百练4152:最佳加法表达式(dp+高精度)

描述

给定n个1到9的数字,要求在数字之间摆放m个加号(加号两边必须有数字),使得所得到的加法表达式的值最小,并输出该值。例如,在1234中摆放1个加号,最好的摆法就是12+34,和为36

输入有不超过15组数据
每组数据两行。第一行是整数m,表示有m个加号要放( 0<=m<=50)
第二行是若干个数字。数字总数n不超过50,且 m <= n-1输出对每组数据,输出最小加法表达式的值样例输入

2
123456
1
123456
4
12345

样例输出

102
579
15

提示要用到高精度计算,即用数组来存放long long 都装不下的大整数,并用模拟列竖式的办法进行大整数的加法。

 

搞了半天的C++高精度,,结果还是可耻地用了java大数。。

 1 import java.math.BigDecimal;
 2 import java.math.BigInteger;
 3 import java.util.Scanner;
 4 
 5 public class Main {
 6     static BigInteger INF = new BigInteger("9999999999999999999999999999999999999999999999999999999");
 7     public static void main(String[] args) {
 8         // TODO Auto-generated method stub
 9         Scanner sc = new Scanner(System.in);
10         BigInteger dp[][] = new BigInteger[55][55];
11         int m,n;
12         BigInteger s;
13         
14         while(sc.hasNext())
15         {
16             m = sc.nextInt();
17             s = sc.nextBigInteger();
18             
19             n = s.toString().length();
20             
21             //BigDecimal num[][] = new BigDecimal[n+1][n+1];//num[i][j]表示从s第i个数到第j个数组成的数字
22             /*
23             for(int i = 1;i<=n;++i)
24                 for(int j = 0;j<=n;++j)
25                 {
26                     if(i<=j)
27                     {
28                         num[i][j] = new BigDecimal(s.toString().substring(i-1,j));
29                     }
30                 }
31                 */
32             
33             dp[0][0] = new BigInteger("0");//没有数字没有加号的最小值是0
34             for(int i = 1;i<=n;++i)
35             {
36                 dp[i][0] =  new BigInteger(s.toString().substring(0,i));//没有加号的情况下,最小值就是数字自己
37             }
38             
39             for(int i = 0;i<=n;++i)
40                 for(int j = 1;j<=m;++j)
41                 {
42                     
43                     dp[i][j] = INF;
44                     if(i>=j+1)//j个加号能插入i个数字中
45                     {
46                         for(int k = j;k<i;++k)
47                         {
48                             dp[i][j] = dp[i][j].min(dp[k][j-1].add(new BigInteger(s.toString().substring(k,i))));
49                         }
50                     }
51                 }
52                 
53             System.out.println(dp[n][m]);    
54         }//while
55     }
56 
57 }

 补一个师傅的C++做法,之后回来补

 1 #include <iostream>
 2 #include <algorithm>
 3 using namespace std;
 4 string Add(string &a, string &b) {
 5     string sum;
 6     int lena = a.length();
 7     int lenb = b.length();
 8     int i = 0;
 9     int j = 0;
10     int carry = 0;
11     int number = 0;
12     while (i < lena || j < lenb) {
13         number = carry;
14         if (i < lena) number += (a[i++] - '0');
15         if (j < lenb) number += (b[j++] - '0');
16         sum += (number % 10 + '0');
17         carry = number / 10;
18     }
19     if (carry == 1) {
20         sum = sum + '1';
21     }
22 
23     return sum;
24 }
25 
26 //1大于 0等于 -1小于
27 int comp(string &a, string &b) {
28     if (a.length() > b.length()) return 1;
29     if (a.length() < b.length()) return -1;
30     for (int i = a.length() - 1; i >= 0; --i) {
31         if (a[i] > b[i])
32             return 1;
33         else if (a[i] < b[i]) {
34             return -1;
35         }
36     }
37     return 0;
38 }
39 
40 string maxSum(vector<vector<string>> &record, string& s, int start, int m) {
41     if (m == 0) return s.substr(start);
42 
43     if (record[start][m] != "") {
44         return record[start][m];
45     }
46 
47     string &rec = record[start][m];
48     string minNumber = s;
49     for (int i = start; i < s.length() - m; ++i) {
50         string sub = s.substr(start, i - start + 1);
51         string remain = maxSum(record, s, i + 1, m - 1);
52         string r = Add(sub, remain);
53         if (comp(minNumber, r) == 1) {
54             minNumber = r;
55         }
56     }
57     return rec = minNumber;
58 }
59 
60 int main()
61 {
62     int m;
63     string s;
64     while (cin >> m) {
65         cin >> s;
66         if (m == 0) {
67             cout << s << endl;
68             continue;
69         }
70         reverse(s.begin(), s.end());
71         vector<vector<string>> record(s.length(), vector<string>(m + 1, ""));
72         string sum = maxSum(record, s, 0, m);
73         reverse(sum.begin(), sum.end());
74         cout << sum << endl;
75     }
76     return 0;
77 }

 

posted @ 2018-08-23 11:08  敲代码不BB  阅读(415)  评论(0编辑  收藏  举报