LightOJ

链接：

https://vjudge.net/problem/LightOJ-1318

题意：

In a country named "Ajob Desh", people play a game called "Ajob Game" (or strange game). This game is actually a game of words. The rules for the game are as follows:

It's an N player game and players are numbered from 1 to N. And the players alternate turns in a circular way. Player 1 starts first. The next turn is for player 2, then player 3 and so on. After the turn for the Nth player, player 1 gets his turn again and the same procedure is continued.
In each turn a player has to propose a pair of words. Each of the words should have length L, and the words should differ in exactly M positions. As their language has K alphabetical symbols, a word is a collection of symbols from these K alphabets.
The pair of words proposed by a player should differ in exactly M positions, it means that there should be exactly M positions where the two words have different symbols, and in other positions they have same symbols. For example, 'abc' and 'abd' differ in exactly 1 position, 'abc' and 'aca' differ in exactly 2 positions, 'abc' and 'cab' differ in exactly 3 positions.
In each turn a player has to propose a new pair of words. Two pairs are different if at least one word is different. Note that here pair refers to unordered pair. Let A, B, C be three different words, then (A, B) and (B, A) are same, but (A, C) and (A, B) are different. For example, if a player already proposed {abc, def}, then none can propose {abc, def} or {def, abc}. But a player can propose {abc, fed} or {abc, abc} or {pqc, abc} etc.
If a player fails to propose a new pair of words, he is treated as the loser of the game. And the game ends.
Let N = 2, K = 2, L = 2, M = 1 and the alphabet is {ab}. All the words of length 2 are: {aa, ab, ba, bb}. Player 1 chooses pair {aa, ab} (differs in 1 position as M = 1) then player 2 chooses pair {ab, bb}. After that player 1 chooses {aa, ba} then player 2 chooses {bb, ba}. And then there is no pair left for player 1, and so, player 1 will lose.

Now this game is played by N players who know this game very well thus they play optimally. You are given N, K, L and M; you have to find the loosing player.

思路:

一对字符串，一边是，(k^l)种，同时另一边需要有m个不同，则是((k-1)^m)
共有(C_l^m*(k-1)^m*k^l)
同时有重复，要除2，考虑2和n不一定互质，当k为偶数时用k/2，否则用(k-1)/2

代码：

// #include<bits/stdc++.h>
#include<iostream>
#include<cstdio>
#include<vector>
#include<string.h>
#include<set>
#include<queue>
#include<algorithm>
#include<math.h>
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
const int MOD = 1e9+7;
const int MAXN = 1e6+10;

LL n, k, l, m;
int pos;
int Pri[MAXN], Isp[MAXN], Cnt[MAXN];

void Init()
{
    pos = 0;
    for (int i = 2;i < MAXN;i++)
    {
        if (Isp[i] == 0)
            Pri[++pos] = i;
        for (int j = 1;j <= pos && 1LL*i*Pri[j] < MAXN;j++)
        {
            Isp[i*Pri[j]] = 1;
            if (i%Pri[j] == 0)
                break;
        }
    }
}

void Upd(LL x, int sta)
{
    for (int i = 1;i <= pos;i++)
    {
        LL tmp = x;
        while(tmp)
        {
            Cnt[i] += sta*(tmp/Pri[i]);
            tmp /= Pri[i];
        }
    }
}

LL PowMod(LL a, LL b, LL p)
{
    LL res = 1;
    while(b)
    {
        if (b&1)
            res = res*a%p;
        a = a*a%p;
        b >>= 1;
    }
    return res;
}

LL C(LL n, LL m, LL p)
{
    memset(Cnt, 0, sizeof(Cnt));
    Upd(l, 1), Upd(m, -1), Upd(l-m, -1);
    LL ans = 1;
    for (int i = 1;i <= pos;i++)
        ans = ans*PowMod(Pri[i], Cnt[i], p)%p;
    return ans;
}

int main()
{
    // freopen("test.in", "r", stdin);
    Init();
    int t, cas = 0;
    scanf("%d", &t);
    while(t--)
    {
        printf("Case %d:", ++cas);
        scanf("%lld%lld%lld%lld", &n, &k, &l, &m);
        LL ans;
        if (m != 0)
        {
            if (k&1)
                ans = C(l, m, n)*PowMod(k, l, n)%n*PowMod(k-1, m-1, n)%n*(k/2)%n;
            else
                ans = C(l, m, n)*PowMod(k, l-1, n)%n*PowMod(k-1, m, n)%n*(k/2)%n;
        }
        else
            ans = PowMod(k, l, n);
        printf(" %lld
", ans+1);
    }

    return 0;
}