下面的程序计算
1-1/3+1/5-1/7 …+1/97-1/99+1/101的值
int main()
{ int i;
double s=0;
for(i=1;i<102;i++)
s+=i%2?1.0/i:-1.0/i;
printf("%f\n",s);
return 0;
}
C语言编写程序求1-1\/3 1\/5-1\/7 …1\/99 1\/101的值
1-1\/3+1\/5-1\/7 …+1\/97-1\/99+1\/101的值 include<stdio.h>int main(){ int i; double s=0; for(i=1;i<102;i++) s+=i%2?1.0\/i:-1.0\/i; printf("%f\\n",s); return 0;}
用C语言编写程序,求1-3+5-7...-99+100的值
include"stdio.h"main(){ int i,s=0,f=1;for (i=1;i<=101;i+=2){ s=s+i*f;f=-f;} printf("%d",s);}
用c语言编写程序,求1–3+5–7+...–99+101的值
int i,j,a,s=0;for(i=1,a=0;i<=101;i=i+2,a++){ j=i*pow(-1,a);s=s+j;} printf("%d",s);}
C语言编写程序,求1-3+5-7+...-99+101的值
int i,j,sum1=0,sum2=0;{ for(i=1;i<=101;i+=4)sum1+=i;for(i=3;i<=99;j+=4)sum2+=j;} printf("sum=%d",sum1-sum2);} 方法二、public class Cds6{ public static void main(String[]args){ int sum=0;for(int i=1;i<=101;i+=4){ sum=sum+i;System.out.prin...
C语言编写程序,求1-3+5-7+...-99+101的值
int i = 1, sum1 = 1, j = 3, sum2 = 3, sum;while (i <= 101){ i = i + 4;sum1 += i;} while (j < 101){ j = j + 4;sum2 += j;} sum2 *= -1;sum = sum1 + sum2;printf("%d", sum);} 用for如下 include<stdio.h> void main(){ int i = 1, ...
C语言编写程序,求1-3+5-7+...-99+101的值 用while和for两种方法编写 计...
sign=-sign; } printf("%d\\n", sum);}用while的 include <stdio.h>int main(){ int i=1, sign=1, sum=0; while(i<=101) { sum+=i*sign; sign=-sign; i+=2; } printf("%d\\n", sum);} ...
C语言求用while语句计算1-3+5-7+...-99+101的值
include<stdio.h> void main(){ int i,j,m,n,sum=0;m=0;n=0;i=1;j=3;while(i<102){ sum=sum+i-j;i+=4;if(j >= 99){ j =0;} else j+=4;} printf("%d\\n",sum);}
c语言的题目:求和运算:1-1\/3+1\/5-1\/7+...+1\/99,
我直接写空的内容：pow(-1,n-1)\/(2*n-1)pow是计算-1的n-1次方的C函数。要包含math.h头文件
试用编程求F=1-\/3+1\/5-1\/7+1\/9-1\/11+1\/13-……1\/99的
java版本的 public class Test{ public static void main(String[] args) { System.out.println(sumA(99)); } public static double sumA(int max) { double sum = 0; \/\/ 符号位 int x = 1; for (int i = 1; i <= max; i += 2) { sum += (1.0...
C语言求 1-3+5-7+...-99+101的值
include<stdio.h>void main(){ int i,sum=0; int j = 1; \/\/ 这里最多循环51 for (i=1; (2*i -1) < 102; i++) { if(i%2!=0) { sum=sum + (2*i -1); printf("+%d", (2*i -1)); } else { sum=sum-(2*i -1); ...
