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當時想到了貪心,但是不知為何舉出了反列。。。。我是逗比,看了點擊開啟連結。才發現我是逗比。
Problem DescriptionThere is a special number sequence which has n+1 integers. For each number in sequence, we have two rules:
● ai ∈ [0,n]
● ai ≠ aj( i ≠ j )
For sequence a and sequence b, the integrating degree t is defined as follows(“⊕” denotes exclusive or):
t = (a0 ⊕ b0) + (a1 ⊕ b1) +···+ (an ⊕ bn)
(sequence B should also satisfy the rules described above)
Now give you a number n and the sequence a. You should calculate the maximum integrating degree t and print the sequence b.
InputThere are multiple test cases. Please process till EOF.
For each case, the first line contains an integer n(1 ≤ n ≤ 105), The second line contains a0,a1,a2,...,an.
OutputFor each case, output two lines.The first line contains the maximum integrating degree t. The second line contains n+1 integers b0,b1,b2,...,bn. There is exactly one space between bi and bi+1
(0 ≤ i ≤ n - 1). Don’t ouput any spaces after bn.
Sample Input
42 0 1 4 3
Sample Output
201 0 2 3 4
Source2014 ACM/ICPC Asia Regional Xi‘an Online
枚舉貪心即可。
#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<cstdlib>#include<map>using namespace std;typedef long long LL;const int maxn=1e5+100;LL a[maxn];LL d[maxn];int main(){ LL n; while(~scanf("%I64d",&n)) { for(LL i=0;i<=n;i++) scanf("%I64d",&a[i]); memset(d,-1,sizeof(d)); LL ans=0; for(LL i=n;i>=0;i--) { LL t=0; if(d[i]==-1) { for(LL j=0;;j++) { if(!(i&(1<<j))) t+=(1<<j); if(t>=i) { t-=(1<<j); break; } } ans+=(i^t)*2; d[i]=t; d[t]=i; } } printf("%I64d\n",ans); for(LL i=0;i<=n;i++) printf(i==n?"%I64d\n":"%I64d ",d[a[i]]); } return 0;}
HDU 5014 Number Sequence(貪心)