Submission #993761


Source Code Expand

#include <algorithm>
#include <climits>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <ctime>
#include <iostream>
#include <sstream>
#include <functional>
#include <map>
#include <string>
#include <cstring>
#include <vector>
#include <queue>
#include <stack>
#include <deque>
#include <set>
#include <list>
#include <numeric>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<ll,ll> P;
const double PI = 3.14159265358979323846;
const double EPS = 1e-12;
const ll INF = 1LL<<29;
const ll mod = 1e9+7;
#define rep(i,n) for(int (i)=0;(i)<(ll)(n);++(i))
#define repd(i,n,d) for(ll (i)=0;(i)<(ll)(n);(i)+=(d))
#define all(v) (v).begin(), (v).end()
#define pb(x) push_back(x)
#define mp(x,y) make_pair((x),(y))
#define mset(m,v) memset((m),(v),sizeof(m))
#define chmin(X,Y) ((X)>(Y)?X=(Y),true:false)
#define chmax(X,Y) ((X)<(Y)?X=(Y),true:false)
#define fst first
#define snd second
#define UNIQUE(x) (x).erase(unique(all(x)),(x).end())
template<class T> ostream &operator<<(ostream &os, const vector<T> &v){int n=v.size();rep(i,n)os<<v[i]<<(i==n-1?"":" ");return os;}

#define N 310
ll dp[N], dp2[N];
int n, m;
ll mod = 1e9+7;

int main(){
	cin>>n>>m;
	dp[0][0] = 1;
	rep(i, m){
		memcpy(dp2, dp, sizeof(dp));
		mset(dp, 0);
		for(int j = 0; j <= n; j++){
			if(j<n-1){
				(dp[j]+=dp2[j]*(j+1))%=mod;
				(dp[j+1]+=dp2[j]*(n-j-1))%=mod;
			} else if(j==n-1){
				(dp[j]+=dp2[j]*j)%=mod;
				(dp[j+1]+=dp2[j])%=mod;
			} else if(j==n){
				(dp[j]+=dp2[j]*n)%=mod;
			}
		}
	}
	cout<<dp[n]<<endl;
	return 0;
}

Submission Info

Submission Time
Task F - Road of the King
User Lepton
Language C++14 (GCC 5.4.1)
Score 0
Code Size 1626 Byte
Status CE

Compile Error

./Main.cpp:44:4: error: conflicting declaration ‘ll mod’
 ll mod = 1e9+7;
    ^
./Main.cpp:27:10: note: previous declaration as ‘const ll mod’
 const ll mod = 1e9+7;
          ^
./Main.cpp: In function ‘int main()’:
./Main.cpp:48:9: error: invalid types ‘ll {aka long long int}[int]’ for array subscript
  dp[0][0] = 1;
         ^