c語(yǔ)言設(shè)計(jì) 分段函數(shù)

#include math.h

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int main()

{

double x,y;

scanf("%lf",x);

if (x0)

y=0.5*(-x);

else

if (x10)

y=exp(x)+3;

else

if(x20)

y=log10(x);

else

if (x30)

y=pow(x,1.5);

else

if (x50)

y=pow (x,0.5)-1;

else

y=3*cos(x);

printf("y=%lf\n",y);

return 0;

}

擴(kuò)展資料

return 0代表程序正常退出。return是C++預(yù)定義的語(yǔ)句，它提供了終止函數(shù)執(zhí)行的一種方式。當(dāng)return語(yǔ)句提供了一個(gè)值時(shí)，這個(gè)值就成為函數(shù)的返回值。

return語(yǔ)句用來(lái)結(jié)束循環(huán)，或返回一個(gè)函數(shù)的值。

1、return 0，說(shuō)明程序正常退出，返回到主程序繼續(xù)往下執(zhí)行。

2、return 1，說(shuō)明程序異常退出，返回主調(diào)函數(shù)來(lái)處理，繼續(xù)往下執(zhí)行。return 0或return 1對(duì)程序執(zhí)行的順序沒(méi)有影響，只是大家習(xí)慣于使用return(0)退出子程序而已。

C語(yǔ)言計(jì)算分段函數(shù)

1. 代碼如下，3）需要實(shí)際運(yùn)行時(shí)輸入測(cè)試

int main(void)

{

double x, y, f;

printf("Please input 2 double number in the form of x y:\n");

scanf("%lf%lf", x, y);

if(x=0 y0)

f = 2*x*x + 3*x +1/(x+y);

else if(x=0 y=0)

f = 2*x*x + 3*x +1/(1+y*y);

else

f = 3*sin(x+y)/(2*x*x) + 3*x + 1;

printf("x=%lf, y=%lf, f(x, y)=%lf\n", x, y, f);

return 0;

}

2.代碼如下

#include stdio.h

#includemath.h

int main(void)

{

double x, y, f;

printf("Please input 2 double number in the form of x y:\n");

scanf("%lf%lf", x, y);

if(x=0)

{

if(y0)

f = 2*x*x + 3*x +1/(x+y);

else

f = 2*x*x + 3*x +1/(1+y*y);

}

else

f = 3*sin(x+y)/(2*x*x) + 3*x + 1;

printf("x=%lf, y=%lf, f(x, y)=%lf\n", x, y, f);

return 0;

}

3.代碼如下

#include stdio.h

int main(void)

{

int score = 0;

printf("Please input a score between 0-100:\n");

scanf("%d", score);

if(score0 || score100)

printf("Wrong input of score!\n");

else if(score=90 score=100)

printf("A\n");

else if(score=80 score=89)

printf("B\n");

else if(score=70 score=79)

printf("C\n");

else if(score=60 score=69)

printf("D\n");

else

printf("E\n");

return 0;

}

怎么用c語(yǔ)言編程一個(gè)分段函數(shù)？

#include

int?main()

{

int?x,y;

scanf("%d",x);

if(0xx10)?y=3*x+2;

else

{if(x=0)?y=0;

else

{if?(x0)?y=x*x;

else?printf("go?die\n");

}

}

printf("%d",y);

return?0;

}該程序的分段函數(shù)如下:

f(x)=3x+2? (0x10)

f(x)=1???????? (x=0)

f(x)?=?x*x??? (x0)

#include stdio.h

#include math.h

void main()

{

float x;

double y;

printf("Please input the value of x:");

scanf("%f",x);

if(x=-10x=4)

{

y=fabs(x-2);

printf("y=%.2f\n",y);

}

else if(x=5x=7)

{

y=x+10;

printf("y=%.2f\n",y);

}

else if(x=8x=12)

{

y=pow(x,4);

printf("y=%.2f\n",y);

}

else

printf("No answer\n");

}