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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
#135221#6637. Perfect StringsDelay_for_five_minutes#ML 0ms0kbC++202.7kb2023-08-05 12:56:512023-08-05 12:56:54

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.
  • [2023-08-05 12:56:54]
  • 评测
  • 测评结果:ML
  • 用时:0ms
  • 内存:0kb
  • [2023-08-05 12:56:51]
  • 提交

answer

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
int n , c;
const int mod = 1e9 + 7;
const int N = 2e7 + 5;
int t[20000005] , rt[20000005];
int st[N + 5] , rst[N + 5];
int fpow(int a,int b)
{
    int ans = 1;
    while(b) {
        if(b & 1) ans = 1LL*ans*a%mod;
        a = 1LL*a*a%mod;b >>= 1;
    }
    return ans;
}
int C(int i,int j)
{
    assert(i >= j);
    return 1LL*t[i]*rt[j] % mod * rt[i - j] % mod;
}
int cata(int n)
{
    if(n == 0) return 1;
    return (C(2*n , n) - C(2*n , n + 1) + mod) % mod;
}
int f[N + 5] , g[N + 5];
void solve()
{
    cin >> n >> c;
    if(c == 1) {
        cout << 1 << '\n' ; return ;
    }

    int ans = 0;
    int t = 1LL*c*fpow(c - 1 , mod - 2) % mod;
    t = (t - 1 + mod) % mod;

  /*  f[0] = 1;
    for(int i = 1;i <= n;i++) {
        for(int j = 0 ; j <i  ;j++) {
            f[i] = (f[i] + 1LL*f[j]*cata(i - j) % mod * t) % mod;
        }
    }

    ans = 1LL*f[n]*fpow(c - 1, n + 1) % mod;
    ans = 1LL*ans*(t + 1) % mod;
    cout << ans << '\n';*/

    int g4 = 16; f[0] = 1 ; f[1] = mod - 2;
    for(int i = 2;i <= n + 2;i++) {
        f[i] = 1LL * g4 * st[2*i - 3] % mod * rst[2 * i] % mod;
        f[i] = mod - f[i];
     //   printf("I %d %d , %d %d\n",i,mod - f[i],g4,st[2*i - 3]);
        g4 = 1LL*g4*4 % mod;
    }
    for(int i = 0;i <= n;i++) f[i] = 1LL*f[i]*(mod - t) % mod;

    f[0] = (f[0] - t + mod) % mod;
    f[1] = (f[1] + 2LL*(1 + t)) % mod;

    int a = 1LL * (mod - t) * fpow(1LL*(1+t)*(1+t)%mod , mod - 2) % mod, fn = 0;
    //a = 1LL*a*fpow(4 , mod - 2) % mod;
    //printf("A %d\n",a);
    if(a == 0) {
        assert(0);
    }
    else{
        int  dt = 1 , sp = mod - fpow(a , mod - 2);
        for(int i = 0;i <= n;i++) {
            fn = (fn + 1LL * dt * f[n - i]) % mod;
            dt = 1LL * dt * sp % mod;
        }
        int d = 2LL*a*(1 + t) % mod * (1 + t) % mod;
        fn = 1LL*fn*fpow(d , mod - 2) % mod;
    }

    ans = 1LL*fn*fpow(c - 1, n + 1) % mod;
    ans = 1LL*ans*(t + 1) % mod;
    cout << ans << '\n';
}
int main()
{
   // freopen("in.txt","r",stdin);
   // freopen("out2.txt","w",stdout);
    ios::sync_with_stdio(false) ; cin.tie(0) ; cout.tie(0);
    t[0] = 1;
    for(int i = 1;i <= N;i++) t[i] = 1LL*t[i - 1]*i % mod;
    rt[N] = fpow(t[N] , mod - 2);
    for(int i = N - 1;i >= 0;i--) rt[i] = 1LL*rt[i + 1]*(i + 1) % mod;

    st[0] = st[1] = 1;
    for(int i = 2;i <= N;i++) st[i] = 1LL*st[i - 2]*i % mod;
    rst[N] = fpow(st[N] , mod - 2) ; rst[N - 1] = fpow(st[N - 1] , mod - 2);
    for(int i = N - 2;i >= 0;i--) {
        rst[i] = 1LL*rst[i + 2]*(i + 2) % mod;
    }

    int T;cin >> T;
    while(T--) solve();
    return 0;
}

详细

Test #1:

score: 0
Memory Limit Exceeded

input:

2
3 1
2 2

output:


result: