UVa 11375 - Matches （递推 JAVA 高精度）

UVA - 11375

Matches

Time Limit: 2000MS Memory Limit: Unknown 64bit IO Format: %lld & %llu

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Description

Problem E: Matches

We can make digits with matches as shown below:

Given N matches, find the number of different numbers representable using the matches. We shall only make numbers greater than or equal to 0, so no negative signs should be used. For instance, if you have 3 matches, then you can only make the numbers 1 or 7. If you have 4 matches, then you can make the numbers 1, 4, 7 or 11. Note that leading zeros are not allowed (e.g. 001, 042, etc. are illegal). Numbers such as 0, 20, 101 etc. are permitted, though.

Input

Input contains no more than 100 lines. Each line contains one integer N (1 ≤ N ≤ 2000).

Output

For each N, output the number of different (non-negative) numbers representable if you have N matches.

Sample Input

34

Sample Output

24

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Problemsetter: Mak Yan Kei

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题意：

用ｎ根火柴能组成多少个非负数？火柴不必用完

f[i]表示恰好ｉ根火柴能够组成的非负数个数，则在f[i]的基础上再加上数字x，那么就变为了用i＋dx[x]根火柴的方案数

对于数字不能有前导零，在状态i=0的时候不能用０，在最后的时候如果i>=6（０需要的火柴数）的时候方案数加上１就行了

需要高精度

import java.util.*;import java.math.*;public class Main {    public static void main(String[] args) {        int maxn = 2000 + 20;        int[] dx = {6, 2, 5, 5, 4, 5, 6, 3, 7, 6};        BigInteger[] f = new BigInteger[maxn];        BigInteger[] sum = new BigInteger[maxn];        for(int i=1; i<maxn; i++) {            f[i] = BigInteger.ZERO;            sum[i] = BigInteger.ZERO;        }        f[0] = BigInteger.ONE;        for(int i=0; i<maxn; i++) {            for(int j=0; j<10; j++) {                if((i!=0 || j!=0) && i + dx[j] < maxn) f[i+dx[j]] = f[i+dx[j]].add(f[i]);            }        }        sum[0] = BigInteger.ZERO;        for(int i=1; i<maxn; i++) sum[i] = sum[i-1].add(f[i]);        Scanner in = new Scanner(System.in);        while(in.hasNextInt()) {            int n = in.nextInt();            if(n < 6) System.out.println(sum[n]);            else System.out.println(sum[n].add(BigInteger.ONE));        }    }}