6 2 3 2 1 4 3 5 3 4
Case #1: 2 2
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <set>
using namespace std;
int const MAX = 1e4 + 5;
int a[MAX], ans[MAX];
set <int> s;
int main()
{
int n, m, ca = 1;
while(scanf("%d %d", &n, &m) != EOF)
{
printf("Case #%d:\n", ca ++);
memset(ans, 0, sizeof(ans));
for(int i = 0; i < n; i++)
scanf("%d", &a[i]);
for(int i = 0; i < n; i++)
{
s.clear();
s.insert(a[i]);
ans[1] ++;
for(int j = i + 1; j < n; j++)
{
set <int> :: iterator it;
it = s.find(a[j]);
if(it != s.end())
break;
s.insert(a[j]);
int st = *s.begin();
it --;
int ed = *it;
int sz = s.size();
if(ed - st + 1 == sz)
ans[sz] ++;
}
}
for(int i = 0; i < m; i++)
{
int k;
scanf("%d", &k);
printf("%d\n", ans[k]);
}
}
}版权声明:本文为博主原创文章,未经博主允许不得转载。
原文地址:http://blog.csdn.net/tc_to_top/article/details/47089085