SRM 512

Mato_No1 — Thu, 14 Jul 2011 02:04:00 GMT

世界上已�l�木有比我还��q��?#8230;…

256�Q�这么水的题竟然搞了30+min�Q�一开始米看见每隔7天就要买同一�U�，以�ؓ是DP�Q�结果测Sample3�c��去，于是重写……�Q�要木有Sample3��杯具了�Q?br />
512�Q�由�?56上卡了太多时��_��q�题在快写完的时候到旉��?#8230;…

1024�Q�看都米�?#8230;…

�l�果�Q�全�?00+�?#8230;…估计下次得掉回DIV2�?#8230;…

Mato_No1 2011-07-14 10:04 发表评论

SRM 511

Mato_No1 — Sat, 02 Jul 2011 18:09:00 GMT

My first SRM……�U�念一�?br />
�q�次��M��感觉�q�不是太差，也算正常发挥�?#8212;—虽然最后还是米有搞�?000�?50�?00两道水题的速度应该�q�可以（从最�l�名�ơ来看）�?br />另外�Q�DIV2全场只有2位神犇搞定了1000……

Orz AHdoc�{�神�?br />————————————————————————————————————
以下�?000题解�Q�看别的��犇的代码搞懂的�Q�：
设F[i][j]为第i轮开始时�Q�还未出牌时�Q�，面对的状态（内存�Q��ؓj�Q�是否必胜。这里设一开始的那一轮�ؓ�W?轮�?br />逆推i。根据or�q�算的性质可以得到�Q�若目前内存为j�Q�某张已�l�出�q�的牌的��gؓK�Q�则K的二�q�制的所�?位在j中对应的位也都是1�Q�也��是j | K = j�Q�，�q�样�Q�扫描每张牌�Q�若其值K | j的��g��{�于j�Q�则该牌不可能出�q�。因此，可以在第i轮出�q�张牌，若至��有一个F[i + 1][K | j]为必败状态则F[i][j]为必胜状态�?br />对于K | j的值等于j的牌�Q�统计它们的张数�Q�设为cnt。易知，前i轮出�q�的i张牌必然都是�q�种牌，因此若cnt>i�Q�且F[i + 1][j]是必败状态，则可以在�W�i轮出一张这��L��牌，必胜�?br />如果上面没有发现一个可以��F[i][j]为必胜状态的�Q�则F[i][j]为必败状态�?br />边界�Q�F[i][511]为必胜状�?0<=i<=N�Q�，F[N][j]为必败状态（0<=j<511�Q�因为第N轮时已经木有牌了�Q��?br />最后，若F[0][0]�Q�初始状态）为必胜状态则先手必胜�Q�否则先手必败�?br />

Mato_No1 2011-07-03 02:09 发表评论

Topcoder注册成功

Mato_No1 — Sun, 15 May 2011 04:20:00 GMT

�l�GYZ、CLJ��牛指点�Q�在N�ơError后终于注册成功了……

Topcoder与NOI的规则区别：
�?】Topcoder的代码不是一个完整的代码�Q�连main()都米有）�Q�而只是一个类�Q�类里面只有一个方法（相当于main()�Q�。不用输入、输出，�pȝ��会将输入数据直接传递到�q�个�Ҏ��的参数里�Q�在�Ҏ��执行完后��返回值直接传递到输出里。类名、方法名、输入参数类型、输出结果类型是在原题中规定的（但参数名、输出结果名可以自定义）。代码中可以�Q�也必须�Q��用C++ STL�?br>�?】捉题的时候，题目描述下面的编码区里可以直接编代码�Q�编好后点下面的Compile�~�译�Q�再点Test��试�Q�可以测试样例和自己的数据）�Q�测试完毕后�Q�点Submit提交。所以，不必向其它OJ一样在IDE里编好再Ctrl+ACV提交�?br>其它的可以参照网上其他�h写的东东�?br>
本沙茶先在里面捉了几题（全是水题�Q�神犇不要鄙视）�Q�代码：

SRM 506 DIV1 250�Q?br>

#include <iostream>
#include <string>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>
#include <time.h>
#include <iomanip>
#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
using namespace std;
#define re(i, n) for (int i=0; i
#define re1(i, n) for (int i=1; i<=n; i++)
#define re2(i, l, r) for (int i=l; i
#define re3(i, l, r) for (int i=l; i<=r; i++)
#define rre(i, n) for (int i=n-1; i>=0; i--)
#define rre1(i, n) for (int i=n; i>0; i--)
#define rre2(i, r, l) for (int i=r-1; i>=l; i--)
#define rre3(i, r, l) for (int i=r; i>=l; i--)
#define debug(x) cout << #x << " = " << x << endl;
#define pb push_back
#define re_t(i,x) for(typeof(x.begin()) i=x.begin();i!=x.end();i++)
#define all(x) x.begin(),x.end()
#define SORT(x) sort(all(x))
#define MP make_pair
typedef pair<int,int> ii;
typedef vector<int> vi;
typedef vi::iterator vit;
typedef set<int> si;
typedef si::iterator sit;
typedef map<int,int> mii;
typedef mii::iterator mit;
typedef long long ll;
typedef unsigned long long ull;
typedef unsigned int uint;
typedef istringstream ISS;
typedef ostringstream OSS;
const int MAXN = 175000, INF = ~0U >> 2;
class MathContest {
    public:
    int countBlack(string s0, int p)
    {
        string s = "";
        bool a[MAXN];
        re(i, p) s += s0;
        int n = s.length(), res = 0;
        re(i, n) a[i] = s[i] == 'B';
        int i = 0, j = s.length() - 1;
        bool reversed = 0, turned = 0, v;
        while (i <= j) {
            if (reversed) {
                v = a[j--];
                if (turned) v = !v;
            } else {
                v = a[i++];
                if (turned) v = !v;
            }
            if (v) {res++; turned = !turned;} else reversed = !reversed;
        }
        return res;
    }
};

SRM 506 DIV2 250�Q?br>

#include <iostream>
#include <string>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>
#include <time.h>
#include <iomanip>
#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
using namespace std;
#define re(i, n) for (int i=0; i
#define re1(i, n) for (int i=1; i<=n; i++)
#define re2(i, l, r) for (int i=l; i
#define re3(i, l, r) for (int i=l; i<=r; i++)
#define rre(i, n) for (int i=n-1; i>=0; i--)
#define rre1(i, n) for (int i=n; i>0; i--)
#define rre2(i, r, l) for (int i=r-1; i>=l; i--)
#define rre3(i, r, l) for (int i=r; i>=l; i--)
#define debug(x) cout << #x << " = " << x << endl;
#define pb push_back
#define re_t(i,x) for(typeof(x.begin()) i=x.begin();i!=x.end();i++)
#define all(x) x.begin(),x.end()
#define SORT(x) sort(all(x))
#define MP make_pair
typedef pair<int,int> ii;
typedef vector<int> vi;
typedef vi::iterator vit;
typedef set<int> si;
typedef si::iterator sit;
typedef map<int,int> mii;
typedef mii::iterator mit;
typedef long long ll;
typedef unsigned long long ull;
typedef unsigned int uint;
typedef istringstream ISS;
typedef ostringstream OSS;
const int MAXN = 10000, INF = ~0U >> 2;
class SlimeXSlimeRancher2 {
    public:
    int train(vector <int> a)
    {
        int n = a.size(), m = -INF, res = 0;
        re(i, n) m = max(m, a[i]);
        re(i, n) res += m - a[i];
        return res;
    }
};

SRM 506 DIV2 500�Q?br>

#include <iostream>
#include <string>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>
#include <time.h>
#include <iomanip>
#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
using namespace std;
#define re(i, n) for (int i=0; i
#define re1(i, n) for (int i=1; i<=n; i++)
#define re2(i, l, r) for (int i=l; i
#define re3(i, l, r) for (int i=l; i<=r; i++)
#define rre(i, n) for (int i=n-1; i>=0; i--)
#define rre1(i, n) for (int i=n; i>0; i--)
#define rre2(i, r, l) for (int i=r-1; i>=l; i--)
#define rre3(i, r, l) for (int i=r; i>=l; i--)
#define debug(x) cout << #x << " = " << x << endl;
#define pb push_back
#define re_t(i,x) for(typeof(x.begin()) i=x.begin();i!=x.end();i++)
#define all(x) x.begin(),x.end()
#define SORT(x) sort(all(x))
#define MP make_pair
typedef pair<int,int> ii;
typedef vector<int> vi;
typedef vi::iterator vit;
typedef set<int> si;
typedef si::iterator sit;
typedef map<int,int> mii;
typedef mii::iterator mit;
typedef long long ll;
typedef unsigned long long ull;
typedef unsigned int uint;
typedef istringstream ISS;
typedef ostringstream OSS;
const int MAXN = 10000, INF = ~0U >> 2;
class SlimeXSlimesCity {
    public:
    int merge(vector <int> a)
    {
        int n = a.size();
        SORT(a);
        ll s = 0, s1;
        int res = 0;
        bool ff;
        re(i, n) {
            s += a[i]; s1 = s; ff = 1;
            re2(j, i + 1, n) {
                if (s1 < a[j]) {ff = 0; break;}
                s1 += a[j];
            }
            res += ff;
        }
        return res;
    }
};

Mato_No1 2011-05-15 12:20 发表评论

欧美视频日韩,国产区精品视频,欧美日韩成人在线观看

SRM 512

SRM 511

Topcoder注册成功