http://acm.hdu.edu.cn/showproblem.php?pid=5014

从小数开始模拟找方法规律，然后推广，尤其敢猜敢尝试，错了一种思路继续猜-----这是一种很重要的方法啊

这道题还是从小数开始模拟，我是根据16以内的找的规律

根据

2^k---2^k-1

2^k+1---2^k-2

...

这样陪下去

当2^k==n的时候，

从2^(k-1)按同样的方法配下去，

WA了很久，是lower_bound用错了 不过没明白为啥  换成upper_bound也可以AC....但还不明白为啥lower_bound  WA......

目的是找以一个<=n的下标
int id=lower_bound(mi,mi+17,n)-mi; 
if(mi[id]>n)id--;
跟
int id=16;
while(mi[id]>n && id>=0)id--;
不一样么？后者AC   前者WA。。

//#pragma comment(linker, "/STACK:102400000,102400000")
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <string>
#include <iostream>
#include <iomanip>
#include <cmath>
#include <map>
#include <set>
#include <queue>
using namespace std;

#define ls(rt) rt*2
#define rs(rt) rt*2+1
#define ll long long
#define ull unsigned long long
#define rep(i,s,e) for(int i=s;i<e;i++)
#define repe(i,s,e) for(int i=s;i<=e;i++)
#define CL(a,b) memset(a,b,sizeof(a))
#define IN(s) freopen(s,"r",stdin)
#define OUT(s) freopen(s,"w",stdout)
const ll ll_INF = ((ull)(-1))>>1;
const double EPS = 1e-8;
const double pi = acos(-1.0);
const int INF = 100000000;

const int MAXN = 1e5+50;
ll a[MAXN],ans[MAXN];
const ll mi[20]={1,2,4,8,16,
            32,64,128,256,512,
            1024,2048,4096,8192,16384,
            32768,65536};

ll n;
void print()
{
    ll out=0;
    out=ans[a[0]]^a[0];
    for(ll i=1;i<=n;i++)
        out+=ans[a[i]]^a[i];
    printf("%I64d\n",out);
    printf("%I64d",ans[a[0]]);

    for(ll i=1;i<=n;i++)
        printf(" %I64d",ans[a[i]]);
    putchar('\n');
}


int main()
{
    //OUT("hdu5014.txt");
    //IN("hdu5014.txt");
    while(~scanf("%I64d",&n))
    {
        for(ll i=0;i<=n;i++)
            scanf("%I64d",&a[i]),ans[i]=i;

        //int id=lower_bound(mi,mi+17,n)-mi;  // id>=1  因为n>=1
        int id=16;
        while(mi[id]>n && id>=0)id--;
        if(id == 0) //说明n == 1
        {
            ans[0]=1;
            ans[1]=0;
            print();
            continue;
        }
        if(mi[id]>n)id--;
        ll cnt=0,last=0,ls=n;
        while(id>=0&& ls>=0)
        {
            ll k;
            for(k=0;mi[id]-k-1>=0 && k+mi[id]<=ls;k++)
            {
                //printf("id=%d  mi[id]+k=%d mi[id]-k-1=%d\n",id,mi[id]+k,mi[id]-k-1);
                swap(ans[mi[id]+k],ans[mi[id]-k-1]);
                //cnt+=2;
                last=mi[id]-k-1;
            }
            ls=last-1;
            id--;
        }
        if(ans[0]==0 && ans[1]==1)swap(ans[0],ans[1]);
        print();
    }
    return 0;
}

hdu 5014 思维题/推理