CF431 C. k-Tree

题目传送门:https://codeforces.com/problemset/problem/431/C

题目大意:
给定\(n,k,d\),构造一个长度任意(假定为\(m\))的序列\(A\),满足\(A_i\in[1,k]\)\(\sum\limits_{i=1}^mA_i=n\),且 \(\exists i\in[1,m],A_i\geqslant d\),问有多少种不同的构造方法?

长度不同,或者\(A_i\neq A_i^`\),即认为是不同的序列


长度任意的话,其实就是一个经典的01背包问题,多开一维记录是否满足\(\exist i\in[1,m],A_i\geqslant d\)即可

/*program from Wolfycz*/
#include<map>
#include<cmath>
#include<cstdio>
#include<vector>
#include<cstring>
#include<iostream>
#include<algorithm>
#define Fi first
#define Se second
#define ll_inf 1e18
#define MK make_pair
#define sqr(x) ((x)*(x))
#define pii pair<int,int>
#define int_inf 0x7f7f7f7f
using namespace std;
typedef long long ll;
typedef unsigned int ui;
typedef unsigned long long ull;
inline char gc(){
	static char buf[1000000],*p1=buf,*p2=buf;
	return p1==p2&&(p2=(p1=buf)+fread(buf,1,1000000,stdin),p1==p2)?EOF:*p1++;
}
template<typename T>inline T frd(T x){
	int f=1; char ch=gc();
	for (;ch<'0'||ch>'9';ch=gc())	if (ch=='-')    f=-1;
	for (;ch>='0'&&ch<='9';ch=gc())	x=(x<<1)+(x<<3)+ch-'0';
	return x*f;
}
template<typename T>inline T read(T x){
	int f=1; char ch=getchar();
	for (;ch<'0'||ch>'9';ch=getchar())	if (ch=='-')	f=-1;
	for (;ch>='0'&&ch<='9';ch=getchar())	x=(x<<1)+(x<<3)+ch-'0';
	return x*f;
}
inline void print(int x){
	if (x<0)	putchar('-'),x=-x;
	if (x>9)	print(x/10);
	putchar(x%10+'0');
}
const int N=1e2,P=1e9+7;
int F[2][N+10];
int main(){
//	freopen(".in","r",stdin);
//	freopen(".out","w",stdout);
	int n=read(0),k=read(0),d=read(0); F[0][0]=1;
	for (int i=1;i<=n;i++){
		for (int j=1;j<=min(i,k);j++){
			F[  1 ][i]=(F[  1 ][i]+F[1][i-j])%P;
			F[j>=d][i]=(F[j>=d][i]+F[0][i-j])%P;
		}
	}
	printf("%d\n",F[1][n]);
	return 0;
}
posted @ 2021-08-06 16:25  Wolfycz  阅读(21)  评论(0编辑  收藏  举报