UVa 12105 Bigger is Better (DP)

题意：用不超过 n 根火柴，组成一个尽可能大的数。

析：很明显的一个DP题，首先不难想到这个dp[i][j] 表示前 i 根火柴，所能拼出的取模 m 为 j 的数，状态转移方程也很好写，

dp[i][j] ==> dp[i+c[k]][(j*10+k)%m] 其中 k 为在后面添加的数，c 数组是用的火柴数，这个方程没问题，就是效率有点低，

因为有一个高精度问题，可以用Java来实现，也能够AC的。

第二种方法就是换一个表示方法，不过确实不太容易想到。dp[i][j] 表示用最多 i 根火柴，取模 m 的数的最大位数。毕竟位数越大，数越大。

状态转移方程也是和上面差不多就是变成位数了而已。

代码如下：

import java.math.BigInteger;

import java.util.*;

public class Main{

	public static final int maxn = 100 + 10;

	public static final int maxm = 3000 + 10;

	public static BigInteger[][] ans;

	public static int[] c;

	public static BigInteger solve(int n, int m){

		for(int i = 0; i <= n; ++i)

			for(int j = 0; j < m; ++j)

				ans[i][j] = BigInteger.valueOf(-1);

		ans[0][0] = BigInteger.ZERO;

		BigInteger res = BigInteger.valueOf(-1);

		for(int i = 0; i <= n; ++i){

			for(int j = 0; j < m; ++j){

				if(ans[i][j].equals(BigInteger.valueOf(-1)))  continue;

				for(int k = 0; k < 10; ++k){

					if(i +c[k] > n)  continue;

					ans[i+c[k]][(j*10+k)%m] = ans[i+c[k]][(j*10+k)%m].max(ans[i][j].multiply(BigInteger.valueOf(10)).add(BigInteger.valueOf(k)));

				}

			}

			if(i > 1 && res.compareTo(ans[i][0]) < 0)  res = ans[i][0];

		}

		return res;

	}

	public static void main(String []args){

		c = new int[15];

		ans = new BigInteger[maxn][maxm];

		c[0] = 6;  c[1] = 2;  c[2] = 5;  c[3] = 5;

		c[4] = 4;  c[5] = 5;  c[6] = 6;  c[7] = 3;

		c[8] = 7;  c[9] = 6;

		Scanner cin = new Scanner(System.in);

		int kase = 0;

		while(cin.hasNext()){

			int n = cin.nextInt();

			if(0 == n)  break;

			int m = cin.nextInt();

			System.out.println("Case " + (++kase) +": " + solve(n, m));

		}

		cin.close();

	}

}

第二种方法：

#pragma comment(linker, "/STACK:1024000000,1024000000")

#include <cstdio>

#include <string>

#include <cstdlib>

#include <cmath>

#include <iostream>

#include <cstring>

#include <set>

#include <queue>

#include <algorithm>

#include <vector>

#include <map>

#include <cctype>

#include <cmath>

#include <stack>

#include <sstream>

#define debug() puts("++++");

#define gcd(a, b) __gcd(a, b)

#define lson l,m,rt<<1

#define rson m+1,r,rt<<1|1

#define freopenr freopen("in.txt", "r", stdin)

#define freopenw freopen("out.txt", "w", stdout)

using namespace std;

typedef long long LL;

typedef unsigned long long ULL;

typedef pair<int, int> P;

const int INF = 0x3f3f3f3f;

const double inf = 0x3f3f3f3f3f3f;

const double PI = acos(-1.0);

const double eps = 1e-5;

const int maxn = 2600 + 10;

const int mod = 1e6;

const int dr[] = {-1, 0, 1, 0};

const int dc[] = {0, 1, 0, -1};

const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};

int n, m;

const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

inline bool is_in(int r, int c){

  return r >= 0 && r < n && c >= 0 && c < m;

}

int dp[110][3010], p[110][3010];

const int c[] = { 6, 2, 5, 5, 4, 5, 6, 3, 7, 6 };

int main(){

  int kase = 0;

  while(scanf("%d %d", &n, &m) == 2 && n){

    printf("Case %d: ", ++kase);

    for(int i = 0; i <= n; ++i)

      for(int j = 0; j < m; ++j){

        int &ans = dp[i][j];

        ans = p[i][j] = -1;

        if(!j)  ans = 0;

        for(int k = 9; k >= 0; --k) if(i >= c[k]){

          int t = dp[i-c[k]][(j*10+k)%m] + 1;

          if(t > 0 && t > ans){

            ans = t;

            p[i][j] = k;

          }

        }

      }

      if(dp[n][0] <= 0)  printf("-1");

      else{

        int i = n, j = 0;

        for(int d = p[i][j]; d >= 0; d = p[i][j]){

          printf("%d", d);

          i -= c[d];

          j = (j*10 + d) % m;

        }

      }

      printf("\n");

  }

  return 0;

}

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UVa 12105 Bigger is Better (DP)

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