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IDProblemSubmitterResultTimeMemoryLanguageFile sizeSubmit timeJudge time
#471429#6698. Flipping Gamechenxinyang2006AC ✓3ms3880kbC++204.0kb2024-07-10 21:07:452024-07-10 21:07:45

Judging History

你现在查看的是最新测评结果

  • [2024-07-10 21:07:45]
  • 评测
  • 测评结果:AC
  • 用时:3ms
  • 内存:3880kb
  • [2024-07-10 21:07:45]
  • 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 C[105][105];
int T;
int n,k,m,inst;
char s[105],t[105];

void solve(){
    scanf("%d%d%d",&n,&k,&m);
    scanf("%s",s + 1);
    scanf("%s",t + 1);
    inst = 0;
    rep(i,1,n) inst += s[i] != t[i];
    Z answer = 0,cur,sum;
    rep(i,0,n){
        cur = sum = 0;
        rep(j,0,min(m,i)){
            if(j & 1) cur -= C[i][j] * C[n - i][m - j];
            else cur += C[i][j] * C[n - i][m - j];
        }
        rep(j,0,min(inst,i)){
            if(j & 1) sum -= C[inst][j] * C[n - inst][i - j];
            else sum += C[inst][j] * C[n - inst][i - j];
        }
        answer += power(cur,k) * sum;
    }
    answer /= power(2,n);
    printf("%d\n",answer.val());
}

int main(){
//    freopen("test.in","r",stdin);
    rep(i,0,100){
        C[i][0] = 1;
        rep(j,1,i) C[i][j] = C[i - 1][j - 1] + C[i - 1][j];
    }
    scanf("%d",&T);
    while(T--) solve();
	return 0;
}

这程序好像有点Bug,我给组数据试试?

Details

Tip: Click on the bar to expand more detailed information

Test #1:

score: 100
Accepted
time: 0ms
memory: 3880kb

input:

3
3 2 1
001
100
3 1 2
001
100
3 3 2
001
100

output:

2
1
7

result:

ok 3 number(s): "2 1 7"

Test #2:

score: 0
Accepted
time: 3ms
memory: 3808kb

input:

1000
8 50 2
11111001
01100001
13 4 5
0010011001101
0000001010010
15 58 12
011111110110100
011010000101000
15 30 2
000101100111101
100010100110000
16 99 15
0111011010111101
1000100101011100
7 73 1
0010010
1010111
1 45 1
1
1
15 64 4
111000001000100
111000110011011
13 16 6
0000001101000
0101001010111
5...

output:

0
0
0
0
0
565123576
0
671397628
866048220
0
0
0
934159397
0
0
0
657964873
0
0
0
297620792
518284447
16636231
1
294524820
259008109
0
0
0
0
0
0
990349196
899244686
0
0
497963164
0
49814547
0
0
0
0
0
529815127
411739397
562040211
0
0
0
0
0
0
0
531433326
0
77531359
703399699
0
0
0
1
0
896329183
0
0
0
0...

result:

ok 1000 numbers