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ID题目提交者结果用时内存语言文件大小提交时间测评时间
#118446#6137. Sub-cycle Graphchenxinyang2006WA 94ms4752kbC++144.0kb2023-07-03 16:02:312023-07-03 16:02:34

Judging History

你现在查看的是最新测评结果

  • [2023-08-10 23:21:45]
  • System Update: QOJ starts to keep a history of the judgings of all the submissions.
  • [2023-07-03 16:02:34]
  • 评测
  • 测评结果:WA
  • 用时:94ms
  • 内存:4752kb
  • [2023-07-03 16:02:31]
  • 提交

answer

#include <bits/stdc++.h>
#define rep(i,j,k) for(int i=(j);i<=(k);i++)
#define per(i,j,k) for(int i=(j);i>=(k);i--)
#define uint unsigned int
#define ll long long
#define ull unsigned long long
#define db double
#define ldb long double
#define pii pair<int,int>
#define pll pair<ll,ll>
#define mkp make_pair
#define eb emplace_back
#define SZ(S) (int)S.size()
//#define mod 998244353
#define mod 1000000007
#define inf 0x3f3f3f3f
#define linf 0x3f3f3f3f3f3f3f3f
using namespace std;

template <class T>
void chkmax(T &x,T y){
    if(x < y) x = y;
}

template <class T>
void chkmin(T &x,T y){
    if(x > y) x = y;
}

inline int popcnt(int x){
    return __builtin_popcount(x);
}

inline int ctz(int x){
    return __builtin_ctz(x);
}

template <int P>
class mod_int
{
    using Z = mod_int;

private:
    static int mo(int x) { return x < 0 ? x + P : x; }

public:
    int x;
    int val() const { return x; }
    mod_int() : x(0) {}
    template <class T>
    mod_int(const T &x_) : x(x_ >= 0 && x_ < P ? static_cast<int>(x_) : mo(static_cast<int>(x_ % P))) {}
    bool operator==(const Z &rhs) const { return x == rhs.x; }
    bool operator!=(const Z &rhs) const { return x != rhs.x; }
    Z operator-() const { return Z(x ? P - x : 0); }
    Z pow(long long k) const
    {
        Z res = 1, t = *this;
        while (k)
        {
            if (k & 1)
                res *= t;
            if (k >>= 1)
                t *= t;
        }
        return res;
    }
    Z &operator++()
    {
        x < P - 1 ? ++x : x = 0;
        return *this;
    }
    Z &operator--()
    {
        x ? --x : x = P - 1;
        return *this;
    }
    Z operator++(int)
    {
        Z ret = x;
        x < P - 1 ? ++x : x = 0;
        return ret;
    }
    Z operator--(int)
    {
        Z ret = x;
        x ? --x : x = P - 1;
        return ret;
    }
    Z inv() const { return pow(P - 2); }
    Z &operator+=(const Z &rhs)
    {
        (x += rhs.x) >= P && (x -= P);
        return *this;
    }
    Z &operator-=(const Z &rhs)
    {
        (x -= rhs.x) < 0 && (x += P);
        return *this;
    }
    Z operator-()
    {
        return -x;
    }
    Z &operator*=(const Z &rhs)
    {
        x = 1ULL * x * rhs.x % P;
        return *this;
    }
    Z &operator/=(const Z &rhs) { return *this *= rhs.inv(); }
#define setO(T, o)                                  \
    friend T operator o(const Z &lhs, const Z &rhs) \
    {                                               \
        Z res = lhs;                                \
        return res o## = rhs;                       \
    }
    setO(Z, +) setO(Z, -) setO(Z, *) setO(Z, /)
#undef setO
    
    friend istream& operator>>(istream& is, mod_int& x)
    {
        long long tmp;
        is >> tmp;
        x = tmp;
        return is;
    }
    friend ostream& operator<<(ostream& os, const mod_int& x)
    {
        os << x.val();
        return os;
    }
};

using Z = mod_int<mod>;
Z power(Z p,ll k){
    Z ans = 1;
    while(k){
        if(k % 2 == 1) ans *= p;
        p *= p;
        k /= 2;
    }
    return ans;
}
Z fact[100005],ifac[100005],P[100005];

Z C(int N,int M){
    if(N < M || M < 0) return 0;
    return fact[N] * ifac[M] * ifac[N - M];
}

void init(int L){
    fact[0] = 1;
    rep(i,1,L) fact[i] = fact[i - 1] * i;
    ifac[L] = 1 / fact[L];
    per(i,L - 1,0) ifac[i] = ifac[i + 1] * (i + 1);
    P[0] = 1;
    rep(i,1,L) P[i] = P[i - 1] / 2;
}
int T,n,m;

void solve(){
    ll tmp;
    scanf("%d%lld",&n,&tmp);
    if(tmp > n){
        printf("0\n");
        return;
    }
    if(tmp == n){
        printf("%d\n",fact[n - 1].val());
        return;
    }
    m = n - tmp;
    Z ans = 0;
    rep(k,0,m) ans += C(m,k) * P[m - k] * C(n - m - 1,m - k - 1);
    ans *= fact[n] * ifac[m];
    printf("%d\n",ans.val());
}

int main(){
    scanf("%d",&T);
    init(100000);
    while(T--) solve();
	return 0;
}

詳細信息

Test #1:

score: 100
Accepted
time: 13ms
memory: 4752kb

input:

3
4 2
4 3
5 3

output:

15
12
90

result:

ok 3 number(s): "15 12 90"

Test #2:

score: -100
Wrong Answer
time: 94ms
memory: 4736kb

input:

17446
3 0
3 1
3 2
3 3
4 0
4 1
4 2
4 3
4 4
5 0
5 1
5 2
5 3
5 4
5 5
6 0
6 1
6 2
6 3
6 4
6 5
6 6
7 0
7 1
7 2
7 3
7 4
7 5
7 6
7 7
8 0
8 1
8 2
8 3
8 4
8 5
8 6
8 7
8 8
9 0
9 1
9 2
9 3
9 4
9 5
9 6
9 7
9 8
9 9
10 0
10 1
10 2
10 3
10 4
10 5
10 6
10 7
10 8
10 9
10 10
11 0
11 1
11 2
11 3
11 4
11 5
11 6
11 7
11...

output:

0
3
3
2
0
6
15
12
6
0
10
45
90
60
24
0
15
105
375
630
360
120
0
21
210
1155
3465
5040
2520
720
0
28
378
2940
13545
35280
45360
20160
5040
0
36
630
6552
42525
170100
393120
453600
181440
40320
0
45
990
13230
114345
643545
2286900
4762800
4989600
1814400
362880
0
55
1485
24750
273735
2047815
10239075
...

result:

wrong answer 1st numbers differ - expected: '1', found: '0'