解法四: 第六章循环结构程序设计 main( Rint i-l; float sum1=0, sum2=0; while(i<=10) i suml=sum1+1.0/i; i=i+2;} while(i<=10) & sum2=sum2+1.0/i i=i+2;} printf(ofn", suml-sum2);第六章循环结构程序设计 11 解法四： ◼ main( ) ◼ {int i=1;float sum1=0,sum2=0; ◼ while(i<=10) ◼ { sum1=sum1+1.0/i; ◼ i=i+2;} ◼ i=2; ◼ while(i<=10) ◼ { sum2=sum2+1.0/i; ◼ i=i+2;} ◼ printf("%f\n",sum1-sum2); ◼ }