CF431C k-Tree dp
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Quite recently a creative student Lesha had a lecture on trees. After the lecture Lesha was inspired and came up with the tree of his own which he called a k-tree.
A k-tree is an infinite rooted tree where:
- each vertex has exactly k children;
- each edge has some weight;
- if we look at the edges that goes from some vertex to its children (exactly k edges), then their weights will equal 1,?2,?3,?...,?k.
The picture below shows a part of a 3-tree.
Help Dima find an answer to his question. As the number of ways can be rather large, print it modulo 1000000007 (109?+?7).
A single line contains three space-separated integers: n, k and d (1?≤?n,?k?≤?100; 1?≤?d?≤?k).
Print a single integer — the answer to the problem modulo 1000000007 (109?+?7).
3 3 2
3
3 3 3
1
4 3 2
6
4 5 2
7
题目描述
最近有一个富有创造力的学生Lesha听了一个关于树的讲座。在听完讲座之后,Lesha受到了启发,并且他有一个关于k-tree(k叉树)的想法。 k-tree都是无根树,并且满足:
- 每一个非叶子节点都有k个孩子节点;
- 每一条边都有一个边权;
- 每一个非叶子节点指向其k个孩子节点的k条边的权值分别为1,2,3,...,k。
当Lesha的好朋友Dima看到这种树时,Dima马上想到了一个问题:“有多少条从k-tree的根节点出发的路上的边权之和等于n,并且经过的这些边中至少有一条边的边权大于等于d呢?” 现在你需要帮助Dima解决这个问题。考虑到路径总数可能会非常大,所以只需输出路径总数 mod 1000000007 即可。(1000000007=10^9+7)
考虑dp[ i ][ 1/0 ]表示总和为i时,最大值是否>=d的方案数;
然后枚举中间状态转移;
注意long long ;
#include<iostream> #include<cstdio> #include<algorithm> #include<cstdlib> #include<cstring> #include<string> #include<cmath> #include<map> #include<set> #include<vector> #include<queue> #include<bitset> #include<ctime> #include<deque> #include<stack> #include<functional> #include<sstream> //#include<cctype> //#pragma GCC optimize("O3") using namespace std; #define maxn 1000005 #define inf 0x3f3f3f3f #define INF 9999999999 #define rdint(x) scanf("%d",&x) #define rdllt(x) scanf("%lld",&x) #define rdult(x) scanf("%lu",&x) #define rdlf(x) scanf("%lf",&x) #define rdstr(x) scanf("%s",x) typedef long long ll; typedef unsigned long long ull; typedef unsigned int U; #define ms(x) memset((x),0,sizeof(x)) const long long int mod = 1e9 + 7; #define Mod 1000000000 #define sq(x) (x)*(x) #define eps 1e-3 typedef pair<int, int> pii; #define pi acos(-1.0) //const int N = 1005; #define REP(i,n) for(int i=0;i<(n);i++) inline ll rd() { ll x = 0; char c = getchar(); bool f = false; while (!isdigit(c)) { if (c == ‘-‘) f = true; c = getchar(); } while (isdigit(c)) { x = (x << 1) + (x << 3) + (c ^ 48); c = getchar(); } return f ? -x : x; } ll gcd(ll a, ll b) { return b == 0 ? a : gcd(b, a%b); } ll sqr(ll x) { return x * x; } /*ll ans; ll exgcd(ll a, ll b, ll &x, ll &y) { if (!b) { x = 1; y = 0; return a; } ans = exgcd(b, a%b, x, y); ll t = x; x = y; y = t - a / b * y; return ans; } */ ll qpow(ll a, ll b, ll c) { ll ans = 1; a = a % c; while (b) { if (b % 2)ans = ans * a%c; b /= 2; a = a * a%c; } return ans; } /* int n, m; int st, ed; struct node { int u, v, nxt, w; }edge[maxn<<1]; int head[maxn], cnt; void addedge(int u, int v, int w) { edge[cnt].u = u; edge[cnt].v = v; edge[cnt].w = w; edge[cnt].nxt = head[u]; head[u] = cnt++; } int rk[maxn]; int bfs() { queue<int>q; ms(rk); rk[st] = 1; q.push(st); while (!q.empty()) { int tmp = q.front(); q.pop(); for (int i = head[tmp]; i != -1; i = edge[i].nxt) { int to = edge[i].v; if (rk[to] || edge[i].w <= 0)continue; rk[to] = rk[tmp] + 1; q.push(to); } } return rk[ed]; } int dfs(int u, int flow) { if (u == ed)return flow; int add = 0; for (int i = head[u]; i != -1 && add < flow; i = edge[i].nxt) { int v = edge[i].v; if (rk[v] != rk[u] + 1 || !edge[i].w)continue; int tmpadd = dfs(v, min(edge[i].w, flow - add)); if (!tmpadd) { rk[v] = -1; continue; } edge[i].w -= tmpadd; edge[i ^ 1].w += tmpadd; add += tmpadd; } return add; } ll ans; void dinic() { while (bfs())ans += dfs(st, inf); } */ int n, k, d; ll dp[200][2]; int main() { //ios::sync_with_stdio(0); //memset(head, -1, sizeof(head)); while (cin >> n >> k >> d) { ms(dp); dp[0][0] = 1; for (int i = 1; i <= n; i++) { for (int j = 1; j <= k; j++) { if (i >= j) { if (j < d) { dp[i][0] = (dp[i][0] + dp[i - j][0]) % mod; dp[i][1] = (dp[i][1] + dp[i - j][1]) % mod; } else { dp[i][1] = (dp[i][1] + dp[i - j][0] + dp[i - j][1]) % mod; } } } } cout << (ll)dp[n][1] << endl; } return 0; }
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