c语言：实验三 循环结构

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1、0052-数值统计时间限制:1000毫秒内存限制:32768K字节总提次数:2789次成功提交次数:1392次判题规则:严格比较问题描述统计给定的n个数中，负数、零和正数的个数。输入输入数据有多组，每组占一行，每行的第一个数是整数n（n<100），表示需要统计的数值的个数，然后是n个实数；如果n=0，则表示输入结束，该行不做处理。输出对于每组输入数据，输出一行a,b和c，分别表示给定的数据中负数、零和正数的个数。输入样列60123-10512340.50输出样例123005officiallyestablishedonJuly1,2013,Yibincity,formerlyknownasth

2、ebus,integratedoriginalrongzhoubuscompanyinYibincityandMetrobuscompany,formedonlyinYibincityofaState-ownedpublictransportenterprises,thecompanyconsistsofoneortwo,thirdDivision.Integrationofpublictransportservicesisnotyetestablished出处ymcofficiallyestablishedonJuly1,2013,Yibincity,formerlyknownasthebu

3、s,integratedoriginalrongzhoubuscompanyinYibincityandMetrobuscompany,formedonlyinYibincityofaState-ownedpublictransportenterprises,thecompanyconsistsofoneortwo,thirdDivision.Integrationofpublictransportservicesisnotyetestablished01.#include02.intmain()03.{04.    inta,b,c,n,i;05.    floatx;06

4、.  07.    while(1)08.    {09.        scanf("%d",&n);10.        if(n==0)11.            break;12.        else13.        a=b=c=0;14.        for(i=1;i<=n;i++)15.        {16.        scanf("%f",&x);17.        if(x<0)18.            a++;19.        elseif(x==0)20.            b++;21.        elseif(x>0)22.    

5、        c++;23.        }24.      printf("%d%d%d",a,b,c);25.    }26.    return0;27.  28.}officiallyestablishedonJuly1,2013,Yibincity,formerlyknownasthebus,integratedoriginalrongzhoubuscompanyinYibincityandMetrobuscompany,formedonlyinYibincityofaState-ownedpublictransportenterprises,thecompanyconsis

6、tsofoneortwo,thirdDivision.Integrationofpublictransportservicesisnotyetestablished0147-阶乘输出时间限制:1000毫秒内存限制:32768K字节总提次数:2214次成功提交次数:1594次判题规则:严格比较问题描述请输出如下所示的阶乘公式3!+4!+5!+6!+7!这个公式表示从3到7的阶乘之和。现在给定两个整数a和b(0

7、样列37输出样例3!+4!+5!+6!+7!出处officiallyestablishedonJuly1,2013,Yibincity,formerlyknownasthebus,integratedoriginalrongzhoubuscompanyinYibincityandMetrobuscompany,formedonlyinYibincityofaState-ownedpublictra