问题 B: a Simple Problem
时间限制: 1 Sec 内存限制: 128 MB
提交: 412 解决: 99
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题目描述
Many people think hh is a diaosi, but hh is a very rich man whose nickname is wenzhoutuhao,and he made a lot of money by buying the stock of neusoft. He bought n diamonds.One day he found that his warehouse is too small to accommodate these diamonds. so he decide to transfer c of the diamonds to another warehouse.He made the n diamonds into a row, with a number written on their positions, the number is the value of the diamond,the unit is billion(oh no so rich man),then,hh tells you to choose c diamonds,which will be sent to other warehouse,he also imposed two conditions.They are:
1.the chosen c diamonds must be formed a contiguous
segment
2.any of the chosen diamond’s value should not be greater than t,because he thought you may be would steal them.Find the number of ways you can choose the c diamonds.
输入
50 group tests,the first line of input will contain three space separated integer n(1<=n<=10^5),t(0<=t<=10^9) and c(1<=c<=n)
the next line will contain n space separated integer,the ith integer is the value of ith diamond,the value will be non-negative and will not be exceed 10^9
输出
print a single integer——the number of ways you can choose the c diamonds
样例输入
4 3 3 2 3 1 1 1 1 1 2
样例输出
2 0
提示
#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
int main()
{
#ifdef CDZSC_OFFLINE
freopen("in.txt","r",stdin);
//freopen("out.txt","w",stdout);
#endif
int n,t,c,i,a[100050],p;
while(scanf("%d%d%d",&n,&t,&c)!=EOF)
{
memset(a,0,sizeof(a));
int sum=0,k=1;
for(i=1; i<=n; i++)
{
scanf("%d",&p);
if(p>t)
{
a[k++]=i;
}
}
a[k++]=n+1;
if(k-2==0)
{
printf("%d\n",n-c+1);
}
else if(k-2==n)
{
printf("0\n");
}
else
{
for(i=1; i<k; i++)
{
if(a[i]-a[i-1]-1>=c)
{
sum+=(a[i]-a[i-1]-1-c+1);
}
}
printf("%d\n",sum);
}
}
return 0;
}