Sereja and Suffixes
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Description
Sereja has an array a, consisting of n integers a1, a2, ..., an. The boy cannot sit and do nothing, he decided to study an array. Sereja took a piece of paper and wrote out m integers l1, l2, ..., lm(1 ≤ li ≤ n). For each number li he wants to know how many distinct numbers are staying on the positions li, li + 1, ..., n. Formally, he want to find the number of distinct numbers among ali, ali + 1, ..., an.?
Sereja wrote out the necessary array elements but the array was so large and the boy was so pressed for time. Help him, find the answer for the described question for each li.
Input
The first line contains two integers n and m(1 ≤ n, m ≤ 105). The second line contains n integers a1, a2, ..., an(1 ≤ ai ≤ 105) — the array elements.
Next m lines contain integers l1, l2, ..., lm. The i-th line contains integer li(1 ≤ li ≤ n).
Output
Print m lines — on the i-th line print the answer to the number li.
Sample Input
10 101 2 3 4 1 2 3 4 100000 9999912345678910
6666654321
题目分析:
这个题目应该是好读懂的,
10 101 2 3 4 1 2 3 4 100000 9999912345678910
10 10
10 这个数列有10个元素
10 有10个数据测试
1 数列从第一个数开始到最后有多少个不同的数 6(自己数一下)
2 数列从第二个数开始到最后有多少个不同的数 6
3 数列从第三个数开始到最后有多少个不同的数 6
.
7 数列从第7 个数开始到最后有 多少个不同的数 4
.
10 数列从第10个数开始到最后有多少个不同的数 1
#include<cstdio>#include<cstring>#include<algorithm>#define maxn 100010using namespace std;int a[maxn],b[maxn],c[maxn];int main(){int n,m;while(~scanf("%d%d",&n,&m)){memset(b,0,sizeof(b));memset(c,0,sizeof(c));for(int i=0;i<n;i++)scanf("%d",&a[i]);int ans=0;for(int i=n-1;i>=0;i--){if(!b[a[i]])ans++;b[a[i]]=1;c[i]=ans;}//for(int i=0;i<n;i++)//printf("%d ",c[i]);//printf("\n");int t;for(int i=0;i<m;i++){scanf("%d",&t);printf("%d\n",c[t-1]);}}return 0;}
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