Question

计算:0.01+0.12+0.23+0.34……0.78+0.89 注意，是循环小数。循环点在最后一位。

注意，是循环小数。循环点在最后一位。
例如：0.011111……
0.122222……

Answer 1

0.01+0.12+0.23+0.34……0.78+0.89
=(0.0+0.1+0.2+...+0.8)+(0.0111...+0.0222...+...+0.0999...)
=(0.0+0.1+0.2+...+0.8)+0.0111...*(1+2+...+9)
=0.1*(1+2+...+8)+1/90*(1+2+...+9)
=3.6+1/90*45
=3.6+0.5
=4.1

Answer 2

希望可以帮助你哈~~~
#include<stdio.h>
#include<stdlib.h>
int main(void)
{
int i = 0, sum = 0;
for(i = 1; i < 90; i += 11)
{
sum += i;
}
printf("最后的结果为:\n");
printf("%f\n", (float)(sum*0.01));
return 0;
}

结果：4.050000

Answer 3

0.01+0.12+0.23+0.34……0.78+0.89
=(0.0+0.1+0.2+...+0.8)+(0.0111...+0.0222...+...+0.0999...)
=(0.0+0.1+0.2+...+0.8)+0.0111...*(1+2+...+9)
=0.1*(1+2+...+8)+1/90*(1+2+...+9)
=3.6+1/90*45
=3.6+0.5
=4.1
---------------此为一解-----顶起-----------------

0.01+0.12+0.23+0.34……0.78+0.89
=(0.01+0.78)+（0.12+0.67）+（0.23+0.56）+（0.34+0.45）+0.89
=0.799999999*4+(0.799999999+0.1)
=0.799999999*5+0.1
=(72/90)*5+0.1=4.1

Answer 4

0.011…+0.122…+0.233…+0.344…+……+0.788…+0.899…
=(0.011…+0.788…)+(0.122…+0.677…)+(0.233…+0.566…)+(0.344…+0.455…)+0.9
=0.8+0.8+0.8+0.8+0.9
=4.1

Answer 5

(1+11+21+31+41+51+61+71+81)/90=369/90=4.1