C++博客-寒月-随笔分类-模拟

UVa 327 - Evaluating Simple C Expressions

wuxu — Tue, 29 Nov 2011 13:19:00 GMT

摘要: 题目链接：http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=104&page=show_problem&problem=263 Code highlighting produced by Actipro CodeHighlighter (freeware)ht... 阅读全文

wuxu 2011-11-29 21:19 发表评论

UVa 11111 - Generalized Matrioshkas

wuxu — Wed, 23 Nov 2011 05:27:00 GMT

题目链接：http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=103&page=show_problem&problem=2052

#include <iostream>
#include <sstream>
#include <stack>
#include <string>
#include <cmath>
#include <cstring>
using namespace std;

struct toy
{
    int x;
    int c;
    toy(int i = 0, int j = 0): x(i), c(j) {}
};
int n;
int a[500000];

bool isMatKa()
{
    stack<toy> st;
    for(int i = 0; i < n; ++i)
    {
        if(a[i] < 0)
            st.push(toy(a[i], a[i]));
        else
        {
            if(st.size() && st.top().x + a[i] == 0)
            {
                st.pop();
                if(st.size())
                {
                    toy t = st.top();
                    st.pop();
                    t.c += a[i];
                    if(t.c >= 0) return false;
                    st.push(t);
                }
                else if(i + 1 < n)
                    return false;
            }
            else return false;
        }
    }
    return st.empty();
}

int main()
{
    string s;
    while(getline(cin, s))
    {
        n = 0;
        stringstream str(s);
        while(str >> a[n]) ++n;
        bool isMK = true;
        isMK = isMatKa();
        if(!isMK)
            cout << ":-( Try again." << endl;
        else
            cout << ":-) Matrioshka!" << endl;
    }
    return 0;
}

wuxu 2011-11-23 13:27 发表评论

UVa 10152 - ShellSort

wuxu — Tue, 22 Nov 2011 08:36:00 GMT

题目链接：http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=103&page=show_problem&problem=1093
首先找到需要移动的字符串，方法如下：以初始序列为准，设初始序列下标为i, 目的序列下标为j, 从n-1开始，如果两下标对应的字符串相等，下标同时减一，否则仅初始序列下标减一。那么目的序列中还未被成功匹配的字符串就是需要移动的字符串。要使移动次数最少，显然应该按未被处理的目的序列中字符串逆序移动（输出）。

#include <iostream>
#include <string>
using namespace std;

int k, n;
string org[210], des[210];

int main()
{
    int i, j;
    cin >> k;
    while(k--)
    {
        cin >> n;
        cin.ignore();
        for(i = 0; i < n; ++i)
        getline(cin, org[i]);
        for(i = 0; i < n; ++i)
        getline(cin, des[i]);
        for(i = j = n - 1; i >= 0; --i)
        {
            if(org[i] == des[j]) --j;
        }
        for( ; j >= 0; --j)
        cout << des[j] << endl;
        cout << endl;
    }
    return 0;
}

wuxu 2011-11-22 16:36 发表评论

UVa 133 - The Dole Queue

wuxu — Sat, 19 Nov 2011 14:12:00 GMT

题目链接：http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=103&page=show_problem&problem=69

#include <iostream>
#include <list>
#include <iterator>
#include <iomanip>
using namespace std;

int N, k, m;
list<int> dole;

int main()
{
    while(cin >> N >> k >> m)
    {
        if(!(N || k || m)) break;
        dole.clear();
        for(int i = 1; i <= N; ++i)
        dole.push_back(i);
        list<int>::iterator is, ie;
        is = dole.begin();
        ie = dole.end();
        --ie;
        while(true)
        {
            for(int i = 1; i < k; ++i)
            {
                ++is;
                if(is == dole.end()) is = dole.begin();
            }
            for(int i = 1; i < m; ++i)
            {
                if(ie == dole.begin())
                {
                    ie = dole.end();
                }
                --ie;
            }
            if(is == ie)
            {
                cout << setw(3) << *is;
                if(dole.size() != 1)
                {
                    cout << ',';
                    is = dole.erase(is);

                    if(is == dole.begin())
                    {
                        ie = dole.end();
                        --ie;
                    }
                    else
                    {
                        ie = is;
                        --ie;
                    }
                    if(is == dole.end()) is = dole.begin();
                }
                else
                {
                    cout << endl;
                    break;
                }
            }
            else
            {
                cout << setw(3) << *is << setw(3) << *ie;
                list<int>::iterator it1 = is;
                list<int>::iterator it2 = ie;
                if(dole.size() != 2)
                {
                    cout << ',';
                    ++is;
                    while(is == it2) ++is;
                    ++ie;
                    while(ie == it1) ++ie;
                    dole.erase(it1);
                    dole.erase(it2);

                    if(is == dole.end()) is = dole.begin();
                    if(ie == dole.begin())
                    {
                        ie = dole.end();
                        --ie;
                    }
                    else --ie;
                }
                else
                {
                    cout << endl;
                    break;
                }
            }
        }
    }
    return 0;
}

wuxu 2011-11-19 22:12 发表评论

UVa 101 - The Blocks Problem

wuxu — Fri, 18 Nov 2011 16:04:00 GMT

题目链接：http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=103&page=show_problem&problem=37

#include <iostream>
#include <sstream>
#include <vector>
#include <iterator>
#include <string>
#include <algorithm>
#include <cstddef>
using namespace std;

int n;
vector<int> block[30];

void move_onto(int ia, int a, int ib, int b)
{
    int tmp;
    tmp = block[ia].back();
    while(tmp != a)
    {
        block[ia].pop_back();
        block[tmp].push_back(tmp);
        tmp = block[ia].back();
    }
    tmp = block[ib].back();
    while(tmp != b)
    {
        block[ib].pop_back();
        block[tmp].push_back(tmp);
        tmp = block[ib].back();
    }
    block[ia].pop_back();
    block[ib].push_back(a);
}

void move_over(int ia, int a, int ib, int b)
{
    int tmp;
    tmp = block[ia].back();
    while(tmp != a)
    {
        block[ia].pop_back();
        block[tmp].push_back(tmp);
        tmp = block[ia].back();
    }
    block[ia].pop_back();
    block[ib].push_back(a);
}

void pile_onto(int ia, int a, int ib, int b)
{
    int tmp;
    tmp = block[ib].back();
    while(tmp != b)
    {
        block[ib].pop_back();
        block[tmp].push_back(tmp);
        tmp = block[ib].back();
    }
    vector<int>::iterator it;
    it = find(block[ia].begin(), block[ia].end(), a);
    block[ib].insert(block[ib].end(), it, block[ia].end());
    block[ia].erase(it, block[ia].end());
}

void pile_over(int ia, int a, int ib, int b)
{
    vector<int>::iterator it;
    it = find(block[ia].begin(), block[ia].end(), a);
    block[ib].insert(block[ib].end(), it, block[ia].end());
    block[ia].erase(it, block[ia].end());
}

void manipulate(const string &s)
{
    string s1, s2;
    int a, b, ia, ib;
    stringstream str(s);
    str >> s1 >> a >> s2 >> b;
    if(a == b) return;
    for(int i = 0; i < n; ++i)
    {
        for(size_t j = 0; j != block[i].size(); ++j)
        {
            if(block[i][j] == a) ia = i;
            if(block[i][j] == b) ib = i;
        }
    }
    if(ia == ib) return;
    if(s1 == "move" && s2 == "onto") move_onto(ia, a, ib, b);
    else if(s1 == "move" && s2 == "over") move_over(ia, a, ib, b);
    else if(s1 == "pile" && s2 == "onto") pile_onto(ia, a, ib, b);
    else pile_over(ia, a, ib, b);
}

int main()
{
    string s;
    cin >> n;
    for(int i = 0; i < n; ++i)
    block[i].push_back(i);
    cin.ignore();
    while(getline(cin, s) && s != "quit")
    manipulate(s);
    for(int i = 0; i < n; ++i)
    {
        cout << i << ':';
        for(size_t j = 0; j != block[i].size(); ++j)
        cout << ' ' << block[i][j];
        cout << endl;
    }
    return 0;
}

wuxu 2011-11-19 00:04 发表评论

UVa127 - "Accordian" Patience

wuxu — Mon, 14 Nov 2011 16:03:00 GMT

题目链接：http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=103&page=show_problem&problem=63
此题若按每发一张牌就处理一次会TLE.

#include <iostream>
#include <cstddef>
#include <string>
#include <vector>
#include <stack>
using namespace std;

inline bool Match(const string &s1, const string &s2)
{
    return s1[0] == s2[0] || s1[1] == s2[1];
}

int main()
{
    string s;
    vector< stack<string> > piles;
    while(cin >> s && s != "#")
    {
        stack<string> tmp;
        tmp.push(s);
        piles.push_back(tmp);
        if(piles.size() == 52)
        {
            while(true)
            {
                size_t i;
                for(i = 0; i != piles.size(); ++i)
                {
                    if(i >=3 && Match(piles[i].top(), piles[i - 3].top()))
                    {
                        piles[i - 3].push(piles[i].top());
                        piles[i].pop();
                        break;
                    }
                    if(i >= 1 && Match(piles[i].top(), piles[i - 1].top()))
                    {
                        piles[i - 1].push(piles[i].top());
                        piles[i].pop();
                        break;
                    }
                }
                if(i == piles.size()) break;
                else if(piles[i].empty())
                {
                    piles.erase(piles.begin() + i);
                }
            }
            cout << piles.size() << (piles.size() == 1 ? " pile remaining:" : " piles remaining:");
            for(size_t i = 0; i != piles.size(); ++i) cout << ' ' << piles[i].size();
            cout << endl;

            piles.clear();
        }
    }
    return 0;
}

wuxu 2011-11-15 00:03 发表评论

UVa 10194 Football (aka Soccer)

wuxu — Fri, 21 Oct 2011 07:27:00 GMT

摘要: 题目链接：http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&category=98&problem=1135&mosmsg=Submission+received+with+ID+9393629 Code highlighting ... 阅读全文

wuxu 2011-10-21 15:27 发表评论

pku1102

wuxu — Fri, 13 Aug 2010 02:52:00 GMT

摘要: 模拟.详细见代码: #include#includeusing namespace std;char a[7][10]={{'-',' ','-','-',' ','-','-','-','-','-'},{'|',' ',' ',' ','|','|',... 阅读全文

wuxu 2010-08-13 10:52 发表评论