C语言编程,判断一个数是否存在于一个数组中

2025-03-16 10:24:18
推荐回答(4个)
回答1:

#define _CRT_SECURE_NO_DEPRECATE
#include
#include
#include
#include

int findarray(int*p,int len,int val);

int main(void)
{
int len; //数组中元素个数
int i=0; //数组元素下标
int val; //要搜索的数
printf("请为一个随机数数组设定元素个数len=");
scanf("%d", &len);
int*p = (int*)malloc(sizeof(int)*(len+1));
for (i = 0; i < len;++i)
{
p[i]=rand()%101; //0到100的随机数
}
printf("已为你生成随机数组为:\n");
for (i = 0; i < len; ++i)
{
printf("%d ", p[i]);
}
printf("\n");

printf("请输入你要搜索的数val=");
scanf("%d", &val);

if (findarray(p, len, val) == len)
printf("数组中无此元素");
else
printf("数组中有这个元素,在第%d个位置", findarray(p, len, val) + 1);

getchar();
getchar();
return 0;
}

int findarray(int*p, int len,int val)
{
p[len] = val;
for (int i = 0; p[i] != val; ++i)
{
if (p[i] == val)
return i;
}
}

回答2:

int tmp,i;
int x=9;//要找的数。。。
int flag=0;
int array[1000];
//array 赋值........
for(i=0; i<1000; i++)
{
if(x==array[i])

{

flag=1;

break;

}

}
if(flag)
printf("在数组中\n");

else
printf("不在数组中\n");

写成一个函数吧!在的话返回1,不在返回0
int IsInArray( int* array, int n, int x)
{
int i;

int flag=0;

for(i=0; i {
if(x==array[i])

{

flag=1;

break;

}

}
return(flag);

}

回答3:

bool fun(array[]){
for(i=0;iif(a==array[i])
return ture;
else return false;}}
array为数组长度,如果为二位数据可以采用双层循环
希望采纳,谢谢

回答4:

一个一个查找数组元素,看有没有相等的就好了

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