root() 根的多项式函数。matlab里面roots和root roots：求多项式的根。例：求P（x）=5x^4+4x^3+3x^2+2x+1的零点。程序如下： P=[5 4 3 2 1]; %多项式各项的系数 roots(p) %求零点，也就是多项式的解 运行后得结果： ans = 0.1378 + 0.6782i 0.1378 - 0.6782i -0.5378 + 0...
int roots(float a,float b,float c);int main() { int T;float a,b,c;printf("T = ");scanf("%d",&T);while(T--) { printf("请输入二次函数的系数a,b,c:");scanf("%f,%f,%f",&a,&b,&c);roots(a,b,c);} return 0;} int roots(float a,float b,float c) { d...
以下是用C语言编写的三个函数,用于计算一元二次方程的根。这三个函数分别为:计算判别式的函数、计算实根的函数和计算虚根的函数。 #include<stdio.h> #include<math.h> //函数声明 doublecalculateDiscriminant(doublea,doubleb,doublec); voidcalculateRealRoots(doublea,doubleb,doublec); voidcalculateComplex...
printf("real roots:x1=%7.2f,x2=%7.2f",x1,x2); }return0; } 输入成绩,若成绩>=90分,输出A;若成绩>=80分,输出B;若成绩>=70分,输出C;若成绩》=60分,输出D;若成绩<60分,输出E。 #include <stdio.h>intmain(){intscore; scanf("%d",&score);switch(score/10) {case:10case9:printf("A\...
using System;class RootOfSquare{ static (double x1, double x2)? Roots(double a, double b, double c) { double delta = b * b - 4 * a * c; if (delta < 0) return null; else return ((-b + Math.Sqrt(delta)) / (2 * a), (-b - Math.Sqrt...
x2=(-b-sqrt(disc))/(2*a);printf("has distincr real roots:%8.4f,%8.4f\n",x1,x2);}else{rea=-b/(2*a);ima=sqrt(-disc)/(2*a);printf("has complex roots:\n");printf("%8.4f+%8.4fi\n",rea,ima);printf("%8.4f-%8.4fi\n",rea,ima);}getch();return 0;}...
int roots; double weight; public: Stump(int r, double w) : roots(r), weight(w) {} }; Stump s1(3, 15.6); // 圆括号 Stump s2{3, 15.6}; // C++11 Stump s3 = {3, 15.6}; // C++11 8、元组 9、后置返回类型(tailing return type) 10、override和final 11、智能指针(smart pointe...
main(){float a,b,c,s1,s2,x1,x2;printf("please input the number:\n");/*scanf("%f,%f,%f",a,b,c);把这句改为下面的语句，输入数字用空格隔开 */ scanf("%f%f%f",&a,&b,&c);/*其实就是去掉了%f,%f之间的逗号，并在a，b，c之前加了区地址符*/ s1=-b+sqrt(b*b-4*a...
printf("the 2 roots of the function are:\n"); for(i=0;i<3;i++) printf("r[%d]=%.2f\n",i,r[i]); for(i=0;i<3;i++) if(fabs(r[i]*r[i]*r[i]+a[0]*r[i]*r[i]+a[1]*r[i]+a[2])<1e-6) j++; if(j==3) printf("congratulations, the answer are all right...
root2=(-b-sqrt(square))/(2*a); //第二个实数根 printf("Two real roots are %.2f ...