Alice's present
Time Limit: 1 Sec
Memory Limit: 256 MB
题目连接
http://acm.hust.edu.cn/vjudge/contest/view.action?cid=95030#problem/A
Description
As a doll master, Alice owns a wide range of dolls, and each of them has a number tip on it's back, the tip can be treated as a positive integer. (the number can be repeated). One day, Alice hears that her best friend Marisa's birthday is coming , so she decides to sent Marisa some dolls for present. Alice puts her dolls in a row and marks them from 1 to n. Each time Alice chooses an interval from i to j in the sequence ( include i and j ) , and then checks the number tips on dolls in the interval from right to left. If any number appears more than once , Alice will treat this interval as unsuitable. Otherwise, this interval will be treated as suitable.
This work is so boring and it will waste Alice a lot of time. So Alice asks you for help .
Input
There are multiple test cases. For each test case:
The first line contains an integer n ( 3≤ n ≤ 500,000) ,indicate the number of dolls which Alice owns.
The second line contains n positive integers , decribe the number tips on dolls. All of them are less than 2^31-1. The third line contains an interger m ( 1 ≤ m ≤ 50,000 ),indicate how many intervals Alice will query. Then followed by m lines, each line contains two integer u, v ( 1≤ u< v≤ n ),indicate the left endpoint and right endpoint of the interval. Process to the end of input.
,Ci，即此题的初始分值、每分钟减少的分值、dxy做这道题需要花费的时间。
Output
For each test case:
For each query, If this interval is suitable , print one line "OK". Otherwise, print one line ,the integer which appears more than once first.
Print an blank line after each case.
Sample Input
5
1 2 3 1 2
3
1 4
1 5
3 5
6
1 2 3 3 2 1
4
1 4
2 5
3 6
4 6
Sample Output
1
2
OK 3
3
3
OK
HINT
题意
查询一个区间是否有重复，以及从右边开始，第一个重复的数是什么
题解：
直接记录pre就好了
然后st区间查询最大值就好了
代码：
#include<stdio.h>
#include<iostream>
#include<cstring>
#include<map>
using namespace std;
#define maxn 500500
int a[];
int dp[maxn][];
int mm[maxn];
int val[maxn];
void initrmp(int n)
{
mm[]=-;
for(int i=;i<=n;i++)
{
mm[i]=((i&(i-))==)?mm[i-]+:mm[i-];
dp[i][]=val[i];
}
for(int j = ;j<=mm[n];j++)
for(int i=;i+(<<j)-<=n;i++)
dp[i][j]=max(dp[i][j-],dp[i+(<<(j-))][j-]);
}
int query(int l,int r)
{
int k = mm[r-l+];
return max(dp[l][k],dp[r-(<<k)+][k]);
}
map<int,int> H;
int main()
{
int n;
while(scanf("%d",&n)!=EOF)
{
memset(mm,,sizeof(mm));
memset(val,,sizeof(val));
memset(dp,,sizeof(dp));
memset(a,,sizeof(a));
H.clear();
//int n;scanf("%d",&n);
for(int i=;i<=n;i++)
scanf("%d",&a[i]);
for(int i=;i<=n;i++)
{
val[i]=H[a[i]];
H[a[i]]=i;
}
initrmp(n);
int m;scanf("%d",&m);
while(m--)
{
int l ,r;
scanf("%d%d",&l,&r);
int p = query(l,r);
if(p<l)printf("OK\n");
else printf("%d\n",a[p]);
}
printf("\n");
}
}