OpenJ_Bailian - 4152 最佳加法表达式 dp

http://bailian.openjudge.cn/practice/4152?lang=en_US

题解 ：dp[i][j]代表前i个字符加j个加号可以得到的最小值，于是dp[i+k[j+1]可以由dp[i][j]得到。具体转移方程看代码。

　　　然后数字是50位所以要用高精度类。自己写了一个

坑：高精度的<和+有bug。一开始的更新方法也在乱写

#define _CRT_SECURE_NO_WARNINGS
#include<stdio.h>
#include<stdlib.h>
#include<iostream>
#include<string>
#include<vector>
#include<algorithm>
#include<string.h>
using namespace std;
const int N =  + ;
 
typedef long long ll;
struct BigInterger {
    static const int BASE = 1e8;
    static const int WIDTH = ;
    vector<int> s;
    BigInterger(long long num = ) { *this = num; }
    BigInterger operator =(long long num) {
        s.clear();
        do {
            s.push_back(num%BASE);
            num /= BASE;
        } while (num > );//num==0
        return *this;
    }
    BigInterger operator =(const string& str) {
        s.clear();
        int x, len = (str.length() - ) / WIDTH + ;//len==width
        for (int i = ; i < len; i++) {
            int end = str.length() - i*WIDTH;
            int start = max(, end - WIDTH);
            sscanf(str.substr(start, end - start).c_str(), "%d", &x);
            s.push_back(x);
        }
        return *this;
    }
 
    BigInterger operator +(const BigInterger& b)const {
        BigInterger c;
        c.s.clear();
        for (int i = , g = ;; i++) {
            if (g ==  && i >= max(s.size(), b.s.size()))break;
            int x = g;
            if (i < s.size())x += s[i];
            if (i < b.s.size()) x += b.s[i];
            c.s.push_back(x%BASE);
            g = x / BASE;
        }
        return c;
    }
    BigInterger operator +=(const BigInterger& b) {
        *this = *this + b; return *this;
    }
 
    bool operator<(const BigInterger& b)const {
        if (s.size() != b.s.size()) return s.size() < b.s.size();
        for (int i = s.size() - ; i >= ; i--) {
            if (s[i] != b.s[i]) return s[i] < b.s[i];//Width 个数字一起比
        }
        return false;//==
    }
    bool operator>(const BigInterger &b)const { return b < *this; }
    bool operator<=(const BigInterger &b)const { return !(b < *this); }
    bool operator>=(const BigInterger &b)const { return !(*this < b); }
    bool operator!=(const BigInterger &b)const { return b < *this || *this<b; }
    bool operator==(const BigInterger &b)const { return!(b < *this) && !(*this<b); }
};
ostream& operator <<(ostream &out, const BigInterger& x) {
    out << x.s.back();
    for (int i = x.s.size() - ; i >= ; i--) {
        char buf[];
        sprintf(buf, "%08d", x.s[i]);
        for (int j = ; j < strlen(buf); j++)cout << buf[j];
    }
    return out;
}
istream&operator>>(istream &in, BigInterger&x) {
    string s;
    if (!(in >> s))return in;
    x = s;
    return in;
}
BigInterger dp[N][N];//前i个数加j个加号的最小值。
int main() {
    int n; BigInterger s; string s1;
    string INF;
    for (int i = ; i < ; i++)INF += "";
    while (cin >> n) {
        cin >> s1;
        int len = s1.length();
        s = s1;
        //if (n == 0) {cout << s<<endl; continue;}
        for (int i = ; i <= len; i++)
            for (int j = ; j <= n; j++) dp[i][j] = INF;
 
        for (int i = ; i <= len; i++)
            for (int j = ; j <= n; j++) {
                if (j == ) { dp[i][j] = s1.substr(, i); }
                else if (i < j + ) dp[i][j] = INF;
                else {
                    for (int k = j; k <= i - ; k++) {
                        BigInterger x;
                        x = s1.substr(k, i - k);
                        dp[i][j] = min(dp[i][j], dp[k][j - ] + x);
                    }
                }
                //cout <<i<<j<<' '<< dp[i][j] << endl;
            }
        cout << dp[len][n] << endl;
 
    }
}