your lucky.
Input
The first line of the input contains an integer T(1<=T<=100) which means the number of test cases. Then T lines follow, each line has a real number Y (fabs(Y) <= 1e10);
Output
For each test case, you should just output one real number(accurate up to 4 decimal places),which is the solution of the equation,or “No solution!”,if there is no solution for the equation between 0 and 100.
Sample Input
2
100
-4
Sample Output
1.6152
No solution!
題目分析:
很明顯,這是一個2分搜索的題目, 但是注意下題目的數據!! 1e10 的實數!! 而且精度是要求在 0.0001 . 所以就算是2分數據量依舊比較大,如果用
通常的遞歸方法嗎很遺憾 , RE了............. 沒辦法, 只能循環了.
下面的是遞歸 RE 的代碼 :
#include <iostream>
#include <cmath>
using namespace std;
#define POW(x) ( (x) * (x) )
#define POW3(x) ( POW(x) * (x) )
#define POW4(x) ( POW(x) * POW(x) )
double y = 0;
bool douEql ( double a,double b )
{
if ( fabs( a - b ) <= 1e-6 )
return true;
return false;
}
double cal ( double n )
{
return 8.0 * POW4(n) + 7 * POW3(n) + 2 * POW(n) + 3 * n + 6 ;
}
double biSearch ( double l, double r )
{
if ( douEql ( l,r ) )
{
if ( douEql ( y, cal ( l ) ) )
return l;
return -1;
}
double mid = ( l + r ) / 2.0;
if ( douEql ( y, cal ( mid ) ) )
return mid;
else if ( cal ( mid ) > y )
return biSearch ( l,mid - 0.0001 );
else
return biSearch ( mid + 0.0001, r );
}
int main ()
{
int T;
scanf ( "%d",&T );
while ( T -- )
{
scanf ( "%lf",&y );
if ( cal(0) >= y && cal(100) <= y )
{
printf ( "No solution!\n" );
continue;
}
double res = biSearch ( 0.0, 100.0 );
if ( res == -1 )
printf ( "No solution!\n" );
else
printf ( "%.4lf\n",res );
}
return 0;
}
AC代碼如下:
MiYu原創, 轉帖請注明 : 轉載自 ______________白白の屋
#include <iostream>
#include <cmath>
using namespace std;
#define POW(x) ( (x) * (x) )
#define POW3(x) ( POW(x) * (x) )
#define POW4(x) ( POW(x) * POW(x) )
double y = 0;
double cal ( double n )
{
return 8.0 * POW4(n) + 7 * POW3(n) + 2 * POW(n) + 3 * n + 6 ;
}
int main ()
{
int T;
scanf ( "%d",&T );
while ( T -- )
{
scanf ( "%lf",&y );
if ( cal(0) > y || cal(100) < y )
{
printf ( "No solution!\n" );
continue;
}
double l = 0.0, r = 100.0,res = 0.0;
while ( r - l > 1e-6 )
{
double mid = ( l + r ) / 2.0;
res = cal ( mid );
if ( res > y )
r = mid - 1e-6;
else
l = mid + 1e-6;
}
printf ( "%.4lf\n",( l + r ) / 2.0 );
}
return 0;
}