[mine518]

The meaning of problems: there is a length n of the sequence of a given number on a m number of positions as the ai have bi, the remaining number may be arbitrarily, so that ci represents ai prefix XOR and seeking $ \ sum_ {i = 1 } ^ {n} is the minimum value of $ CI, 1 <= m, bi <= n <= 1e9,0 <= ai <= 1e9

 

Vacancies in two ways: 1 after the position or vacancy, the vacancy will be able greedy 0; 2 position after already know, this time to consider each and every period of exclusive or 1 if you want the back, that is. s0 + 1 and determines the size of s1 (si represented by a binary number of i in the segment), to greedy (note that the first position before no gaps)

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 struct ji{
 4     int id,k;
 5     bool operator < (const ji &a){
 6         return id<a.id;
 7     }
 8 }a[100005];
 9 int n,m,s[2][101];
10 long long ans;
11 int main(){
12     scanf("%d%d",&n,&m);
13     for(int i=1;i<=m;i++)scanf("%d%d",&a[i].id,&a[i].k);
14     sort(a+1,a+m+1);
15     bool flag=(a[1].id==1);
16     a[++m].id=n+2;
17     for(int i=1;i<=m;i++){
18         if (a[i].id==a[i-1].id+1)a[i].k^=a[i-1].k;
19         else{
20             for(int j=0;j<31;j++){
21                 if (flag)ans+=s[1][j]*(1LL<<j);
22                 else ans+=min(s[0][j]+1,s[1][j])*(1LL<<j);
23                 s[0][j]=s[1][j]=0;
24             }
25             printf("%lld ",ans);
26             flag=0;
27         }
28         for(int j=0;j<31;j++)s[((a[i].k&(1<<j))>0)][j]++;
29     }
30     printf("%lld",ans);
31 }

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