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hdu 5014 Number Sequence(意淫题)
Number Sequence
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 545 Accepted Submission(s): 258
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
题解及题解:
瞎搞就搞出来了,最后的结果是n*(n+1),手算一下前5个就能看出来。剩下的输出从大开始向下遍历,输出2^(len)-1-i就可以了,边计算边标记(len为当前数的2进制表示形式的位数)。
#include <iostream> #include <cstdio> #include <cstring> using namespace std; int len(int n) { int ans=0; while(n) { n>>=1; ans++; } return ans; } int s[100010]; int t[100010]; int main() { int n,le,v; __int64 d; while(scanf("%d",&n)!=EOF) { memset(s,-1,sizeof(s[0])*(n+5)); for(int i=0;i<=n;i++) { scanf("%d",&t[i]); } for(int i=n;i>=0;i--) if(s[i]<0) { le=len(i); v=1<<le; v--; s[i]=v-i; s[v-i]=i; } __int64 m=(__int64)n; d=m*(m+1); printf("%I64d\n",d); for(int i=0;i<n;i++) { printf("%d ",s[t[i]]); } printf("%d\n",s[t[n]]); } return 0; }
hdu 5014 Number Sequence(意淫题)
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