#include <stdio.h>
#include <stdlib.h>
int main()
{
int a[101], b[101];
int n, m, i, j, flag;
while(scanf("%d", &n) != EOF)
{
//接收输入数组
for(i = 0; i < n; i ++)
{
scanf("%d", a + i);
}
//接收查找数组
scanf("%d", &m);
for(j = 0; j < m; j ++)
{
scanf("%d", b + j);
}
//判断查找存在
for(j = 0; j < m; j ++)
{
flag = 0;
for(i = 0; i < n; i ++)
{
if(b[j] == a[i])
{
flag = 1;
break;
}
}
if(flag)
{
printf("YES\n");
}else
{
printf("NO\n");
}
}
}
return 0;
}
#include <stdio.h>
#include <stdlib.h>
int partition(int *A, int left, int right);
void quicksort(int *A, int begin, int end);
int main()
{
int i, j, n, k;
int a[1001];
while(scanf("%d",&n) != EOF)
{
//接受stdin输入数据
for(i = 0; i < n; i ++)
{
scanf("%d",a + i);
}
scanf("%d",&k);
//快速排序
quicksort(a, 0, n - 1);
//输出第k小的数
for(i = 0, j = 0; i < n && j < k; i ++)
{
if(a[i] != a[i + 1])
{
if(j == k - 1)
{
printf("%d\n",a[i]);
break;
}else
{
j ++;
}
}
}
}
return 0;
}
void quicksort(int *A, int begin, int end)
{
int pivot;
if(begin < end)
{
pivot = partition(A, begin, end);
quicksort(A, begin, pivot - 1);
quicksort(A, pivot + 1, end);
}
}
int partition(int *A, int left, int right)
{
int stand = A[left];
while(left < right)
{
while(left < right && A[right] >= stand)
{
right --;
}
if(left < right)
{
A[left ++] = A[right];
}
while(left < right && A[left] <= stand)
{
left ++;
}
if(left < right)
{
A[right --] = A[left];
}
}
A[left] = stand;
return left;
}
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main()
{
char a[101];
char b[101];
char ch, key;
int i, lena, lenb, flag;
int count[11];
while(scanf("%s",a) != EOF)
{
//接收回车符
ch = getchar();
//接收出牌
scanf("%s",b);
ch = getchar();
//长度
lena = strlen(a);
lenb = strlen(b);
//初始化
memset(count,0,sizeof(count));
//遍历手牌
for(i = 0; i < lena; i ++)
{
count[a[i] - '0'] ++;
}
//检测对方出牌
switch(lenb)
{
case 1:
case 2:
case 3:
case 4:
flag = 0;
for(key = b[0] - '0' + 1; key <= 9; key ++)
{
if(count[key] >= lenb)
{
flag = 1;
break;
}
}
break;
case 5:
flag = 0;
for(key = b[0] - '0' + 1; key < 9; key ++)
{
if(count[key] > 0 && count[key + 1] > 0 && count[key + 2] > 0 && count[key + 3] > 0 && count[key + 4] > 0)
{
flag = 1;
break;
}
}
break;
}
//打印输出
if(flag)
{
printf("YES\n");
}else
{
printf("NO\n");
}
}
return 0;
}
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