HDU 4747 Mex
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Mex
Time Limit: 15000/5000 MS (Java/Others) Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 1414 Accepted Submission(s): 464
Total Submission(s): 1414 Accepted Submission(s): 464
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.
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
30 1 351 0 2 0 10
Sample Output
524HintFor 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
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liuyiding
这题真是做了很长时间啊，但是还是没有过，代码已经出来了，自几测试了上百组数据都过了，交上去就是wa，求数据啊，求数据，求数据，急切的求数据。
终于发现错误了，其实三四天前的代码，只要改一下输出就对了，用long long ,要用cout，否则会错，三四天前的代码如果改成cout<<ans<<endl; 就过了，只不过那时候我怀疑用map映射的时候，映射10^9 ,会出错，就没有向输出考虑，于是就把代码改了，改的时候忽略了一个问题，我把大于200000的没有加到数组中，当时只是以为大于200000的没用，其实还是很有用的，我没仔细想想，就以为他一点用没有，导致这三四天来，反复的改代码都wa。哎，慎重，一定要慎重啊！
三四天前改后AC的代码：
这题真是做了很长时间啊，但是还是没有过，代码已经出来了，自几测试了上百组数据都过了，交上去就是wa，求数据啊，求数据，求数据，急切的求数据。
终于发现错误了，其实三四天前的代码，只要改一下输出就对了，用long long ,要用cout，否则会错，三四天前的代码如果改成cout<<ans<<endl; 就过了，只不过那时候我怀疑用map映射的时候，映射10^9 ,会出错，就没有向输出考虑，于是就把代码改了，改的时候忽略了一个问题，我把大于200000的没有加到数组中，当时只是以为大于200000的没用，其实还是很有用的，我没仔细想想，就以为他一点用没有，导致这三四天来，反复的改代码都wa。哎，慎重，一定要慎重啊！
#include <iostream>#include <cstdio>#include <cstring>#include <cmath>#define N 200010#define INF 0x7ffffffusing namespace std;struct num{ int l,r; __int64 col,Max,sum;}a[4*N];int pt[N],b[N],tag[N],pos1,pos2;int q[N];int main(){ //freopen("data.in","r",stdin); //freopen("data1.out","w",stdout); void build(int k,int l,int r); void del(int k,int val); void find(int k,int val); void update(int k,int l,int r,int val); int n; while(scanf("%d",&n)!=EOF) { if(n==0) { break; } int Top=1; for(int i=1;i<=n;i++) { scanf("%d",&b[Top]); if(b[Top]<=200000) //错误之处 { Top++; } } n = Top-1; if(n==0) { printf("0\n"); continue; } memset(q,0,sizeof(q)); int res = 0; __int64 ans=0; for(int i=1;i<=n;i++) { q[b[i]]=1; while(q[res]!=0) { res++; } tag[i] = res; ans+=(__int64)res; } memset(pt,0,sizeof(pt)); memset(q,0,sizeof(q)); for(int i=1;i<=n;i++) { int x=b[i]; if(q[x]!=0) { pt[q[x]]=i; } q[x]=i; } build(1,1,n); for(int i=1;i<=n;i++) { del(1,i); pos1 = INF; int val=b[i]; if(a[1].Max<val) { ans+=a[1].sum; continue; } find(1,val); pos2 = pt[i]-1; if(pos2 ==-1) { pos2 = n; } if(pos1>pos2) { ans+=a[1].sum; continue; } update(1,pos1,pos2,val); ans+=a[1].sum; } cout<<ans<<endl; } return 0;}void pushup(int k){ a[k].Max = max(a[k<<1].Max,a[k<<1|1].Max); a[k].sum = a[k<<1].sum+a[k<<1|1].sum; a[k].col= 1;}void build(int k,int l,int r){ a[k].l = l; a[k].r = r; if(l==r) { a[k].sum = tag[l]; a[k].Max = tag[l]; a[k].col = 0; return; } int mid = (l+r)>>1; build(k<<1,l,mid); build(k<<1|1,mid+1,r); pushup(k);}void del(int k,int val){ if(a[k].l==a[k].r) { a[k].col=0; a[k].sum=0; return ; } if(a[k].col==0) { a[k<<1].col=0; a[k<<1].Max = a[k].Max; a[k<<1].sum = a[k].Max*(__int64)(a[k<<1].r-a[k<<1].l+1); a[k<<1|1].col=0; a[k<<1|1].Max = a[k].Max; a[k<<1|1].sum = a[k].Max*(__int64)(a[k<<1|1].r-a[k<<1|1].l+1); } int mid = (a[k].l+a[k].r)>>1; if(mid>=val) { del(k<<1,val); }else { del(k<<1|1,val); } pushup(k);}void find(int k,int val){ if(a[k].col==0&&a[k].Max>val) { pos1 = a[k].l; return ; } if(a[k].col==0) { return; } if(a[k].l==a[k].r&&a[k].Max>val) { pos1 = a[k].l; return ; } if(a[k<<1].Max>val) { find(k<<1,val); }else if(a[k<<1|1].Max>val) { find(k<<1|1,val); }}void update(int k,int l,int r,int val){ if(a[k].l==l&&a[k].r==r) { a[k].Max = val; a[k].sum = (__int64)(a[k].r -a[k].l + 1)*(__int64)val; a[k].col=0; return ; } if(a[k].col==0) { a[k<<1].col=0; a[k<<1].Max = a[k].Max; a[k<<1].sum = a[k].Max*(__int64)(a[k<<1].r-a[k<<1].l+1); a[k<<1|1].col=0; a[k<<1|1].Max = a[k].Max; a[k<<1|1].sum = a[k].Max*(__int64)(a[k<<1|1].r-a[k<<1|1].l+1); } int mid = (a[k].l+a[k].r)>>1; if(r<=mid) { update(k<<1,l,r,val); }else if(l>mid) { update(k<<1|1,l,r,val); }else { update(k<<1,l,mid,val); update(k<<1|1,mid+1,r,val); } pushup(k);}AC的代码：#include <iostream>#include <cstdio>#include <cstring>#include <cmath>#define N 200010#define INF 0x7ffffffusing namespace std;struct num{ int l,r; __int64 col,Max,sum;}a[4*N];int pt[N],b[N],tag[N],pos1,pos2;int q[N];int main(){ //freopen("data.in","r",stdin); //freopen("data1.out","w",stdout); void build(int k,int l,int r); void del(int k,int val); void find(int k,int val); void update(int k,int l,int r,int val); int n; while(scanf("%d",&n)!=EOF) { if(n==0) { break; } for(int i=1;i<=n;i++) { scanf("%d",&b[i]); } memset(q,0,sizeof(q)); int res = 0; __int64 ans=0; for(int i=1;i<=n;i++) { if(b[i]<=200000) { q[b[i]]=1; } while(q[res]!=0) { res++; } tag[i] = res; ans+=(__int64)res; } memset(pt,0,sizeof(pt)); memset(q,0,sizeof(q)); for(int i=1;i<=n;i++) { int x=b[i]; if(x>200000) { continue; } if(q[x]!=0) { pt[q[x]]=i; } q[x]=i; } build(1,1,n); for(int i=1;i<=n;i++) { del(1,i); pos1 = INF; int val=b[i]; if(a[1].Max<val) { ans+=a[1].sum; continue; } find(1,val); pos2 = pt[i]-1; if(pos2 ==-1) { pos2 = n; } if(pos1>pos2) { ans+=a[1].sum; continue; } update(1,pos1,pos2,val); ans+=a[1].sum; } cout<<ans<<endl; } return 0;}void pushup(int k){ a[k].Max = max(a[k<<1].Max,a[k<<1|1].Max); a[k].sum = a[k<<1].sum+a[k<<1|1].sum; a[k].col= 1;}void build(int k,int l,int r){ a[k].l = l; a[k].r = r; if(l==r) { a[k].sum = tag[l]; a[k].Max = tag[l]; a[k].col = 0; return; } int mid = (l+r)>>1; build(k<<1,l,mid); build(k<<1|1,mid+1,r); pushup(k);}void del(int k,int val){ if(a[k].l==a[k].r) { a[k].col=0; a[k].sum=0; return ; } if(a[k].col==0) { a[k<<1].col=0; a[k<<1].Max = a[k].Max; a[k<<1].sum = a[k].Max*(__int64)(a[k<<1].r-a[k<<1].l+1); a[k<<1|1].col=0; a[k<<1|1].Max = a[k].Max; a[k<<1|1].sum = a[k].Max*(__int64)(a[k<<1|1].r-a[k<<1|1].l+1); } int mid = (a[k].l+a[k].r)>>1; if(mid>=val) { del(k<<1,val); }else { del(k<<1|1,val); } pushup(k);}void find(int k,int val){ if(a[k].col==0&&a[k].Max>val) { pos1 = a[k].l; return ; } if(a[k].col==0) { return; } if(a[k].l==a[k].r&&a[k].Max>val) { pos1 = a[k].l; return ; } if(a[k<<1].Max>val) { find(k<<1,val); }else if(a[k<<1|1].Max>val) { find(k<<1|1,val); }}void update(int k,int l,int r,int val){ if(a[k].l==l&&a[k].r==r) { a[k].Max = val; a[k].sum = (__int64)(a[k].r -a[k].l + 1)*(__int64)val; a[k].col=0; return ; } if(a[k].col==0) { a[k<<1].col=0; a[k<<1].Max = a[k].Max; a[k<<1].sum = a[k].Max*(__int64)(a[k<<1].r-a[k<<1].l+1); a[k<<1|1].col=0; a[k<<1|1].Max = a[k].Max; a[k<<1|1].sum = a[k].Max*(__int64)(a[k<<1|1].r-a[k<<1|1].l+1); } int mid = (a[k].l+a[k].r)>>1; if(r<=mid) { update(k<<1,l,r,val); }else if(l>mid) { update(k<<1|1,l,r,val); }else { update(k<<1,l,mid,val); update(k<<1|1,mid+1,r,val); } pushup(k);}三四天前改后AC的代码：
#include <iostream>#include <cstdio>#include <cstring>#include <cmath>#include <map>#define N 200010using namespace std;struct num{ long long int l,r,color,Max,sum;}a[5*N];map<int,int>q;int pt[N],b[N],tag[N],pos1,pos2;int main(){ //freopen("data.in","r",stdin); void build(int k,int l,int r); void del(int k,int val); void find(int k,int val); void update(int k,int l,int r,int val); int n; while(scanf("%d",&n)!=EOF) { if(n==0) { break; } for(int i=1;i<=n;i++) { scanf("%d",&b[i]); } q.clear(); int res = 0; long long int ans=0; for(int i=1;i<=n;i++) { q[b[i]]=1; while(q[res]!=0) { res++; } tag[i] = res; ans+=(long long int)res; } q.clear(); memset(pt,0,sizeof(pt)); for(int i=1;i<=n;i++) { int x=b[i]; if(q[x]!=0) { pt[q[x]]=i; } q[x]=i; } build(1,1,n); for(int i=1;i<=n;i++) { del(1,i); int val=b[i]; find(1,val); pos2 = pt[i]-1; if(pos2 ==-1) { pos2 = n; } if(pos1>pos2||a[1].Max<val) { ans+=a[1].sum; continue; } update(1,pos1,pos2,val); ans+=a[1].sum; } cout<<ans<<endl; } return 0;}void pushup(int k){ a[k].Max = max(a[k<<1].Max,a[k<<1|1].Max); a[k].sum = a[k<<1].sum+a[k<<1|1].sum; a[k].color = 1;}void build(int k,int l,int r){ a[k].l = l; a[k].r = r; if(l==r) { a[k].sum = tag[l]; a[k].Max = tag[l]; a[k].color = 0; return; } int mid = (l+r)>>1; build(k<<1,l,mid); build(k<<1|1,mid+1,r); pushup(k);}void del(int k,int val){ if(a[k].l==a[k].r) { a[k].color=0; a[k].sum=0; return ; } if(a[k].color==0) { a[k<<1].color=0; a[k<<1].Max = a[k].Max; a[k<<1].sum = a[k].Max*(long long int)(a[k<<1].r-a[k<<1].l+1); a[k<<1|1].color=0; a[k<<1|1].Max = a[k].Max; a[k<<1|1].sum = a[k].Max*(long long int)(a[k<<1|1].r-a[k<<1|1].l+1); } int mid = (a[k].l+a[k].r)>>1; if(mid>=val) { del(k<<1,val); }else { del(k<<1|1,val); } pushup(k);}void find(int k,int val){ if(a[k].color==0&&a[k].Max>val) { pos1 = a[k].l; return ; } if(a[k].color==0) { return; } if(a[k].l==a[k].r&&a[k].Max>val) { pos1 = a[k].l; return ; } if(a[k<<1].Max>val) { find(k<<1,val); }else if(a[k<<1|1].Max>val) { find(k<<1|1,val); }}void update(int k,int l,int r,int val){ if(a[k].l==l&&a[k].r==r) { a[k].Max = val; a[k].sum = (a[k].r -a[k].l + 1)*val; a[k].color=0; return ; } if(a[k].color==0) { a[k<<1].color=0; a[k<<1].Max = a[k].Max; a[k<<1].sum = a[k].Max*(a[k<<1].r-a[k<<1].l+1); a[k<<1|1].color=0; a[k<<1|1].Max = a[k].Max; a[k<<1|1].sum = a[k].Max*(a[k<<1|1].r-a[k<<1|1].l+1); } int mid = (a[k].l+a[k].r)>>1; if(r<=mid) { update(k<<1,l,r,val); }else if(l>mid) { update(k<<1|1,l,r,val); }else { update(k<<1,l,mid,val); update(k<<1|1,mid+1,r,val); } pushup(k);}- hdu 4747 Mex
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