2017 Multi-University Training Contest 5 solutions 1011 Rikka with Competition
2017-08-10 09:27
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临时加的签到题。把 aiai 从大到小排序,那么第 ii 强人要获胜,最优情况下是最强的人输给了第二强的人,第二强的人输给了第三强的人,以此类推。因此只需要判断排序后 maxj<i(aj−aj+1)maxj<i(aj−aj+1) 和 KK 的大小关系即可。
时间复杂度 O(nlogn)O(nlogn)。
#include<cstdio>
#include<algorithm>
using namespace std;
int a[100000+10];
int main()
{
int n,k,t;
scanf("%d",&t);
while(t--)
{
scanf("%d%d",&n,&k);
for(int i=0;i<n;i++)
scanf("%d",&a[i]);
sort(a,a+n);
int s=1;
for(int i=n-1;i>=1;i--)
if(a[i]-a[i-1]<=k)
s++;
else
break;
printf("%d\n",s);
}
return 0;
}
时间复杂度 O(nlogn)O(nlogn)。
#include<cstdio>
#include<algorithm>
using namespace std;
int a[100000+10];
int main()
{
int n,k,t;
scanf("%d",&t);
while(t--)
{
scanf("%d%d",&n,&k);
for(int i=0;i<n;i++)
scanf("%d",&a[i]);
sort(a,a+n);
int s=1;
for(int i=n-1;i>=1;i--)
if(a[i]-a[i-1]<=k)
s++;
else
break;
printf("%d\n",s);
}
return 0;
}
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