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  1. #include <stdio.h>
  2. #include <math.h>
  3. int main(void) {
  4. 	// your code goes here
  5. 	int i=0;
  6. 	float a=0,t=0,b=0;
  7. 	for(i=0;i<=10;i++){
  8. 		t=0.1*i;
  9. 		a=-2.5*exp(-2*t);
  10. 		b=-2.5+t*5*exp(-2*t);
  11. 		printf("指数関数：e(%f)=%f\n数値解：b(%f)=%f\n",t,a,b);}
  12. 	return 0;
  13. }
  14.  
Success #stdin #stdout 0s 5700KB
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Standard input is empty
stdout
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指数関数：e(0.000000)=-2.500000
数値解：b(-2.500000)=0.000000
指数関数：e(0.100000)=-2.046827
数値解：b(-2.090635)=0.000000
指数関数：e(0.200000)=-1.675800
数値解：b(-1.829680)=0.000000
指数関数：e(0.300000)=-1.372029
数値解：b(-1.676783)=0.000000
指数関数：e(0.400000)=-1.123322
数値解：b(-1.601342)=0.000000
指数関数：e(0.500000)=-0.919699
数値解：b(-1.580301)=0.000000
指数関数：e(0.600000)=-0.752985
数値解：b(-1.596417)=0.000000
指数関数：e(0.700000)=-0.616492
数値解：b(-1.636911)=0.000000
指数関数：e(0.800000)=-0.504741
数値解：b(-1.692414)=0.000000
指数関数：e(0.900000)=-0.413247
数値解：b(-1.756155)=0.000000
指数関数：e(1.000000)=-0.338338
数値解：b(-1.823324)=0.000000
https://ideone.com/J5kC6i
language:
C (gcc 8.3)
created:
2 years ago
visibility:
public

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