Question

按下面的公式求s的近似值C语言 s=1+1/(3*3)+1/(5*5)+1/(7*7)……+1/（n*n） n的值由键盘输入

Answer 1

#include<stdio.h>
int main()
{
int i, j, n;
float s = 0;
scanf("%d",&n);
for(i = 1; i <= n; i+=2)
{
j = i*i;
s += 1.0/j;
}
printf("%lf\n",s);
return 0;
}

Answer 2

#include <iostream>
#include <cstdio>
using namespace std;
double formula(unsigned short n){
//按下面的公式求s的近似值C语言 s=1+1/(3*3)+1/(5*5)+1/(7*7)……+1/（n*n） n的值由键盘输入
double sum=0.0;
unsigned int i;
for(i=1;i<=n;i+=2){
sum+=1.0/(i*i);
}
return sum;

}
int main()
{
unsigned short n;
printf("输入s:");
scanf("%u",&n);
printf("计算得:%.16f",formula(n));
return 0;
}=============
输出
输入s:3
计算得:1.1111111111111112
输入s:7
计算得:1.1715192743764173
输入s:50
计算得:1.2237018827237791

Answer 3

#include "Stdio.h"
int main(void)
{
long sum=1;
int i,n;
scanf("%d",&n);
for(i=3;i<=n;i+=2)
sum+=1.0/(i*i);
printf("%ld",sum);
getch();
return 0;
}
我想那是可以的

Answer 4

#include <stdio.h>
int main()
{
double s;
int n;
int i;
printf("input n:\n");
scanf("%d",&n);
s = 0;
for(i=1;i<=n;i++,i++)
{
s += 1.0/(i*i);
}
printf("s = %f\n",s);
return 0;
}