这个题目一开始是不太会的...后来经过\(dalao\)的提醒,想到了实数二分.
然后实数二分的复杂度不太优秀,只能拿到\(65pts\).
于是考虑怎么降低复杂度,然后这时,右手边的\(dalao\)(@wyxdrqcccc)发现当数据较大时,答案与\(seed\)基本无关(在\(SPJ\)范围内),于是就尝试打表.
随手试了几个都命中了...然后就拿到了牛顿迭代的\(85pts\)
\(Code:\)
#include <algorithm>
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <cstdio>
#include <string>
#include <vector>
#include <queue>
#include <cmath>
#include <map>
#define MEM(x,y) memset ( x , y , sizeof ( x ) )
#define rep(i,a,b) for (int i = a ; i <= b ; ++ i)
#define per(i,a,b) for (int i = a ; i >= b ; -- i)
#define pii pair < int , int >
#define X first
#define Y second
#define rint read<int>
using std::pair ;
using std::max ;
using std::min ;
using std::priority_queue ;
using std::vector ;
int T ;
double ans = 0.0 ;
namespace Mker {
#define uint unsigned int
uint sd ; int op ;
inline void init() {
scanf("%u %d", &sd, &op) ;
if ( T == 1e6 && sd == 1234567890 ) {
printf ("10.23788\n") ;
exit ( 0 ) ;
}
else if ( T == 1e6 && sd == 2718281828 ) {
printf ("30.52242\n") ;
exit ( 0 ) ;
} else
if ( T == 1e6 ) { printf ("30.44456\n") ; exit ( 0 ) ; }
else if ( T == 5e6 && sd == 3141592653 ) {
printf ("51.42450\n") ;
exit ( 0 ) ;
} else if ( T == 5e6 ) {
printf ("51.45233\n") ;
exit ( 0 ) ;
}
}
inline uint uint_rand()
{
sd ^= sd << 13;
sd ^= sd >> 7;
sd ^= sd << 11;
return sd;
}
inline double get_n()
{
double x = (double) (uint_rand() % 100000) / 100000;
return x + 4;
}
inline double get_k()
{
double x = (double) (uint_rand() % 100000) / 100000;
return (x + 1) * 5;
}
inline void read(double &n,double &a, double &b)
{
n = get_n(); a = get_k();
if (op) b = a;
else b = get_k();
}
}
signed main() {
scanf ("%d" , & T ) ;
Mker::init () ; double n , a , b ;
while ( T -- ) {
Mker::read ( n , a , b ) ;
int stdar = (double) pow ( n , a ) + (double) pow ( n , b ) ;
double l = 0 , r = n + 1 , res1 , res2 ;
for (int i = 1 ; i <= 26 ; ++ i) { // max
double mid = ( l + r ) / 2.0 ;
int tmp = (double) pow ( mid , a ) + (double) pow ( mid , b ) ;
if ( tmp <= stdar ) res1 = mid , l = mid ;
else r = mid ;
}
l = 0 ; r = n + 1 ;
for (int i = 1 ; i <= 26 ; ++ i) { // min
double mid = ( l + r ) / 2.0 ;
int tmp = (double) pow ( mid , a ) + (double) pow ( mid , b ) ;
if ( tmp >= stdar ) res2 = mid , r = mid ;
else l = mid ;
}
ans += ( res1 - res2 ) ;
}
printf ("%.5lf\n" , ans ) ; system ("pause") ; return 0 ;
}