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#include <stdio.h>
int
main(
const int argc,
const char *argv[]
)
{
double dEnd = 0.01;
double sum = 0.0;
double value;
int idx = 1;
while (1) {
value = 1.0 / idx;
printf("1/%d", idx);
if (value - dEnd < 0.00000000001) {
break;
}
printf("%c", (idx % 3 == 0 ? '+' : '-'));
sum += value * (idx % 2 == 0 ? -1 : 1);
idx += 2;
}
printf("=%f\n", sum);
return 0;
}
#include <stdio.h>
#include <stdlib.h>
int main( )
{
int flag;
double sum=0.0;
double n=1;
int num=0;
while(1/n>0.00001)
{
num++;
flag = num%2;
if (flag==0)
{
sum -= 1/n;
}
else
{
sum += 1/n;
}
n += 2;
}
printf("\n%f\n", sum);
return 1;
}
#include <stdio.h>
#include <math.h>
void main()
{
// C语言求1-1/3+1/5-1/7...直到最后项的绝对值<10的-5次方
double dResult = 0.0;
long dDeno = 1;
int i = 0;
int tag = 1;
double dValueN = 0.0;
do
{
dValueN = 1.0 / dDeno;
dResult += tag *dValueN;
dDeno += 2;
tag = -tag;
}while(!(fabs(dValueN) < 10e-5));
printf("1-1/3+1/5-1/7... = %lf\n", dResult);
}