Can you solve this equation?
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Now please try your lucky.
2100-4
1.6152No solution!
思路:题目中的关于x的式子是单调递增的,当<f(0) || >f(100)时不存在,所以可以采用二分法不断缩小范围,但要注意二分法的终止条件
#include <stdio.h>
#include <math.h>
double f(double x)
{
return (8.0*x*x*x*x+7.0*x*x*x+2*x*x+3*x+6);
}
int main()
{
//freopen("e:\\in.txt","r",stdin);
int T;
double re;
scanf("%d",&T);
while(T--)
{
scanf("%lf",&re);
if(f(0)>re||f(100)<re)
printf("No solution!\n");
else
{
double low=0.0,high=100.0,middle=50.0;
while(fabs(f(middle)-re)>1e-5)
{
if(f(middle)>re)
high=middle;
else if(f(middle)<re)
low=middle;
middle=(low+high)/2;
}
printf("%.4lf\n",middle);
}
}
return 0;
}
要点:1.二分法的终止条件:fabs()<1e-5,意义为当其小于0时,(二分法的终止条件要根据题目来判断)
2.从键盘上输入的是整数,但如果它的类型是double,则实际是浮点型数字。
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