SRM506

Summary:
Don't know how to solve level three, also the programming speed is too slow.
My rate has been improved, also I am a green man now... Happy.

Div 1, Level 1

Div 1, Level 2

Div 1, Level 3

Div 2, Level 1

Div 2, Level 2

Div 2, Level 3

Tutorials:

Division One - Level Three:

Solution

Source Code:

Division One - Level Two:

Solution

Source Code:

Division One - Level One:

Solution

Same to the Div2, Level Two.

Source Code:

Division Two - Level Three:

Solution

Source Code:

Division Two - Level Two:

Solution

Sort, find any combination of part of array, the sum would no less than the other part. Just go through each element, then check it. PS: 1st, add the one if they are the same. 2nd, add the one if it's less than the upgraded element.

Source Code:

//Wed May 11 19:58:39 CDT 2011
#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cctype>
#include <string>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <ctime>

using namespace std;

class SlimeXSlimesCity {
public:
    int merge(vector<int> population) {
        int N = population.size();
        long long sum = 0;
        for (int i = 0; i < N; i++) {
            v.push_back(make_pair(population[i], i));
            sum += population[i];
        }
        sort(v.begin(), v.end());
        int count = 0;
        for (int i = 0; i < N; i++) {
            int p;
            for (p = 0; p < N; p++) {
                if (v[p].second == i)
                    break;
            }
            long long tmp = 0;
            for (int k = 0; k < N; k++) {
                if (tmp >= v[k].first || v[k].first <= v[p].first)
                    tmp += v[k].first;
                else
                    break;
            }
            if (tmp * 2 >= sum)
                count++;
        }
        return count;
    }
private:
    vector<pair<int, int> > v;
};

Division Two - Level One:

Solution

Only operations are stay the same or increase some amount of number. Thus, sort the array, then sum all the difference between the maximum and each element.

Source Code:

//Wed May 11 19:58:39 CDT 2011
#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cctype>
#include <string>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <ctime>

using namespace std;

class SlimeXSlimeRancher2 {
public:
    int train(vector<int> attributes) {
        sort(attributes.begin(), attributes.end());
        int sum = 0;
        for(int i=0; i<attributes.size(); i++)
            sum += attributes[attributes.size()-1] - attributes[i];
        return sum;
    }
};