HDU5014:Number Sequence(贪心)
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Number Sequence
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 2442 Accepted Submission(s): 977
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
42 0 1 4 3
Sample Output
201 0 2 3 4
Source
2014 ACM/ICPC Asia Regional Xi'an Online
题意:找出符合题意的序列b,使得b0^a0 + b1^a1 + b2^a2 +... bn^an最大,其中bi<=n,输出最大值并输出b序列。
思路:异或最大,找出二进制互补的数字即可,要从大到小找。
写法一:
# include <stdio.h># include <string.h># define MAXN 100000int a[MAXN+1], b[MAXN+1];int fun(int num){ int ans = 0; while(num) { num >>= 1; ++ans; } return ans;}int main(){ long long sum; int n; while(~scanf("%d",&n)) { sum = 0; memset(b, -1, sizeof(b)); for(int i=0; i<=n; ++i) scanf("%d",&a[i]); for(int i=n; i>=0; --i) { if(b[i] == -1) { int length = fun(i); int tmp = ((1<<length)-1)^i; b[i] = tmp; b[tmp] = i; sum += (i^tmp)<<1; } } printf("%lld\n",sum); for(int i=0; i<n; ++i) printf("%d ",b[a[i]]); printf("%d\n",b[a[n]]); } return 0;}
写法二:
# include <stdio.h># include <string.h># define MAXN 100000int a[MAXN+1], b[MAXN+1];int main(){ long long n; while(~scanf("%lld",&n)) { memset(b, -1, sizeof(b)); for(int i=0; i<=n; ++i) scanf("%d",&a[i]); for(int i=n; i>=0; --i) { int ans=0, t=1, s=i; if(b[i] == -1) { while(s) { ans += t*((s&1)^1); t <<= 1; s >>= 1; } b[i] = ans; b[ans] = i; } } printf("%lld\n",n*n+n); for(int i=0; i<n; ++i) printf("%d ",b[a[i]]); printf("%d\n",b[a[n]]); } return 0;}
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