【Codeforces 474D】Flowers

【链接】 我是链接,点我呀:)
【题意】

让你吃东西 B食物一次必须要吃连续k个 但是对A食物没有要求 问你有多少种吃n个食物的方法(吃的序列)

【题解】

设f[i]表示长度为i的吃的序列且符合要求的方法 有两种转移方法 一种是吃一个A食物 一种是吃k个食物 f[i] = f[i-1]+f[i-k] f[0] = 1 然后做一个前缀和输出区间和就好

【代码】

import java.io.*;
import java.util.*;

public class Main {
	
	
	static InputReader in;
	static PrintWriter out;
		
	public static void main(String[] args) throws IOException{
		//InputStream ins = new FileInputStream("E:\rush.txt");
		InputStream ins = System.in;
		in = new InputReader(ins);
		out = new PrintWriter(System.out);
		//code start from here
		new Task().solve(in, out);
		out.close();
	}
	
	static int N = (int)1e5;
	static int MOD = (int)1e9+7;
	static class Task{
		public void solve(InputReader in,PrintWriter out) {
			int f[] = new int[N+10];
			f[0] = 1;
			int t,k;
			t = in.nextInt();k = in.nextInt();
			for (int i = 1;i <= N;i++) {
				f[i] = f[i-1];
				if (i-k>=0) f[i] = (f[i]+f[i-k])%MOD;
			}
			for (int i = 1;i <= N;i++) {
				f[i] = (f[i]+f[i-1])%MOD;
			}
			
			for (int i = 1;i <= t;i++) {
				int a,b;
				a = in.nextInt();b = in.nextInt();
				out.println((f[b]-f[a-1]+MOD)%MOD);
			}
		}
	}

	

	static class InputReader{
		public BufferedReader br;
		public StringTokenizer tokenizer;
		
		public InputReader(InputStream ins) {
			br = new BufferedReader(new InputStreamReader(ins));
			tokenizer = null;
		}
		
		public String next(){
			while (tokenizer==null || !tokenizer.hasMoreTokens()) {
				try {
				tokenizer = new StringTokenizer(br.readLine());
				}catch(IOException e) {
					throw new RuntimeException(e);
				}
			}
			return tokenizer.nextToken();
		}
		
		public int nextInt() {
			return Integer.parseInt(next());
		}
	}
}
原文地址:https://www.cnblogs.com/AWCXV/p/10363876.html