[NOIP2001 提高组] 一元三次方
- 题目描述
- 思路
- code
tag：数学，二分查找
题目链接：洛谷P1024: [NOIP2001 提高组] 一元三次方
题目描述
思路
先对三次函数求导得到二次函数
3
a
x
2
+
2
b
x
+
c
=
0
3ax^2 + 2bx + c = 0
3ax2+2bx+c=0
然后利用求根公式计算出两个极值点
x
1
x1
x1
x
2
x2
x2
将整个区间划分为三个单调区间
[
−
100
,
x
1
]
,
[
x
1
,
x
2
]
,
[
x
2
,
100
]
[-100, x1],[x1,x2],[x2,100]
[−100,x1],[x1,x2],[x2,100]，在这三个区间中分别使用二分。
令
f
(
x
)
=
该一元三次函数
f(x) = 该一元三次函数
f(x)=该一元三次函数
每次判断
f
(
l
e
f
t
)
f(left)
f(left) *
f
(
m
i
d
)
f(mid)
f(mid)
f
(
l
e
f
t
)
f(left)
f(left) *
f
(
m
i
d
)
<
=
0
:
f(mid)<= 0 :
f(mid)<=0: 说明此时两个函数值异号，让
r
i
g
h
t
=
m
i
d
right=mid
right=mid
否则说明两个函数值同号，让
l
e
f
t
=
m
i
d
left = mid
left=mid
code
#include <bits/stdc++.h>
using namespace std;
const double eps = 1e-4;
double a,b,c,d;
double calc(double x) {
return a*x*x*x + b*x*x + c*x + d;
}
void bsearch(double l, double r) {
while (r - l > eps) {
double mid = (l + r) / 2;
if (calc(mid) * calc(l) <= 0) r = mid;
else l = mid;
}
printf("%.2lf ", l);
}
int main() {
scanf("%lf%lf%lf%lf",&a,&b,&c,&d);
// 3ax^2 + 2bx + c
double x1 = (-2*b - sqrt(4*b*b - 12*a*c)) / (6 * a);
double x2 = (-2*b + sqrt(4*b*b - 12*a*c)) / (6 * a);
// printf("%lf %lf\n", x1, x2);
bsearch(-100, x1);
bsearch(x1, x2);
bsearch(x2, 100);
return 0;
}
