2016多校第六场1001-1003（hdu5793&hdu5794&hdu5795）

这场就做出一道题，怎么会有窝这么辣鸡的人呢？

1001

很复杂的公式，打表找的规律，最后是m^0+m^1+...+m^n，题解直接是(m^(n+1)-1)/(m-1)，长姿势，原来还能化简……做出来的题不写题解了。

1002

一个棋盘，走棋的姿势要满足(x1-x2)^2+(y1-y2)^2==5,也就是以“日”字走，且只能向右下走。

其中有一些障碍不能经过，注意障碍有可能在终点，求从(1,1)走到(n,m)的路径数。

容斥＋组合数 这题的简单版CF559C（对啊，我做过这题，我还是没做出来，wa了十多次，啦啦啦）

over是防止重点的，感觉不会有重点，但是比赛时是实在没辙了=_=#

#include <cstdio>
#include <cstring>
#include <vector>
#include <algorithm>
#include <iostream>
using namespace std;
typedef long long ll;
const ll MOD = 110119;
const int MAX_P = 2000005;

ll powMod(ll a, ll b, ll mod)
{
    ll res = 1;
    while (b) {
        if (b & 1) res = res * a % mod;
        a = a * a % mod;
        b >>= 1;
    }
    return res;
}

ll fac[MAX_P];
void getFact()
{
    fac[0] = 1;
    for (int i = 1; i <= 2000000; ++i)
        fac[i] = fac[i - 1] * i % MOD;
}

ll Lucas(ll n, ll m, ll p)
{
    ll res = 1;
    while (n && m) {
        ll a = n % p;
        ll b = m % p;
        if (a < b) return 0;
        res = res * fac[a] % p * powMod(fac[b] * fac[a - b] % p, p - 2, p) % p;
        n /= p;
        m /= p;
    }
    return res;
}

struct Point {
    ll x, y;
    Point(ll x, ll y):x(x),y(y){}
    Point(){}
    bool operator<(const Point a) const
    {
        if (x == a.x) return y < a.y;
        return x < a.x;
    }
    bool operator==(const Point a) const
    {
        if (x == a.x && y == a.y) return true;
        return false;
    }
} p[2005];

ll cal(Point a, Point b)
{
	ll dx = b.x - a.x;
	ll dy = b.y - a.y;
	ll r, c;
	if ((dx*2-dy)%3 || (dy*2-dx)%3) return 0;
	ll inv = powMod(3, MOD-2, MOD);
	r = (2 * dx - dy) / 3;
	c = (2 * dy - dx) / 3;
	if (r < 0 || c < 0) return 0;
	if (c == 0 || r == 0) return 1;
	return Lucas(r+c, r, MOD);
}

ll ans[2005];
bool over[2005];
int main()
{
    //freopen("in", "r", stdin);
    ll m, n, r;
    getFact();
    int cas = 0;
    while (cin >> m >> n >> r) {
    	printf("Case #%d: ", ++cas);
        Point s(1, 1);
        for (int i = 0; i < r; ++i) scanf("%lld%lld", &p[i].x, &p[i].y);
        p[r].x = m, p[r].y = n;
        sort(p, p + r);
        if (p[r-1].x == m && p[r-1].y == n) {
        	printf("0\n");
        	continue;
        }
        memset(over, false, sizeof over);
        for (int i = 1; i < r; ++i) {
        	if (p[i] == p[i-1]) over[i] = true;
        }
        for (int i = 0; i <= r; ++i) {
            if (over[i]) continue;
            ans[i] = cal(s, p[i]);
            for (int j = 0; j < i; ++j) {
            	if (over[j]) continue;
                if (p[j].x < p[i].x && p[j].y < p[i].y) {
                    ans[i] = ((ans[i] - cal(p[j], p[i]) * ans[j] % MOD) % MOD + MOD) % MOD;
                }
            }
        }
        ans[r] = (ans[r] + MOD) % MOD;
        printf("%lld\n", ans[r]);
    }
    return 0;
}

1003

nim博弈变形

有n堆糖，每次拿走一堆的任意个， 或者把一堆分成三堆。

对一堆求SG函数值，然后打表找规律，感觉不难，但是没想明白，后来有一个网友提醒我（额 虽然比赛时这样不太好……），终于开始写，其实一开始我是蒙蔽的，写写的突然清晰了，但是错了一个数……sg[2]应该是2，我竟然随手写成了0，妈蛋T^T

#include <cstdio>
#include <cstring>
#include <vector>
#include <algorithm>
#include <iostream>
using namespace std;

#define PF(x) cout << "debug: " << x << " ";
#define EL cout << endl;
#define PC(x) puts(x);

typedef long long ll;

const int N = 1000005;
const int MOD = 1e9+7;

int a[N];
int sg[N];

int get_sg(int x)
{
    if (x == 0) return 0;
    if (x == 1) return 1;
    if (x == 2) return 2;

    if (sg[x] != -1) return sg[x];

    int mex[200000] = {0};

    for (int i = 1; i < x; ++i) {
        for (int j = i; j < x-i; ++j) {
            int k = x - i - j;
            if (k <= 0) break;
            int tmp = get_sg(i) ^ get_sg(j) ^ get_sg(k);
            mex[tmp] = 1;
        }
    }

    for (int i = 0; i < x; ++i) {
        mex[get_sg(i)] = 1;
    }

    for (int i = 0; ; i++) {
        if (!mex[i]) return sg[x] = i;
    }
}

int main(int argc, char const *argv[])
{
    // memset(sg, -1, sizeof sg);

    // for (int i = 0; i < 100; ++i) {
    //     cout << i << " " << get_sg(i) << endl;
    // }

    //freopen("in", "r", stdin);
    int T;
    scanf("%d", &T);
    while (T--) {
        int n;
        scanf("%d", &n);
        int ans = 0;
        for (int i = 0; i < n; ++i) {
            scanf("%d", a+i);
            if (a[i] % 8 == 0) a[i]--;
            else if (a[i] % 8 == 7) a[i]++;
            ans ^= a[i];
        }
        printf("%s\n", ans ? "First player wins.":"Second player wins.");
    }
    return 0;
}

怎么会有窝这么弱的人呢？