[c(7,2)×c(23,3)]/c(30,5)=

#includeintmain(){doublesum=1;doublean=1;for(inti=1;i<=1000;i++){if(i%2==0)an=(double)i/(i+1);elsean=(double)(i+1)/i;sum*=an;printf("%f\t",2*sum);//输出每次计算的近似值}printf("\n所求结果的近似值为：%f\n",2*sum);return0;}运行