HDU 4869 Turn the pokers(思维+逆元)

考试的时候没有做出来。。。

想到了答案一定是一段连续的区间，一直在纠结BFS判断最后的可行1数。

原来直接模拟一遍就可以算出来最后的端点。。。

剩下的就是组合数取模了，用逆元就行了。。。

# include <cstdio>

# include <cstring>

# include <cstdlib>

# include <iostream>

# include <vector>

# include <queue>

# include <stack>

# include <map>

# include <set>

# include <cmath>

# include <algorithm>

using namespace std;

# define lowbit(x) ((x)&(-x))

# define pi 3.1415926535

# define eps 1e-

# define MOD

# define INF

# define mem(a,b) memset(a,b,sizeof(a))

# define FOR(i,a,n) for(int i=a; i<=n; ++i)

# define FO(i,a,n) for(int i=a; i<n; ++i)

# define bug puts("H");

# define lch p<<,l,mid

# define rch p<<|,mid+,r

# define mp make_pair

# define pb push_back

typedef pair<int,int> PII;

typedef vector<int> VI;

# pragma comment(linker, "/STACK:1024000000,1024000000")

typedef long long LL;

int Scan() {

    int res=, flag=;

    char ch;

    if((ch=getchar())=='-') flag=;

    else if(ch>=''&&ch<='') res=ch-'';

    while((ch=getchar())>=''&&ch<='')  res=res*+(ch-'');

    return flag?-res:res;

}

void Out(int a) {

    if(a<) {putchar('-'); a=-a;}

    if(a>=) Out(a/);

    putchar(a%+'');

}

const int N=;

//Code begin...

LL f[N];

LL pow_mod(LL a, LL n, LL mod){

    LL ret=, tmp=a%mod;

    while (n) {

        if (n&) ret=ret*tmp%MOD;

        tmp=tmp*tmp%MOD;

        n>>=;

    }

    return ret;

}

LL inv(LL a, LL mod){return pow_mod(a,mod-,mod);}

void init(){

    f[]=;

    FO(i,,N) f[i]=(f[i-]*i)%MOD;

}

int main ()

{

    int n, m, x, l, r, tmpl, tmpr;

    LL ans;

    init();

    while (~scanf("%d%d",&n,&m)) {

        l=r=;

        ans=;

        FOR(i,,n) {

            scanf("%d",&x);

            if (l>=x) tmpl=l-x;

            else if(r>=x) tmpl=((l%)==(x%))?:;

            else tmpl=x-r;

            if (r+x<=m) tmpr=r+x;

            else if(l+x<=m) tmpr=(((l+x)%)==(m%)?m:m-);

            else tmpr=*m-l-x;

            l=tmpl; r=tmpr;

        }

        for (int i=l; i<=r; i+=) ans=(ans+(f[m]*inv(f[i]*f[m-i]%MOD,MOD))%MOD)%MOD;

        printf("%lld\n",ans);

    }

    return ;

}