东大OJ-一元三次方程解的个数

1043: Fixed Point

时间限制: 5 Sec 内存限制: 128 MB
提交: 26 解决: 5
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题目描述

In mathematics, a fixed point (sometimes shortened to fixpoint, also known as an invariant point) of a function is a point that is mapped to itself by the function. That is to say, x is a fixed point of the function f if and only if f(x) = x. For example, if f is defined on the real numbers by f(x)=x*x-3*x+4.,then 2 is a fixed point of f, because f(2) = 2.

Not all functions have fixed points: for example, if f is a function defined on the real numbers as f(x) = x + 1, then it has no fixed points, since x is never equal to x + 1 for any real number. In graphical terms, a fixed point means the point (x, f(x)) is on the line y = x, or in other words the graph of f has a point in common with that line. The example f(x) = x + 1 is a case where the graph and the line are a pair of parallel lines.

------ http://en.wikipedia.org/wiki/Fixed_point_%28mathematics%29

Our problem is,for a defined on real number function:

f(x)=a*x*x*x+b*x*x+c*x+d,how many different fixed points does it have?

输入

There are multiple test cases.For each test cases, there are four integers a, b, c and d in a single line.
You can assume that -2¹³<=a<=2¹³, -2¹³<=b<=2¹³, -2¹³<=c<=2¹³, -2¹³<=d<=2¹³,and the number of the result is countable.

输出

For each test case, output the answer in a single line.

样例输入

3 111 793 -3456
5 -135 811 0
-1 9 10 -81

样例输出

3
3
3

#include<iostream>
#include<math.h>
using namespace std;
int main(){freopen("in.txt", "r", stdin);double a3, b3, c3, d3;double a2, b2, c2;while (cin >> a3 >> b3 >> c3 >> d3){c3--;a2 = 3 * a3; b2 = 2 * b3; c2 = c3;if (a3 == 0){if (b3 == 0)//一次方程{if (c3 == 0){//只剩常数cout << 0 << endl;}else cout << 1 << endl;}else{//二次方程double delta = c3*c3 - 4 * b3*d3;if (delta == 0)cout << 1 << endl;else if (delta>0)cout << 2 << endl;else cout << 0 << endl;}}else{//三次方程double delta = b2*b2 - 4 * a2*c2;if (delta <= 0)//单调递增cout << 1 << endl;else{double x1 = (-b2 - sqrt(delta)) / (2 * a2);double x2 = (-b2 + sqrt(delta)) / (2 * a2);double y1 = a3*x1*x1*x1 + b3*x1*x1 + c3*x1 + d3;double y2 = a3*x2*x2*x2 + b3*x2*x2 + c3*x2 + d3;double see = y1*y2;if (see == 0)cout << 2 << endl;else if (see > 0)cout << 1 << endl;else cout << 3 << endl;}}}return 0;
}

转载于:https://www.cnblogs.com/weiyinfu/p/5013901.html