2017多校第7场 HDU 6129 Just do it 找规律

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=6129

题意：求a序列后m次xor前缀和

解法：

手动对1位置对每个位置的贡献打表发现

第一次 贡献为 1 1 1 1 1 1 1 1 1 1 1

第二次 贡献为 1 0 1 0 1 0 1 0 1 0 1 0

第四次 贡献为 1 3个0 1 3个0 1 3个0 1 3个0

第八次 贡献为 1 7个0 1 7个0 1 7个0 1 7个0

...

这是比赛之后才知道的，看着比赛的时候通过了200+人，被虐记。

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int maxn = 2e5+10;
struct FastIO
{
    static const int S = 1310720;
    int wpos;
    char wbuf[S];
    FastIO() : wpos(0) {}
    inline int xchar()
    {
        static char buf[S];
        static int len = 0, pos = 0;
        if(pos == len)
            pos = 0, len = fread(buf, 1, S, stdin);
        if(pos == len)
            exit(0);
        return buf[pos ++];
    }
    inline unsigned long long xuint()
    {
        int c = xchar();
        unsigned long long x = 0;
        while(c <= 32)
            c = xchar();
        for(; ‘0‘ <= c && c <= ‘9‘; c = xchar())
            x = x * 10 + c - ‘0‘;
        return x;
    }
    inline long long xint()
    {
        long long s = 1;
        int c = xchar(), x = 0;
        while(c <= 32)
            c = xchar();
        if(c == ‘-‘)
            s = -1, c = xchar();
        for(; ‘0‘ <= c && c <= ‘9‘; c = xchar())
            x = x * 10 + c - ‘0‘;
        return x * s;
    }
    inline void xstring(char *s)
    {
        int c = xchar();
        while(c <= 32)
            c = xchar();
        for(; c > 32; c = xchar())
            * s++ = c;
        *s = 0;
    }
    inline double xdouble()
    {
        bool sign = 0;
        char ch = xchar();
        double x = 0;
        while(ch <= 32)
            ch = xchar();
        if(ch == ‘-‘)
            sign = 1, ch = xchar();
        for(; ‘0‘ <= ch && ch <= ‘9‘; ch = xchar())
            x = x * 10 + ch - ‘0‘;
        if(ch == ‘.‘)
        {
            double tmp = 1;
            ch = xchar();
            for(; ch >= ‘0‘ && ch <= ‘9‘; ch = xchar())
                tmp /= 10.0, x += tmp * (ch - ‘0‘);
        }
        if(sign)
            x = -x;
        return x;
    }
    inline void wchar(int x)
    {
        if(wpos == S)
            fwrite(wbuf, 1, S, stdout), wpos = 0;
        wbuf[wpos ++] = x;
    }
    inline void wint(long long x)
    {
        if(x < 0)
            wchar(‘-‘), x = -x;
        char s[24];
        int n = 0;
        while(x || !n)
            s[n ++] = ‘0‘ + x % 10, x /= 10;
        while(n--)
            wchar(s[n]);
    }
    inline void wstring(const char *s)
    {
        while(*s)
            wchar(*s++);
    }
    inline void wdouble(double x, int y = 6)
    {
        static long long mul[] = {1, 10, 100, 1000, 10000, 100000, 1000000, 10000000, 100000000, 1000000000, 10000000000LL, 100000000000LL, 1000000000000LL, 10000000000000LL, 100000000000000LL, 1000000000000000LL, 10000000000000000LL, 100000000000000000LL};
        if(x < -1e-12)
            wchar(‘-‘), x = -x;
        x *= mul[y];
        long long x1 = (long long) floorl(x);
        if(x - floor(x) >= 0.5)
            ++x1;
        long long x2 = x1 / mul[y], x3 = x1 - x2 * mul[y];
        wint(x2);
        if(y > 0)
        {
            wchar(‘.‘);
            for(size_t i = 1; i < y && x3 * mul[i] < mul[y]; wchar(‘0‘), ++i);
            wint(x3);
        }
    }
    ~FastIO()
    {
        if(wpos)
            fwrite(wbuf, 1, wpos, stdout), wpos = 0;
    }
} io;

int a[maxn];
int main()
{
    int T,n,m;
    T = io.xint();
    while(T--)
    {
        n = io.xint();
        m = io.xint();
        for(int i=1; i<=n; i++) a[i] = io.xint();
        for(int k=0; (1<<k)<=m; k++){
            if(m&(1<<k)){
                for(int j=1; j<=n; j++){
                    if((long long)j+(1<<k)>(long long)n) break;
                    a[j+(1<<k)] ^= a[j];
                }
            }
        }
        for(int i=1; i<n; i++) io.wint(a[i]), io.wchar(‘ ‘);
        io.wint(a[n]), io.wchar(‘\n‘);
    }
    return 0;
}