void main()
{
int i;
int j;
int s=0;
for(i=1;i<102;i+=4)
{
s+=1/i;
}
for(j=3;j<100;j+=4)
{
s-=1/j;
}
printf("1-1/3+1/5+.....+1/101=%d",s);
}本回答被网友采纳
C语言 29 计算 s=1-1\/3 + 1\/5- 1\/7+…1\/101 的值并输出。
while(v<=101){ s+=1.0\/v*flag;flag=-flag;v+=2;} printf("\\n%f",s);}
C语言编程求和S=1-1\/3+1\/5-1\/7+…+1\/n (1\/n<0.0001)
if(i%4==1)y+=1.0\/i;else y-=1.0\/i;printf("%f\\n",y);return 0;}
求1-1\/3+1\/5-1\/7+……1\/n的值。c语言怎么写
s+=i%4==1?1.0\/i:-1.0\/i; printf("%f\\n",s); return 0;}
用C语言编写程序计算并输出数列1-1\/3+1\/5-1\/7+1\/9-1\/11+···+1\/101...
}while(s <= 101);printf("1-1\/3+1\/5-1\/7+1\/9-1\/11+···+1\/101=%f\\n", result);system("pause");return 1;}
C语言帮我看一下这个程序,实现求s=1\/1-1\/3+1\/5-1\/7+1\/9-1\/11+1\/13+...
除了没有定义n，t之外没有问题
C语言程序编写 求s\/4=1-1\/3+1\/5-1\/7+1\/9-1\/11+….,当某项的绝对值小于0...
这个表达式是用来计算圆周率π的近似值的。include <stdio.h>#include <math.h>void main(){double i,k=1,m=0,flag=1;for(i=1;abs(k)>=1e-6;i++){k=flag\/(i*2-1);m+=k;flag=-flag;}printf("s=%f",m*4);}
怎样用C语言求1-1\/3+1\/5-1\/7...的和
include<stdio.h>int main(){int n,i; double s=0; scanf("%d",&n); for(i=1;i<=n;i+=2) if(i%4==1)s+=1.0\/i; else s-=1.0\/i; printf("%lf\\n",s); return 0;}
c语言问题(算1-1\/3+1\/5-1\/7+1\/9)
sum=sum+1\/(2*i-1)*k;改为：sum=sum+1.0\/(2*i-1)*k;
C语言计算数列1 - 1\/3 + 1\/5 - 1\/7 + 1\/9 - 1\/11 + …的前n项之和...
){long n;while (scanf_s("%ld", &n) != -1){double s = 0.0;int i, a = 1;for (i = 1; i <= 2 * n - 1; i += 2){if (((i + 1) \/ 2) % 2 == 1)s += 1.0 \/ i;elses -= 1.0 \/ i;}printf("%.5f\\n", s); \/\/这里改了}return 0;} ...
...编写程序:计算如下分数序列的和,1\/1,1\/3,1\/5,1\/7,..., 1\/101...
float sum=0;int i;for(i=1;i<=101;i=i+2){ sum=sum+1.0\/i;} printf("%f",sum);\/\/在1楼基本上加了两字符，结果就变了 1\/i 和 1.0\/i 是不同的 一个整除一个是符点除法
