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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.
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.
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
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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：
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#include
#include
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 , t); for(__int64 i = 0; i < n; i++) printf(%I64d , a[i]); printf(%I64d , a[n]); } return 0; }
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