XidianOJ 1175: count

2024-04-30 23:03:32

题目描述

There is a sequence with n integers, your task is finding there are how many consecutive subsequences have two characters:
1.The length of the subsequence is k.
2.Every number in this subsequence is different.

输入

The input contains multiple test cases.
The first line of each case contains two integers n and k (1≤k≤n≤200000). Then second line contains n positive integers a[i](1≤a[i]≤1000000000).

输出

For each case ,print the number of consecutive subsequences in the only line.

样例输入

10 3
1 2 3 4 3 2 1 1 2 2

样例输出

动态规划

设dp[i]是以第i个数字结尾的满足不重复数字的最长长度，则dp[i] = dp[i-1] + 1(a[i]未出现过),dp[i] = i - m[a[i]] (a[i]出现过，使用map容器储存上个a[i]的下标)

#include <cstdio>
#include <cctype>
#include <iostream>
#include <cstring>
#include <algorithm>
#include <string>
#include <vector>
#include <map>
#include <set>
#include <cmath>
using namespace std;
typedef long long LL;LL n,k;
LL a[200005];
LL dp[200005] = {0};
map<LL,LL> m;int main(){//freopen("test.in","r",stdin);while (scanf("%lld %lld",&n,&k) != EOF){m.clear();memset(dp,0,sizeof(dp));LL i,total = 0;for (i=1;i<=n;i++){scanf("%lld",&a[i]);}dp[1] = 1; m[a[1]] = 1;for (i=2;i<=n;i++){if (m[a[i]] == 0){dp[i] = dp[i-1]+1; m[a[i]] = i;}else {dp[i] = i - m[a[i]];// printf("%lld %lld\n",i,dp[i]);
        m.clear();for (int j=i,k=0;k<dp[i];k++){m[a[j-k]] = j-k;}if (i == 5){//printf("%lld\n",m[a[2]]);
        }}}//printf("%lld = k\n",k);for (i=k;i<=n;i++){if (dp[i] >= k) {//  printf("%lld\n",i);total ++;}}printf("%lld\n",total);}return 0;
}

View Code

转载于:https://www.cnblogs.com/ToTOrz/p/6735997.html

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