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 = 0;
    int j = 0;
    int carry = 0;
    int number = 0;
    while (i < lena || j < lenb) {
        number = carry;
        if (i < lena) number += (a[i++] - '0');
        if (j < lenb) number += (b[j++] - '0');
        sum += (number % 10 + '0');
        carry = number / 10;
    }
    if (carry == 1) {
        sum = sum + '1';
    }

    return sum;
}

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

string maxSum(vector<vector<string>> &record, string& s, int start, int m) {
    if (m == 0) 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 + 1);
        string remain = maxSum(record, s, i + 1, m - 1);
        string r = Add(sub, remain);
        if (comp(minNumber, r) == 1) {
            minNumber = r;
        }
    }
    return rec = minNumber;
}

int main()
{
    int m;
    string s;
    while (cin >> m) {
        cin >> s;
        if (m == 0) {
            cout << s << endl;
            continue;
        }
        reverse(s.begin(), s.end());
        vector<vector<string>> record(s.length(), vector<string>(m + 1, ""));
        string sum = maxSum(record, s, 0, m);
        reverse(sum.begin(), sum.end());
        cout << sum << endl;
    }
    return 0;
}