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。。

 import java.math.BigDecimal;
 import java.math.BigInteger;
 import java.util.Scanner;

 public class Main {
     static BigInteger INF = new BigInteger("9999999999999999999999999999999999999999999999999999999");
     public static void main(String[] args) {
         // TODO Auto-generated method stub
         Scanner sc = new Scanner(System.in);
         BigInteger dp[][] = new BigInteger[55][55];
         int m,n;
         BigInteger s;

         while(sc.hasNext())
         {
             m = sc.nextInt();
             s = sc.nextBigInteger();

             n = s.toString().length();

             //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
             /*
             for(int i = 1;i<=n;++i)
                 for(int j = 0;j<=n;++j)
                 {
                     if(i<=j)
                     {
                         num[i][j] = new BigDecimal(s.toString().substring(i-1,j));
                     }
                 }
                 */

             dp[0][0] = new BigInteger("0");//没有数字没有加号的最小值是0
             for(int i = 1;i<=n;++i)
             {
                 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
             }

             for(int i = 0;i<=n;++i)
                 for(int j = 1;j<=m;++j)
                 {

                     dp[i][j] = INF;
                     if(i>=j+1)//j个加号能插入i个数字中
                     {
                         for(int k = j;k<i;++k)
                         {
                             dp[i][j] = dp[i][j].min(dp[k][j-1].add(new BigInteger(s.toString().substring(k,i))));
                         }
                     }
                 }

             System.out.println(dp[n][m]);
         }//while
     }

 }

补一个师傅的C++做法，之后回来补

 #include <iostream>
 #include <algorithm>
 using namespace std;
 string Add(string &a, string &b) {
     string sum;
     int lena = a.length();
     int lenb = b.length();
     int i = ;
     int j = ;
     int carry = ;
     int number = ;
     while (i < lena || j < lenb) {
         number = carry;
         if (i < lena) number += (a[i++] - '');
         if (j < lenb) number += (b[j++] - '');
         sum += (number %  + '');
         carry = number / ;
     }
     if (carry == ) {
         sum = sum + '';
     }

     return sum;
 }

 //1大于 0等于 -1小于
 int comp(string &a, string &b) {
     if (a.length() > b.length()) return ;
     if (a.length() < b.length()) return -;
     for (int i = a.length() - ; i >= ; --i) {
         if (a[i] > b[i])
             return ;
         else if (a[i] < b[i]) {
             return -;
         }
     }
     return ;
 }

 string maxSum(vector<vector<string>> &record, string& s, int start, int m) {
     if (m == ) return s.substr(start);

     if (record[start][m] != "") {
         return record[start][m];
     }

     string &rec = record[start][m];
     string minNumber = s;
     for (int i = start; i < s.length() - m; ++i) {
         string sub = s.substr(start, i - start + );
         string remain = maxSum(record, s, i + , m - );
         string r = Add(sub, remain);
         if (comp(minNumber, r) == ) {
             minNumber = r;
         }
     }
     return rec = minNumber;
 }

 int main()
 {
     int m;
     string s;
     while (cin >> m) {
         cin >> s;
         if (m == ) {
             cout << s << endl;
             continue;
         }
         reverse(s.begin(), s.end());
         vector<vector<string>> record(s.length(), vector<string>(m + , ""));
         string sum = maxSum(record, s, , m);
         reverse(sum.begin(), sum.end());
         cout << sum << endl;
     }
     return ;
 }