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hdu5014 Number Sequence(异或运算)
题目链接:
huangjing
题意:
这个题目的意思是给出0~n的排列,然后找出与这个序列的配对使(a0 ⊕ b0) + (a1 ⊕ b1) +·+ (an ⊕ bn)最大。。
思路:
从大到小遍历每个数,然后找到与这个数二进制位数互补的数,那么他们的抑或值必定是pow(2,n)-1,,肯定是最大的。。。。
题目:
Number Sequence
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 713 Accepted Submission(s): 337
Special Judge
Problem Description
There 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.
● ai ∈ [0,n]
● ai ≠ aj( i ≠ j )
For sequence a and sequence b, the integrating degree t is defined as follows(“⊕” denotes exclusive or):
(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.
Input
There 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.
For each case, the first line contains an integer n(1 ≤ n ≤ 105), The second line contains a0,a1,a2,...,an.
Output
For 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
4 2 0 1 4 3
Sample Output
20 1 0 2 3 4
Source
2014 ACM/ICPC Asia Regional Xi‘an Online
代码:
#include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<map> #include<vector> #include<cmath> #include<string> #include<queue> #define eps 1e-9 #define ll long long #define INF 0x3f3f3f3f using namespace std; const int maxn=100000+10; int a[maxn],n,ans[maxn]; bool vis[maxn]; int gao(int k) { int cnt=0; while(k) { k>>=1; cnt++; } return cnt; } int main() { while(~scanf("%d",&n)) { memset(vis,false,sizeof(vis)); for(int i=0;i<=n;i++) scanf("%d",&a[i]); for(int i=n;i>=0;i--) { if(vis[i]) continue; int move=gao(i); int temp=(((1<<move)-1)^i);//找到和i二进制表示互补的数 ans[temp]=i; ans[i]=temp; vis[i]=vis[temp]=true; } ll sum=0; for(int i=0;i<=n;i++) sum+=i^ans[i]; printf("%I64d\n",sum); for(int i=0;i<=n;i++) { if(i==n) printf("%d\n",ans[a[i]]); else printf("%d ",ans[a[i]]); } } return 0; }
hdu5014 Number Sequence(异或运算)
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