C语言:π/4=1-1/3+1/5-…求π的近似值,直到最后一位的绝对值小于10^(-5),并统计累加项数,下面哪里错了#include#includevoid main(){int n=1,count=1,k=0;double sum,term;sum=0;term=1;do{term=count/n;sum=sum+term;n=n+

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C语言:π/4=1-1/3+1/5-…求π的近似值,直到最后一位的绝对值小于10^(-5),并统计累加项数,下面哪里错了#include#includevoid main(){int n=1,count=1,k=0;double sum,term;sum=0;term=1;do{term=count/n;sum=sum+term;n=n+
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C语言:π/4=1-1/3+1/5-…求π的近似值,直到最后一位的绝对值小于10^(-5),并统计累加项数,下面哪里错了#include#includevoid main(){int n=1,count=1,k=0;double sum,term;sum=0;term=1;do{term=count/n;sum=sum+term;n=n+
C语言:π/4=1-1/3+1/5-…求π的近似值,直到最后一位的绝对值小于10^(-5),并统计累加项数,下面哪里错了
#include
#include
void main()
{
int n=1,count=1,k=0;
double sum,term;
sum=0;
term=1;
do
{
term=count/n;
sum=sum+term;
n=n+2;
count=-count;
k++;
} while(n

C语言:π/4=1-1/3+1/5-…求π的近似值,直到最后一位的绝对值小于10^(-5),并统计累加项数,下面哪里错了#include#includevoid main(){int n=1,count=1,k=0;double sum,term;sum=0;term=1;do{term=count/n;sum=sum+term;n=n+
把count定义成double类型的.如下:
#include<stdio.h>
#include<math.h>
void main()
{
    int n=1,k=0;
    double count=1,sum,term;
    sum=0;
    term=1;
    do
    {
        term=count/n;
        sum=sum+term;
        n=n+2;
        count=-count;
        k++;
    } while(n<100000);
    printf("pi=%f,k=%d\n",sum*4,k);
}