• hdu 4747(DP?线性递推)


                                                   Mex

    Time Limit: 15000/5000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)
    Total Submission(s): 3876    Accepted Submission(s): 1290

    Problem Description

    Mex is a function on a set of integers, which is universally used for impartial game theorem. For a non-negative integer set S, mex(S) is defined as the least non-negative integer which is not appeared in S. Now our problem is about mex function on a sequence.

    Consider a sequence of non-negative integers {ai}, we define mex(L,R) as the least non-negative integer which is not appeared in the continuous subsequence from aL to aR, inclusive. Now we want to calculate the sum of mex(L,R) for all 1 <= L <= R <= n.

    Input

    The input contains at most 20 test cases.
    For each test case, the first line contains one integer n, denoting the length of sequence.
    The next line contains n non-integers separated by space, denoting the sequence.
    (1 <= n <= 200000, 0 <= ai <= 10^9)
    The input ends with n = 0.

     Output

    For each test case, output one line containing a integer denoting the answer.

     Sample Input

    3
    0 1 3 
    5 
    1 0 2 0 1 0
    

    Sample Output

    5

    24

    Hint

    For the first test case: mex(1,1)=1, mex(1,2)=2, mex(1,3)=2, mex(2,2)=0, mex(2,3)=0,mex(3,3)=0. 1 + 2 + 2 + 0 +0 +0 = 5.

    Source

    2013 ACM/ICPC Asia Regional Hangzhou Online

     Recommend

    liuyiding

    题意:给n个数,求所有区间内没有出现的最小非负整数和。

    思路:有一个性质,对于mex(i,j),i固定,j增加,则mex(i,j)非严格递增。

           记F[i]为以i结束的所有区间的值的和,则我们可以从F[i]推到F[i+1];记第i+1位的数位A[i+1],则我们只用考虑大于等于A[i+1]的数对F[i+1]=F[i]+?产生的影响(注意add的值,因为<A[I+1]-1的数不产生增量,A[I+1]-1的数之后会得到统计)
          记S[i]表示0到i的所有数都出现的最小位置,如 数列 7 3 2 1 4 0 5 6 s[0]=6 s[1]=4 s[2]=3 s[3]=2 s[4]=2 s[5]=2 s[6]=2 s[7]=1 (倒着往前的)
    如果j(j>=a[i+1])出现在1~i区间中,比较j最晚出现的位置与覆盖完全的1~j-1的最小位置的较小位置k,那么区间j的前一次出现的位置到k位置这个区间内所有点到i位置的值都+1。
           为什么是j最晚出现的位置:例子是 8 8 8 6 7 3 2 1 4 0 5 6,做6的时候计算一下

    #include<iostream>
    #include<cstring>
    #include<cmath>
    #include<cstdlib>
    #include<cstdio>
    #include<algorithm>
    #include<string>
    #include<map>
    #include<queue>
    #include<vector>
    #include<stack>
    #define ll long long
    #define maxn 4001000
    #define lson l,m,rt<<1
    #define rson m+1,r,rt<<1|1
    #define rep(i,a,b) for(int i=a;i<=b;i++)
    const ll P=1e9+7;
    using namespace std;
    map<int,int> p;
    int n,a[201000],pre[201000],s[201000];
    int main()
    {
        while(scanf("%d",&n)!=EOF)
        {
        	if(n==0) break;
        	for(int i=1;i<=n;i++) scanf("%d",&a[i]);
        	memset(pre,0,sizeof(pre));
        	memset(s,0,sizeof(s));
        	ll ans=0,add=0;
    		for(int i=1;i<=n;i++,ans+=add)
    		if(a[i]<=n)
    		{
    			int temp=pre[a[i]];
    			pre[a[i]]=i;
    			for(int j=a[i];j<=n;j++)
    			{
    				s[j]=j==0?pre[j]:min(s[j-1],pre[j]);
    				if(s[j]>temp) add+=s[j]-temp;
    				else break;
    			}
    		}
    		printf("%lld
    ",ans); 
    	}
    }
  • 相关阅读:
    String Matching Patterns(字符串匹配模式)
    脚本元素定位实例学习
    vue 实现逻辑分页 以及动态表格自动转行
    vue 打印功能(js)
    vue 打包时使用/deep/ 报错
    vue 去掉后缀名
    VSCode: 提高工作效率的快捷键
    修改 input/textarea 标签属性 placeholder 文本的颜色
    H5超级播放器+FFmpeg实现摄像头在线查看
    Lucene.net常用功能说明
  • 原文地址:https://www.cnblogs.com/The-Pines-of-Star/p/9878822.html
Copyright © 2020-2023  润新知