异或密码---hdu5968（CCPC合肥，二分）

 题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5968

思路：先把所有的连续异或值保存起来，排序，然后用二分找到距离x最近的那个点，判断即可；

 

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <algorithm>
#include <bitset>
#include <iostream>
#include <time.h>

typedef long long LL;

using namespace std;

const int N = ;
const double eps = 1e-;
const int INF = 0x3f3f3f3f;
const int mod = ;

struct node
{
    int len, num;
}c[N*N];

int a[N], b[N], k, c1[N*N], c2[N*N];

bool cmp(node p, node q)
{
    if(p.num!=q.num)
        return p.num<q.num;
    return p.len < q.len;
}

int Find(int num)
{
    int pos = upper_bound(c1, c1+k, num)-c1;
    return c2[pos-];
}

int solve(int x)///Find x in the c[];
{
    int pos = lower_bound(c1, c1+k, x)-c1;

    if(pos ==  || c1[pos] == x)///如果数组中含有x，那就找到异或值为x的最大长度;
        return Find(c1[pos]);

    int p = abs(c1[pos]-x);
    int q = abs(c1[pos-]-x);
    int ans1 = Find(c1[pos]);
    int ans2 = Find(c1[pos-]);

    if(p < q || (p == q && ans1>ans2))
        return ans1;
    return ans2;
}

int main()
{
    int T, n, m;

    scanf("%d", &T);

    while(T --)
    {
        scanf("%d", &n);

        memset(a, , sizeof(a));
        memset(b, , sizeof(b));
        memset(c, , sizeof(c));

        for(int i=; i<=n; i++)
        {
            scanf("%d", &a[i]);
            b[i] = b[i-]^a[i];
        }

        k = ;

        for(int i=; i<=n; i++)/// the length is i;
        {
            for(int j=; i+j-<=n; j++)/// from j to j+i;
            {
                int num = b[j+i-]^b[j-];
                c[k].len = i;
                c[k++].num = num;
            }
        }

        sort(c, c+k, cmp);

        for(int i=; i<k; i++)
        {
            c1[i] = c[i].num;
            c2[i] = c[i].len;
        }

        scanf("%d", &m);

        for(int i=; i<=m; i++)
        {
            int x;

            scanf("%d", &x);

            int ans = solve(x);

            printf("%d\n", ans);
        }
        printf("\n");
    }
    return ;
}