SOJ-4075（n个数异或第k大的数）

/******************************************************************************************************
 ** Copyright (C) 2011.07.01-2013.07.01
 ** Author: famousDT <[email protected]>
 ** Edit date: 2011-10-03
******************************************************************************************************/
#include <stdio.h>
#include <stdlib.h>//abs,atof(string to float),atoi,atol,atoll
#include <math.h>//atan,acos,asin,atan2(a,b)(a/b atan),ceil,floor,cos,exp(x)(e^x),fabs,log(for E),log10
#include <vector>
#include <queue> 
#include <map>
#include <time.h>
#include <set>
#include <stack>
#include <string>
#include <iostream>
#include <assert.h>
#include <string.h>//memcpy(to,from,count
#include <ctype.h>//character process:isalpha,isdigit,islower,tolower,isblank,iscntrl,isprll
#include <algorithm>
using namespace std;

typedef long long ll;

#define MY_PI acos(-1)
#define MY_MAX(a, b) ((a) > (b) ? (a) : (b))   
#define MY_MIN(a, b) ((a) < (b) ? (a) : (b))
#define MY_MALLOC(n, type) ((type *)malloc((n) * sizeof(type)))
#define MY_ABS(a) (((a) >= 0) ? (a) : (-(a)))
#define MY_INT_MAX 0x7fffffff

/*==========================================================*\
| 高斯消元+
| 把n个数拿去高斯消元，于是这n个数的所有异或结果就可以用那一组
| 线性基来表示了。 特判掉0以后，将K分解二进制，对应的是1的位的
| 那个基向量拿去异或就是答案
\*==========================================================*/

ll d[10000 + 1];
ll n;
ll base = 0;

void gauss(const ll m = 62)
{
    base =0;
    ll i, j;
    for (i = m; i >= 0; --i) {
        int flag = 0;
        for (j = base; j < n; ++j) {
            if ((d[j] >> i) & 1) {
                swap(d[base], d[j]);
                flag = 1;
                break;
            }
        }
        if (flag) {
            for (j = 0; j < n; ++j) {
                if (j != base && (d[j] & ((ll)1 << i)))
                    d[j] ^= d[base];
            }
            ++base;
        }
    }
}

int main()
{
    ll cases;
    ll i, j;
    ll q;
    ll ask;
    cin>>cases;
    ll index = 1;
    while (cases--) {
        scanf("%lld", &n);
        for (i = 0; i < n; ++i) {
            scanf("%lld", &d[i]);
        }
        gauss();
        printf("Case #%lld:\n", index++);
        scanf("%lld", &q);
        for (i = 0; i < q; ++i) {
            scanf("%lld", &ask);
            if (base < n) {
                if (ask == 1) {
                    printf("0\n");
                    continue;
                }
                else ask -= 1;
            }
            if (ask >= ((ll)1 << base)) {
                printf("-1\n");
                continue;
            }
            ll ans = 0;
            for (j = 0; j < 63; ++j) {
                if ((ask >> j) & 1)
                    ans ^= d[base - j - 1];
            }
            printf("%lld\n", ans);
        }
    }
    return 0;
}