Codeforce C. Pearls in a Row

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 ai.

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/printfinstead 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 ai (1 ≤ ai ≤ 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 lj, rj (1 ≤ lj ≤ rj ≤ 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".

Sample test(s)

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

题意：给你一串数字，要你分段，要求每个段中至少有两个相同的数字，要你求最多能分多少段。
思路：贪心，从前面开始循环将出现的数字记录，然后遇到相同的就跳出，这就是一段，然后将前面记录的数字清空，重复前面的操作
这样符合最优原则，遇到有两个了相同的就分为一段。要注意最后一段。

#include<stdio.h>
#include<algorithm>
#include<iostream>
#include<string.h>
#include<cstdio>
#include<climits>
#include<math.h>
#include<queue>
#include<string.h>
#include<stack>
#include<vector>
#include<map>
#include<ext/hash_map>
#define sc(x) scanf("%I64d",&x)
#define pr(x) printf("%I64d",x);
#define prr(x) printf("%I64d
",x);
#define prrr(x) printf(" %I64d",x);
#define FOR(i,p,q) for(int i=p;i<=q;i++)
int cmp(const void*p,const void*q);
using namespace std;
typedef long long ll;
typedef struct pp
{
    int x;
    int y;
    int flag;
} ss;
ss bb[3*100005];
ss aa[6*100005];
ss dd[3*100005];
map<int,int>my;//用map记录是否前面出现了某个数字
ss rk[3*100005];
int main(void)
{
    int  n,i,j,k,p,q;
    while(scanf("%d",&k)!=EOF)
    {my.clear();
        for(i=0; i<k; i++)
        {
            scanf("%d",&bb[i].x);
            bb[i].x;
            bb[i].y=i;
        }
        qsort(bb,k,sizeof(ss),cmp);//这里的离散化可以不需要，直接映射就可以了。
        int cnt=0;
        dd[bb[0].y].x=cnt;
        dd[bb[0].y].y=bb[0].y;
        for(i=1; i<k; i++)
        {
            if(bb[i].x!=bb[i-1].x)
            {
                cnt++;
            }
            dd[bb[i].y].x=cnt;
            dd[bb[i].y].y=bb[i].y;
        }
        int fr=0;
        int sum=0;
        for(i=0; i<k; i++)
        {
            if(my[dd[i].x]==0)
            {
                my[dd[i].x]=1;
            }
            else if(my[dd[i].x])
            {
                sum++;
                rk[sum].x=fr;
                rk[sum].y=i;
                fr=i+1;
                my.clear();//有一段后清空映射，因为段不能相交
            }
        }
        if(sum==0)
        {
            printf("-1
");
        }
        else
        {if(rk[sum].y!=k-1)//最后一段要特别注意，如果没到端尾，要到端尾
        {
            rk[sum].y=k-1;
        }
            printf("%d
",sum);
            for(i=1; i<=sum; i++)
            {
                printf("%d %d
",rk[i].x+1,rk[i].y+1);
            }
        }
    }
    return 0;
}
int cmp(const void*p,const void*q)
{
    ss*n=(ss*)p;
    ss*m=(ss*)q;
    return n->x-m->x;
}