Create J(状压dp)

Author: angrymilk

Training_With_Contest/PSTraining Camp 2016 Day 9/J(状压dp)

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+//Solution 1：
+//dp[105][1<<10][35]:表示算到第i个点，顺序状态是sta，%=35的情况总数。
+#include <string.h>
+#include <algorithm>
+#include <iostream>
+#include <cstdio>
+#include <cstdlib>
+#define rep(i,n) for(int i = 0; i < n; i++)
+using namespace std;
+typedef long long ll;
+
+const int MX = 105, MM = 35, n10 = 1<<10;
+
+// Mod int
+const int mod = 1000000007;
+struct mint{
+	ll x;
+	mint():x(0){}
+	mint(ll x):x(x){}
+	mint operator+=(mint a){ if((x+=a.x)>=mod) x-=mod; return *this;}
+	mint operator-=(mint a){ if((x+=mod-a.x)>=mod) x-=mod; return *this;}
+	mint operator*=(mint a){ (x*=a.x)%=mod; return *this;}
+	mint operator+(mint a){ return mint(*this) += a;}
+	mint operator-(mint a){ return mint(*this) -= a;}
+	mint operator*(mint a){ return mint(*this) *= a;}
+};
+//
+
+int n, m, k;
+int a[MX];
+mint dp[MX][n10][MM];
+
+int f(int x){
+	int res = 0, pre = -1;
+	rep(i,k){
+		if((x&1) != pre) res++;
+		pre = (x&1); x >>= 1;
+	}
+	return res-1;
+}
+
+
+int main() {
+	scanf("%d%d%d",&n,&m,&k);
+	rep(i,n-1) scanf("%d",&a[i+1]);
+	dp[0][0][0] = 1;
+	rep(i,n)rep(j,1<<k)rep(r,m){
+		int nxtm = (r+f(j)*a[i])%m;
+		dp[i+1][j][nxtm] += dp[i][j][r];
+		rep(l,k){
+			if(j>>l&1) continue;
+			dp[i+1][j|1<<l][nxtm] += dp[i][j][r];
+		}
+	}
+	
+	cout << dp[n][(1<<k)-1][0].x << endl;
+	return 0;
+}
+
+//Solution 2:
+//dp[105][k][k][m][2]:表示有k1个1°点，k2个2°点。