| Time Limit: 5000MS | Memory Limit: 65535KB | 64bit IO Format: %I64d & %I64u |
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.
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.
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
Source
首先O(n)复杂度求出 mex(1,i)，可以知道 mex(i,i),mex(i,i+1)到 mex(i,n)是递增的。
然后使用线段树维护，需要不断删除前面的数。
若删掉第一个数 a[1]，那么在下一个 a[1]出现前的大于 a[1]的 mex 值都要变成 a[1]
然后需要线段树实现区间修改和区间求和。
1 #include<iostream> 2 #include<algorithm> 3 #include<cstring> 4 #include<cmath> 5 #include<cstdio> 6 #define ls l,mid,rt<<1 7 #define rs mid+1,r,rt<<1|1 8 #define warning keep silent 9 using namespace std; 10 const int mxn=200020; 11 struct node{ 12 long long sum; 13 int max; 14 int mark; 15 }t[mxn<<2]; 16 int a[mxn]; 17 int n; 18 long long ans=0; 19 int cnt[mxn]; 20 int mex[mxn],pos[mxn]; 21 int nxt[mxn]; 22 void pushdown(int l,int r,int rt){ 23 if(t[rt].mark!=-1){ 24 t[rt<<1].mark=t[rt<<1|1].mark=t[rt].mark; 25 t[rt<<1].max=t[rt<<1|1].max=t[rt].mark; 26 int mid=(l+r)>>1; 27 t[rt<<1].sum=(long long)t[rt].mark*(mid-l+1); 28 t[rt<<1|1].sum=(long long)t[rt].mark*(r-mid); 29 t[rt].mark=-1; 30 } 31 return; 32 } 33 void Build(int l,int r,int rt){ 34 t[rt].mark=-1; 35 if(l==r){ 36 t[rt].max=t[rt].sum=mex[l]; 37 return; 38 } 39 int mid=(l+r)>>1; 40 Build(ls);Build(rs); 41 t[rt].sum=t[rt<<1].sum+t[rt<<1|1].sum; 42 t[rt].max=max(t[rt<<1].max,t[rt<<1|1].max); 43 return; 44 } 45 void change(int v,int L,int R,int l,int r,int rt){ 46 if(L<=l && r<=R){ 47 t[rt].mark=v; 48 t[rt].max=v; 49 t[rt].sum=(long long)v*(r-l+1); 50 return; 51 } 52 pushdown(l,r,rt); 53 int mid=(l+r)>>1; 54 if(L<=mid)change(v,L,R,ls); 55 if(R>mid)change(v,L,R,rs); 56 t[rt].sum=t[rt<<1].sum+t[rt<<1|1].sum; 57 t[rt].max=max(t[rt<<1].max,t[rt<<1|1].max); 58 return; 59 } 60 int posi(int v,int l,int r,int rt){ 61 if(l==r)return l; 62 int mid=(l+r)>>1; 63 pushdown(l,r,rt); 64 if(t[rt<<1].max>v)return posi(v,ls); 65 else return posi(v,rs); 66 } 67 int main(){ 68 while(scanf("%d",&n) && n){ 69 memset(t,0,sizeof t); 70 memset(cnt,0,sizeof cnt); 71 int i,j; 72 for(i=1;i<=n;i++){scanf("%d",&a[i]);if(a[i]>n)a[i]=n;}; 73 int last=0; 74 for(i=1;i<=n;i++){ 75 cnt[a[i]]=1; 76 while(cnt[last])last++; 77 mex[i]=last; 78 } 79 for(i=0;i<=n;i++)pos[i]=n+1;// 80 for(i=n;i;i--){//预处理，找到相同值的下一个出现位置 81 nxt[i]=pos[a[i]]; 82 pos[a[i]]=i; 83 } 84 Build(1,n,1); 85 ans=0; 86 for(i=1;i<=n;i++){ 87 ans+=t[1].sum; 88 change(0,i,i,1,n,1); 89 if(a[i]<t[1].max){ 90 int l=posi(a[i],1,n,1); 91 int r=nxt[i]-1; 92 if(l<=r)change(a[i],l,r,1,n,1); 93 } 94 } 95 printf("%lld ",ans); 96 } 97 return 0; 98 }