CodeForces - 620C Pearls in a Row 贪心 STL

C. Pearls in a Row

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

There are
n pearls in a row. Let's enumerate them with integers from
1 to
n from the left to the right. The pearl number
i has the type
a_i.

Let's call a sequence of consecutive pearls a
segment. Let's call a segment
good if it contains two pearls of the same type.

Split the row of the pearls to the maximal number of good segments. Note that each pearl should appear in exactly one segment of the partition.

As input/output can reach huge size it is recommended to use fast input/output methods: for example, prefer to use

scanf/printf instead of
cin/cout in C++, prefer to use
BufferedReader/PrintWriter instead of
Scanner/System.out in
Java.

Input

The first line contains integer
n (1 ≤ n ≤ 3·10⁵)
— the number of pearls in a row.

The second line contains
n integers
a_i (1 ≤ a_i ≤ 10⁹)
– the type of the
i-th pearl.

Output

On the first line print integer
k — the maximal number of segments in a partition of the row.

Each of the next
k lines should contain two integers
l_j, r_j
(1 ≤ l_j ≤ r_j ≤ n)
— the number of the leftmost and the rightmost pearls in the
j-th segment.

Note you should print the correct partition of the row of the pearls, so each pearl should be in exactly one segment and all segments should contain two pearls of the same type.

If there are several optimal solutions print any of them. You can print the segments in any order.

If there are no correct partitions of the row print the number "-1".

Examples

Input

5
1 2 3 4 1

Output

1
1 5

Input

5
1 2 3 4 5

Output

-1

Input

7
1 2 1 3 1 2 1

Output

2
1 3
4 7

这题找规律，从第一个数开始向后搜索，当有一个数字重复时则记录该位置，再从下一个数字开始向后重新搜索。可使用STL中的set函数

AC代码

#include<stdio.h>
#include<set>
using namespace std;
int pos[300005];
set<int>a;
int main()
{
	int n,m=0,b;
	scanf("%d",&n);
	for(int i=1;i<=n;i++)
	{
		scanf("%d",&b);
		if(a.count(b)==0)//检查b是否出现过
		{
			a.insert(b);//在末尾插入b
		}
		else
		{
			pos[m]=i;
			m++;
			a.clear();//清除a中数据
		}

	}
	if(m==0)
	{
		printf("-1\n");
	}
	else
	{
		pos[m-1]=n;
		printf("%d\n",m);
		printf("1 %d\n",pos[0]);
		for(int i=0;i<m-1;i++)
		{
			printf("%d %d\n",pos[i]+1,pos[i+1]);
		}
	}
	return 0;
} 

最近刚学的STL，本来想用循环来搜索重复数字的，结果超时，不过可作为思路

#include<stdio.h>
int a[300005],pos[300005];
int main()
{
	int n,m=0;
	pos[m]=0;
	scanf("%d",&n);
	for(int i=1;i<=n;i++)
	{
		scanf("%d",&a[i]);
		for(int j=pos[m]+1;j<i;j++)
		{
			if(a[j]==a[i])
			{
				m++;
				pos[m]=i;
			}
		}
	}
	if(m==0)
	{
		printf("-1\n");
	}
	else
	{
		pos[m]=n;
		printf("%d\n",m);
		printf("1 %d\n",pos[1]);
		for(int i=1;i<m;i++)
		{
			printf("%d %d\n",pos[i]+1,pos[i+1]);
		}
	}
	return 0;
}