用sstream处理输入的问题

• (题目还未解决，先mark一下，当作笔记 - DP求解)

Question

Please solve the following problem in Java.
You are given a list (n <= 100) of positive intergers (value <= 500), 
your program divides these intergers into two groups, such that the difference 
between the sums of these two groups is minimized.

Example 1:
Input: 1 1 1 2 2
Output: 1

Example 2:
Input: 3 9 2 3 5
Output: 0

思路：多重背包问题，令j = sum / 2, 模仿yxc学长的思路

#include <iostream>
#include <algorithm>
#include <vector>
#include <sstream>
//#include <string>

using namespace std;

const int N = 110, M = 510;
//int v[M];
int f[N][N]; // 最多N个v相加

int main()
{
    string line;
    int temp;
    vector<int> v;
    while (getline(cin, line))
    {
        stringstream ss(line);
        while(ss >> temp)
        {
            v.push_back(temp);
        }
    }

    int sum = 0;
    for (int i = 0; i < v.size(); i ++ ) 
        sum += v[i];
        //cout << v[i] << " " ;
        //cout << endl;
    for (int i = 1; i <= v.size(); i ++ )
        for (int j = 0; j < sum / 2; j ++ )
            for (int k = 0; k <= 100; k ++ )
                f[i][j] = max(f[i][j], f[i - 1][j - k * v[i]] + k * v[i]);

    int ans  = abs(f[v.size()][sum / 2] - (sum - f[v.size()][sum / 2]));
    cout << ans << endl;

    return 0;
}

方法一： 用 vector

#include <iostream>
#include <vector>
#include <string>
#include <sstream>
using namespace std;
int main()
{
    string str;
    while (getline(cin, str))
    {
        vector<int> num;
        int temp;
        istringstream iss(str);
        while (iss >> temp)
        {
            num.push_back(temp);
        }
        // 输出显示
        for (int i = 0; i < num.size(); i++)
            cout << num[i] << " ";
        cout << endl;
    }
}

方法二 ： 常规数组

#include <iostream>
#include <algorithm>
#include <sstream>
#include <vector>
#include <string>

using namespace std;

const int N = 110, M = 510;

int n, m;
int s[N];
int v[M];
int f[N][N];

int main()
{
    string line;
    int n = 0;
    while (getline(cin, line))
    {
        stringstream ss(line);
        while (ss >> v[n]) n ++ ;
    }
    cout << n << endl;
    for (int i = 0; i < n; i ++ ) cout << v[i] << endl;

    return 0;
}