QOJ.ac

QOJ

IDProblemSubmitterResultTimeMemoryLanguageFile sizeSubmit timeJudge time
#225327#5661. Multi-LaddersBUET_Twilight#AC ✓1ms3892kbC++232.3kb2023-10-24 14:58:202023-10-24 14:58:20

Judging History

你现在查看的是最新测评结果

  • [2023-10-24 14:58:20]
  • 评测
  • 测评结果:AC
  • 用时:1ms
  • 内存:3892kb
  • [2023-10-24 14:58:20]
  • 提交

answer

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp> 
#include <ext/pb_ds/detail/standard_policies.hpp>
using namespace __gnu_pbds;
using namespace std;
#define getbit(n, i) (((n) & (1LL << (i))) != 0) 
#define setbit0(n, i) ((n) & (~(1LL << (i)))) 
#define setbit1(n, i) ((n) | (1LL << (i))) 
#define togglebit(n, i) ((n) ^ (1LL << (i))) 
#define lastone(n) ((n) & (-(n))) 
char gap = 32;
#define int long long
#define ll long long 
#define lll __int128_t
#define pb push_back
// #define double long double
#define pii pair<int,int>
typedef tree<
int,
null_type,
less<int>,
rb_tree_tag,
tree_order_statistics_node_update>
ordered_set;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
ll hashPrime = 1610612741;
typedef pair<double, double> pdd;
const double EPS = 1e-12;


typedef vector<vector<int>> matrix;

const int mod = 1e9 + 7;

matrix mul(matrix a, matrix b){
    int n = a.size();
    matrix c(n, vector<int>(n));
    for(int i=0;i<n;i++){
        for(int j=0;j<n;j++){
            for(int k=0;k<n;k++){
                c[i][j] += a[i][k]*b[k][j] % mod;
                c[i][j] %= mod;
            }
        }
    }
    return c;
}

long long binpow(long long a, long long b, long long m) {
    a %= m;
    long long res = 1;
    while (b > 0) {
        if (b & 1)
            res = res * a % m;
        a = a * a % m;
        b >>= 1;
    }
    return res;
}

matrix power(matrix a, int b){
    int n = a.size();
    matrix res(n, vector<int>(n));
    for(int i=0;i<n;i++) res[i][i] = 1;
    while(b){
        if(b&1) res = mul(res, a);
        a = mul(a, a);
        b >>= 1;
    }
    return res;
}

void solve(){
    int n, k, l;
    cin >> n >> k >> l;    

    matrix b = {
        {0, 1},
        {l-1, l-2},    
    };
    matrix a = {
        {l},
        {0},
    };

    matrix res = mul(power(b, k-1), a);
    int val = res[1][0];

    int e = (l * l - 3*l + 3) % mod;
    e = binpow(e, n-1, mod);
    e = binpow(e, k, mod);
    cout << (e * val) % mod << "\n";



}
signed main(){

    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    int tc;
    cin>>tc;
    while(tc--)
    solve();
    
    return 0;
}

Details

Tip: Click on the bar to expand more detailed information

Test #1:

score: 100
Accepted
time: 1ms
memory: 3644kb

input:

1
2 3 3

output:

162

result:

ok single line: '162'

Test #2:

score: 0
Accepted
time: 0ms
memory: 3892kb

input:

20
2 3 3
1 3 3
10 3 0
10 3 2
1 21 2
1 22 0
2000 15000 2000
12000 30000 200000
1000000000 3 3
2 1000000000 3
2 3 100000000
1000000000 1000000000 10
1000000000 3 100000000
2 1000000000 100000000
1 1000000000 10
1 1000000000 100000000
1 1000 100000000
1000000000 1000000000 0
1000000000 1000000000 1
100...

output:

162
6
0
0
0
0
349400141
243010659
52489881
53690844
176686901
218103365
558243892
991895211
693053429
883715672
80402569
0
0
311752813

result:

ok 20 lines