HDU 5014 Number Sequence (Greedy), hdu5014

HDU 5014 Number Sequence (Greedy), hdu5014

At that time, I thought of greed, but I don't know why I cited the anti-column .... I clicked to open the link. I found out that I am a funny girl.

Problem DescriptionThere is a special number sequence which has n + 1 integers. For each number in sequence, we have two rules:

● Ai in [0, n]
● Ai = aj (I = j)

For sequence a and sequence B, the integrating degree t is defined as follows ("writable" denotes exclusive or ):

T = (a0 rjb0) + (a1 rjb1) + · + (an 1_bn)

(Sequence B shoshould also satisfy the rules described abve)

Now give you a number n and the sequence a. You shoshould calculate the maximum integrating degree t and print the sequence B.
 
InputThere 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,...,.
 
OutputFor 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

 
Source2014 ACM/ICPC Asia Regional Xi 'an Online
Enumerate greedy.

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<cstdlib>#include<map>using namespace std;typedef long long LL;const int maxn=1e5+100;LL a[maxn];LL d[maxn];int main(){    LL n;    while(~scanf("%I64d",&n))    {        for(LL i=0;i<=n;i++)            scanf("%I64d",&a[i]);        memset(d,-1,sizeof(d));        LL ans=0;        for(LL i=n;i>=0;i--)        {            LL t=0;            if(d[i]==-1)            {                for(LL j=0;;j++)                {                    if(!(i&(1<<j)))  t+=(1<<j);                    if(t>=i)                    {                        t-=(1<<j);                        break;                    }                }                ans+=(i^t)*2;                d[i]=t;                d[t]=i;            }        }        printf("%I64d\n",ans);        for(LL i=0;i<=n;i++)        printf(i==n?"%I64d\n":"%I64d ",d[a[i]]);    }    return 0;}

Number sequence

Computing time limit: 2000/1000 Ms
1 <= n <= 100,000,000
When n is large, it times out.

This part of the program uses periodicity to end the operation ahead of schedule.
F [I] = () mod 7 can only be 0, 1, 2, 3, 4, 5, 6

R [8] [8] Put in the array Lattice f [I-1], f [I-2]} f [I]
When there is a number in the same grid, it is a periodicity.
With a period, you can calculate the number that is equal to the start of the period when n is reached. Is the result. The period must be less than 8*8.

Hangdian ACM1005 Number Sequence

/*
1 1 3
2
1 2 10
5
0 0 0
Press any key to continue
*/
# Include <stdio. h> int main (void) {int a1 = 1, a2 = 1, an; int A, B, n, I; while (1) {scanf ("% d", & A, & B, & n ); if (A = 0 & B = 0 & n = 0) break; a1 = 1; a2 = 1; for (I = 3; I <= n; ++ I) {an = (A * a2 + B * a1) % 7; a1 = a2; a2 = an;} printf ("% d \ n ", an);} return 0 ;}