C语言 求负数的个数

#includevoidmain(){inta[10],i,count1=0,count2=0,sum1=0,sum2=0;floataverage1,average2;for(i=0;i

mian(){floats1=0,s2=0,x;intn1=0,n2=0,i;for(i=0;i0){s1+=x;n1++;}if(x

修改如下：#include"stdio.h"intmain(){ints[40],i=0,j,num=0,a=0,b=0,c=0;for(i=0;i

#include"stdio.h"#defineN10//N个数中最大数的.voidmain(){inti,a,max;\x09scanf("%d",&a);\x09max=min=a;\x09for

#include"stdio.h"main(){floata[20],avg=0.00;inti;printf("Pleaseinputdatas!\n");for(i=0;i

#includeintmain(){\x09intnum;\x09intsum=0;\x09intcount=0;\x09while(scanf("%d",&num)&&num!=-1)//一直接受输

#include"stdio.h"intispal(intn){\x09intm=0;\x09intt=n;\x09while(t)\x09{\x09\x09m=m*10+t%10;\x09\x09t

#includevoidmain(){\x05inti,a=0,b=0,c=0,sum=0;intx[20]={-1,2,-3,4,0,23,145,-123,42,123,145,12,0,-24,

a%b中输出结果的正负只与a也就是分子有关,所以当a是负数时你把它转成正的就可以了#includeintmain(){inta,b;scanf("%d%d",&a,&b);printf("%d",a>

#include#defineRow50//行数最大值voidmain(){inti=0;intj=0;//j为临时变量intarray[Row][101];//二维数组//输入数组do{printf

#includevoidmain(){inta,b,c;intn;doubleimport;while(scanf("%d",&n)!=EOF&&n!=0){//instala=b=c=0;while

#includeintmain(){\x09inta=0,b=0,c=0,n,m;//分别表示正数,负数,零\x09printf("请输入数字的个数\n");\x09scanf("%d",&n);\x

//下面的程序经测试是符合你要求的.#includeintmain(){inta,i=0;floatavg=0.0;scanf("%d",&a);if(a==-1)return0;while(1){a

#includeintmain(){intn,i;doublesum=0,temp;scanf("%d",&n);for(i=0;i

#includeintmain(){printf("Pleaseinput10number.\n");inta;scanf("%d",&a);intmax=a,min=a;for(inti=0;i

#include#defineN100intmain(){inta[N],i,j,count1,count2;floatave1,ave2;count1=count2=ave1=ave2=0;for(

#includemain(){inta,b,c,d,n=0;scanf("%d%d%d%d",&a,&b,&c,&d);n=(a+b+c+d)/4;printf("%d\n",n);}再问：不是

最大公约数基本都采用辗转法intgongyueshu(intx,inty){inta;while(x%y){a=x%y;x=y;y=a;}returna;}也就是说用第一个数不断的取模第二个数如果除尽

先取绝对值求余,再按照被除数的符号,如(-10)%3等于-1,10%(-3)等于1

int精度限制,int数计算是有最大值,超过这个最大值就会地址溢出,计算的结果就是错误的再问：����Ϊ˫����Ҳ����再答：˫����Ҳ�з�Χ���㳬�����Ҳ����?���Ū�����Χ