HDUOJ2199 Can you solve this equation?
Canyousolve
this
equation
?
TimeLimit: 2000 / 1000 MS(Java / Others)MemoryLimit: 32768 / 32768 K(Java / Others)
TotalSubmission(s): 322 AcceptedSubmission(s): 148
ProblemDescription
Now,giventheequation 8 * x ^ 4 + 7 * x ^ 3 + 2 * x ^ 2 + 3 * x + 6 == Y,canyoufinditssolutionbetween 0 and 100 ;
Nowplease try yourlucky.
Input
ThefirstlineoftheinputcontainsanintegerT( 1 <= T <= 100 )whichmeansthenumberoftestcases.ThenTlinesfollow,eachlinehasarealnumberY(fabs(Y) <= 1e10);
Output
Foreachtest case ,youshouldjustoutputonerealnumber(accurateupto 4 decimal places),which is thesolutionoftheequation,or“Nosolution ! ”, if there is nosolution for theequationbetween 0 and 100 .
SampleInput
2
100
- 4
SampleOutput
1.6152
Nosolution !
TimeLimit: 2000 / 1000 MS(Java / Others)MemoryLimit: 32768 / 32768 K(Java / Others)
TotalSubmission(s): 322 AcceptedSubmission(s): 148
ProblemDescription
Now,giventheequation 8 * x ^ 4 + 7 * x ^ 3 + 2 * x ^ 2 + 3 * x + 6 == Y,canyoufinditssolutionbetween 0 and 100 ;
Nowplease try yourlucky.
Input
ThefirstlineoftheinputcontainsanintegerT( 1 <= T <= 100 )whichmeansthenumberoftestcases.ThenTlinesfollow,eachlinehasarealnumberY(fabs(Y) <= 1e10);
Output
Foreachtest case ,youshouldjustoutputonerealnumber(accurateupto 4 decimal places),which is thesolutionoftheequation,or“Nosolution ! ”, if there is nosolution for theequationbetween 0 and 100 .
SampleInput
2
100
- 4
SampleOutput
1.6152
Nosolution !
题目分析:
很明显,这是一个2分搜索的题目, 但是注意下题目的数据!! 1e10 的实数!! 而且精度是要求在 0.0001 . 所以就算是2分数据量依旧比较大,如果用
通常的递归方法吗很遗憾 , RE了............. 没办法, 只能循环了.
下面的是递归 RE 的代码 :
通常的递归方法吗很遗憾 , RE了............. 没办法, 只能循环了.
下面的是递归 RE 的代码 :
#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代码如下: