编写程序,计算1+1/3+1/5+...+1/2007的值

public double cal()

double sum=0;

for(i=1;i<2007;i+=2)

{

sum+=1.0/i

}

return sum;

}

//编写程序,计算1+1/3+1/5+...+1/2007的值 #include <stdio.h>

int main( void ) { int count = 1; float sum,num; sum = 0; num = 1; while( num <= 2007 ) { sum += 1/num; printf( "sum%d = %2.2f\n", count, sum); count++; num = num + 2; } printf( "一共执行了%d次且sum = %2.2f\n", count-1, sum); return 0; }