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题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5014
Number Sequence
Problem Description
There is a special number sequence which has n+1 integers. For each number in sequence, we have two rules:
● a i ∈ [0,n]
● a i ≠ a j( i ≠ j )
For sequence a and sequence b, the integrating degree t is defined as follows(“⊕” denotes exclusive or):
t = (a 0 ⊕ b 0) + (a 1 ⊕ b 1) +···+ (a n ⊕ b n)
(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.
● a i ∈ [0,n]
● a i ≠ a j( 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 ≤ 10 5), The second line contains a 0,a 1,a 2,...,a n.
For each case, the first line contains an integer n(1 ≤ n ≤ 10 5), The second line contains a 0,a 1,a 2,...,a n.
Output
For each case, output two lines.The first line contains the maximum integrating degree t. The second line contains n+1 integers b 0,b 1,b 2,...,b n. There is exactly one space between b i and b i+1 (0 ≤ i ≤ n - 1). Don’t ouput any spaces after b n.
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坑爹爆long long，换了__int64过了。想法很简单，把两个数二进制的0和1尽量补全，优先满足大的数就可以了。不过要找到区间。
代码：
#include <iostream>
#include <cstdio>
using namespace std;__int64 n, a[100010];
struct right
{__int64 s, r, l;
}rt[1000];
__int64 getNear(__int64 x)
{__int64 z = 1;while(x){x >>= 1;z <<= 1;}return z-1;
}
int main()
{while(~scanf("%I64d", &n)){__int64 m = n;rt[0].r = m;rt[0].s = getNear(m);rt[0].l = rt[0].s-rt[0].r;//cout << rt[0].l << " " << rt[0].r << " " << rt[0].s << endl;__int64 cnt = 0;while(1){m = rt[cnt].l-1;if(m < 0) break;cnt++;rt[cnt].r = m;rt[cnt].s = getNear(m);rt[cnt].l = rt[cnt].s-rt[cnt].r;//cout << rt[cnt].l << " " << rt[cnt].r << " " << rt[cnt].s << endl;}for(__int64 i = 0; i <= n; i++)scanf("%I64d", &a[i]);//a[i] = i;__int64 t = 0;for(__int64 i = 0; i <= n; i++)for(__int64 j = 0; j <= cnt; j++){if(a[i] >= rt[j].l && a[i] <= rt[j].r){//cout << rt[j].l << " " << rt[j].r << " " << rt[j].s << endl;//printf("%d ", rt[j].s-a[i]);a[i] = rt[j].s-a[i];t += rt[j].s;break;}}printf("%I64d\n", t);for(__int64 i = 0; i < n; i++)printf("%I64d ", a[i]);printf("%I64d\n", a[n]);}return 0;
}
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