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HDU 5014 Number Sequence(异或 进制问题)
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5014
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):
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
思路:
尽可能找2^x-1,智商真是捉急啊!
最终可以全部异或为11111……(二进制位全是1)
所以最终的异或的和就是n*(n+1);
代码如下:
#include <cstdio> #include <cstring> #define MAXN 100017 typedef __int64 LL; int a[MAXN], vis[MAXN]; int main() { LL n; while(~scanf("%I64d",&n)) { memset(vis,-1,sizeof(vis)); for(int i = 0; i <= n; i++) { scanf("%d",&a[i]); } int m = 1; while(m < n)//2^x-1 { m = m*2+1; } for(int i = n; i >= 0; i--) { if(i <= m/2) m/=2; if(vis[i] == -1) { vis[i] = i^m; vis[i^m] = i; } } LL ans = n*(n+1); printf("%I64d\n",ans); for(int i = 0; i < n; i++) { printf("%d ",vis[a[i]]); } printf("%d\n",vis[a[n]]); } return 0; }
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
HDU 5014 Number Sequence(异或 进制问题)
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