#include
int main()
{ int A,B;
while(scanf("%d%d",&A,&B)!=EOF)
{
int l,m,i,j;
l=0;
m=0;
long long int s[A],r[B];
for(i=m;i<A;i++)
{
scanf("%lld",&s[i]);
}
for(i=m;i<B;i++)
{
scanf("%lld",&r[i]);
}
for(j=0;j<B;j++)
{
for(i=m;i<A;i++)
{
if(r[j]==s[i])
{
l=l+1;
m=i+1;
break;
}
else continue;
}
}
if(l==B)
printf("yes");
else
printf("no");
}
return 0;
}
这么做对嘛
使用两个for循环。外部循环挑选出arr2[]的每一个元素赋值为value;然后再内部循环中,搜索在arr1[]中是否存在value。如果arr2[]中的每一个元素都可以在arr1[]中被找到,则返回true,否则返回false。具体代码如下:
boolean isSubset(int arr1[],
int arr2[], int m, int n)
{
int i = 0;
int j = 0;
for (i = 0; i < n; i++)
{
for (j = 0; j < m; j++)
if(arr2[i] == arr1[j])
break;
/* If the above inner loop
was not broken at all then
arr2[i] is not present in
arr1[] */
if (j == m)
return false;
}
/* If we reach here then all
elements of arr2[] are present
in arr1[] */
return true;
作者:Danny_姜
来源:CSDN
原文:https://blog.csdn.net/zxm317122667/article/details/83999828