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August 4, 2016 15:10
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SRM 694 Div2 Hard.
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#include <bits/stdc++.h> | |
using namespace std; | |
typedef pair<int,int> pii; | |
typedef long long ll; | |
#define ITR(i,c) for(auto i=begin(c);i!=end(c);i++) | |
#define FORE(x,arr) for(auto &x:arr) | |
#define FOR(i,a,n) for(int i=a;i<(int)(n);i++) | |
#define REP(i,n) FOR(i,0,n) | |
#define ALL(c) begin(c),end(c) | |
const int DX[4]={0,1,0,-1}, DY[4]={-1,0,1,0}; | |
const int INF = 1e9; | |
template<class T,class U>ostream&operator<<(ostream &os,const pair<T,U> &p){ | |
os<<"("<<p.first<<","<<p.second<<")";return os;} | |
template<class T>ostream&operator<<(ostream &os,const vector<T> &v){ | |
ITR(i,v)os<<*i<<(i==end(v)-1?"":" ");return os;} | |
//----------------------------------------------------------- | |
static const int MOD = 1000000007; | |
struct UpDownNess { | |
ll dp[52][5001]; | |
int count(int N, int K) { | |
dp[1][0]=1; | |
FOR(i,1,N+1) { | |
REP(j,i+1) { | |
REP(k,K+1) { | |
if(k+(j*(i-j))<=K) | |
dp[i+1][k+(j*(i-j))] += (dp[i][k] % MOD); | |
} | |
} | |
} | |
return (dp[N][K] % MOD); | |
} | |
}; |
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