爬格子呀--IEEE極限程式設計大賽留念

10.14，坐标：電子科技大學

24h，不間斷的程式設計，感覺還是很爽的。

排名一般，但是這是開始，未來還很遠。

題目舉例1：

廣袤的非洲大草原上，獅子居住在一個個的網格裡，他們的勢力範圍會以曼哈頓路程的的方式給出，求出哪個獅子重疊在其他獅子領地裡的次數最少。

代碼如下：

#include<stdio.h>
#include<algorithm>
#include<iostream>
#include<vector>
#include<map>
#include<set>
#include<utility>
using namespace std;
int n, m, k, ri, ci, di;
int sign[][];

void draw() {
    int i, j;
    for (i = ri - di + ; i < ri+di; i++) {
        for (j = ci - di + ; j < ci+di; j++) {
            if (i <  || j <  || i >= n || j >= m)
                continue;
            if (i == ri&&j == ci) {
                sign[i][j]--;
            }
            sign[i][j]++;
        }
    }
    int row1, row2, col1, col2;
    if ((row1 = ri - di) >= ) sign[row1][ci]++;
    if ((row2 = ri + di) < n) sign[row2][ci]++;
    if ((col1 = ci - di) >= ) sign[ri][col1]++;
    if ((col2 = ci + di) < m) sign[ri][col2]++;
}

int main() {
    vector<int>row, col;
    map<int, vector<int>>s;
    cin >> n >> m >> k;
    int i = ;
    while (i++ < k) {
        cin >> ri >> ci >> di;
        ri--; ci--;
        row.push_back(ri);
        col.push_back(ci);
        draw();
    }
    i = ;
    vector<int>::iterator it = row.begin(), ittt = col.begin();
    set<int>ss;
    while (i++ < k) {
        s.insert(make_pair(sign[*it][*ittt], i));
        it++;
        ittt++;
    }
    map<int, vector<int>>::iterator tar = s.end();
    advance(tar, -);
    sort(tar->second.begin(), tar->second.end());
    cout << *(tar->second).begin() << " " << tar->first;
    return ;
}

題目舉例2：

有一堆人，每個人都有身高，将人按身高排序，同樣身高的放到一行，同時按名字排序，每一行的最後輸出這一行人的始編号和末編号，

代碼如下：

#include<stdio.h>
#include<iostream>
#include<map>
#include<vector>
#include<string>
#include<utility>
#include<algorithm>
using namespace std;
multimap<int, string>num;
int n;
int main() {
    cin >> n;
    int t = , height;
    string s;
    while (t < n) {
        cin >> s >> height;
        num.insert(make_pair(height, s));
        t++;
    }
    vector<string>store[];
    multimap<int, string>::iterator it = num.begin(), itt = num.begin();
    int k = ;
    for (; it != num.end(); it++) {
        if (it->first == itt->first) {
            store[k].push_back(it->second);
        }
        else {
            k++;
            itt = it;
            store[k].push_back(it->second);
        }
    }
    vector<string>::iterator iet;
    int time = , init = ;
    for (int j = ; j <= k; j++) {
        sort(store[j].begin(), store[j].end());
        for (iet = store[j].begin(); iet != store[j].end(); iet++) {
            cout << *iet << " ";
            time++;
        }
        printf("%d %d\n", init, time);
        init = time+;
    }
    return ;
}

題目舉例3：

有一個n*n的地圖，你從（0，0）出發，目的地是（n,n)，一路上隻能像下走或者向右走，每個格子都有血藥或者毒藥，走上去會按藥的數量扣血或加血，計算開始時的應該有的最低血量。

代碼如下：

#include<stdio.h>
#include<iostream>
#include<vector>
using namespace std;
int r, c, health;
int rev_div[][] = { {-,},{,-} };
//vector<vector<int>>step;

struct {
    int potion;
    bool state = false;
}unit[][];

void init() {
    int i = , j = ;
    while (i < r) {
        unit[i][].potion += unit[i + rev_div[][]][].potion;
        i++;
    }
    while (j < c) {
        unit[][j].potion += unit[][j + rev_div[][]].potion;
        j++;
    }
}

int main() {
    cin >> r >> c;
    for (int i = ; i < r; i++) {
        for (int j = ; j < c; j++) {
            cin >> unit[i][j].potion;
        }
    }
//  struct type position;
    init();
    int mid = ;
    while () {
        int i = , j = ;
        while (i < r) {
            while (j < c) {
                mid = unit[i + rev_div[][]][j + rev_div[][]].potion > unit[i + rev_div[][]][j + rev_div[][]].potion ? unit[i + rev_div[][]][j + rev_div[][]].potion : unit[i + rev_div[][]][j + rev_div[][]].potion;
                unit[i][j].potion += mid;
                j++;
            }
            j = ;
            i++;
        }
        health = (unit[r-][c-].potion >  ?  :  - unit[r-][c-].potion);
        break;
    }
    cout << health;
    return ;
}

題目舉例4：

這道題就是生命遊戲的實作，要求左邊界與右邊界相接，上邊界與下邊界相接，輸出疊代n次後的結果。

代碼如下：

1：啰裡啰唆版：

#include<cstdio>
#include<iostream>
using namespace std;
char cell[][], new_cell[][];
int n, m, c;

int get_neigh(int row,int col) {
    int count=, i, j;
    //左上
    if (row ==  && col == ) {
        if (cell[][] == '*') count++;
        if (cell[][] == '*') count++;
        if (cell[][] == '*') count++;
        if (cell[n-][] == '*') count++;
        if (cell[n-][] == '*') count++;
        if (cell[n-][m-] == '*') count++;
        if (cell[][m-] == '*') count++;
        if (cell[][m-] == '*') count++;
        return count;

    }
    //左下
    else if (row == n- && col == ) {
        if (cell[n-][] == '*') count++;
        if (cell[n-][] == '*') count++;
        if (cell[n-][] == '*') count++;
        if (cell[n - ][m-] == '*') count++;
        if (cell[n - ][m-] == '*') count++;
        if (cell[][] == '*') count++;
        if (cell[][] == '*') count++;
        if (cell[][m - ] == '*') count++;
        return count;
    }
    //右上
    else if (row ==  && col == m - ) {
        if (cell[][] == '*') count++;
        if (cell[][] == '*') count++;
        if (cell[][m - ] == '*') count++;
        if (cell[][m - ] == '*') count++;
        if (cell[][m - ] == '*') count++;
        if (cell[n - ][] == '*') count++;
        if (cell[n - ][m - ] == '*') count++;
        if (cell[n - ][m - ] == '*') count++;
        return count;
    }
    //右下
    else if (row == n- && col == m-) {
        if (cell[][] == '*') count++;
        if (cell[n-][m-] == '*') count++;
        if (cell[n-][m - ] == '*') count++;
        if (cell[n-][m - ] == '*') count++;
        if (cell[n-][] == '*') count++;
        if (cell[n - ][] == '*') count++;
        if (cell[][m - ] == '*') count++;
        if (cell[][m - ] == '*') count++;
        return count;
    }
    //首行
    else if (row == ) {
        for (i = row; i <= row + ; i++) {
            for (j = col - ; j <= col + ; j++) {
                if (cell[i][j] == '*')
                    count++;
            }
        }
        for (j = col - ; j <= col + ; j++) {
            if (cell[n-][j] == '*')
                count++;
        }
        if (cell[row][col] == '*')
            count--;
        return count;
    }
    //末行
    else if (row == n - ) {
        for (i = row-; i <= row; i++) {
            for (j = col - ; j <= col + ; j++) {
                if (cell[i][j] == '*')
                    count++;
            }
        }
        for (j = col - ; j <= col + ; j++) {
            if (cell[][j] == '*')
                count++;
        }
        if (cell[row][col] == '*')
            count--;
        return count;
    }
    //首列
    else if (col == ) {
        for (i = row-; i <= row + ; i++) {
            for (j = col ; j <= col + ; j++) {
                if (cell[i][j] == '*')
                    count++;
            }
        }
        for (i = row - ; i <= row + ; i++) {
            if (cell[i][m-] == '*')
                count++;
        }
        if (cell[row][col] == '*')
            count--;
        return count;
    }
    //末列
    else if (col == m) {
        for (i = row - ; i <= row + ; i++) {
            for (j = col-; j <= col ; j++) {
                if (cell[i][j] == '*')
                    count++;
            }
        }
        for (i = row - ; i <= row + ; i++) {
            if (cell[i][] == '*')
                count++;
        }
        if (cell[row][col] == '*')
            count--;
        return count;
    }
    //一般
    else {
        for (i = row - ; i <= row + ; i++) {
            for (j = col - ; j <= col + ; j++) {
                if (cell[i][j] == '*')
                    count++;
            }
        }
        if (cell[row][col] == '*')
            count--;
        return count;
    }
}

int main() {
    cin >> n >> m >> c;
    int i, j;
    for (i = ; i < n; i++) {
        for (j = ; j < m; j++) {
            cin >> cell[i][j];
        }
    }
    int kase = ;
    while (kase < c) {
        for (i = ; i < n; i++)
        {
            for (j = ; j < m; j++)
            {
                switch (get_neigh(i, j))
                {
                case :
                case :
                case :
                case :
                case :
                case :
                case :
                    new_cell[i][j] = '-';
                    break;
                case :
                    new_cell[i][j] = cell[i][j];
                    break;
                case :
                    new_cell[i][j] = '*';
                    break;
                }
            }
        }
        kase++;
    }
    cout << endl;
    for (i = ; i < n; i++) {
        for (j = ; j < m; j++) {
            cout << cell[i][j];
        }
        cout << endl;
    }
    return ;
}

代碼舉例2：

這個最後改的挺多的，其實按照第一個的大思路然後把拿這個的鄰居計算思路就行。

#include<cstdio>
#include<iostream>
#include<vector>
using namespace std;

int n, m, c;
//擷取鄰居數；
int get_neigh(int row, int col, char (*cell)[]) {
    int count = , i, j;
    for (i = -; i <= ; i++) {
        for (j = -; j <= ; j++) {
            if (cell[(row + i + n) % n][(col + j + m) % m] == '*')
                count++;
        }
    }
    if (cell[row][col] == '*')
        count--;
    return count;
}

void change(char (*mat1)[], char (*mat2)[]) {
    int i, j, num;
    for (i = ; i < n; i++)
    {
        for (j = ; j < m; j++)
        {
            num = get_neigh(i, j, mat1);
            if (num == )  mat2[i][j] = mat1[i][j];
            else if (num == ) {
                mat2[i][j] = '*';
            }
            else {
                mat1[i][j] = '-';
            }
        }
    }
}

int main() {
    char cell[][], new_cell[][] = {};
    cin >> n >> m >> c;
    int i, j;
    for (i = ; i < n; i++) {
        for (j = ; j < m; j++) {
            cin >> cell[i][j];
        }
    }
    int kase = ;
    while (kase <= c) {
        if (kase %  != )
            change(cell, new_cell);
        else
            change(new_cell, cell);
        kase++;
    }
    for (i = ; i < n; i++) {
        for (j = ; j < m; j++) {
            cout << new_cell[i][j];
        }
        cout << endl;
    }
    return ;
}

題目舉例5：有好多個長方形，每個長方形中間都會包一個小的長方形，大長方形已經内部已經一個一個小格的依次編号，題目會給出你小長方形的兩個對角線頂點，計算每個長方形在小長方形之外的所有數字的異或和

代碼如下：

#include<stdio.h>
#include<iostream>
#include<set>
#include<algorithm>
#include<cmath>
using namespace std;
long long xor1;
int t;
set<int>s;

struct {
    int l, h, d1, d2;
    long long val;
    int length, width, position;
}rec[];

int main() {
    cin >> t;
    int i = , j = ;
    while (i < t) {
        cin >> rec[i].l >> rec[i].h >> rec[i].val >> rec[i].d1 >> rec[i].d2;
        int a1, a2, b1, b2, min1, min2;
        a1 = (rec[i].d1 - rec[i].val + ) % rec[i].l;
        a2 = (rec[i].d2 - rec[i].val + ) % rec[i].l;
        b1 = (rec[i].d1 - rec[i].val + ) / rec[i].l;
        b2 = (rec[i].d2 - rec[i].val + ) / rec[i].l;
        min1 = a1 > a2 ? a2 : a1;
        min2 = b1 > b2 ? b2 : b1;
        rec[i].position = rec[i].l*min2 + min1;
        rec[i].length = abs(a1 - a2) + ;
        rec[i].width = abs(b1 - b2) + ;
        i++;
    }
    set<int>::iterator it ,itt;
    for (i = ; i < t; i++) {
        j = ;
        while (j < rec[i].l*rec[i].h) {
            s.insert(rec[i].val++);
            j++;
        }
        it = s.begin();
        advance(it, rec[i].position-);
        itt = it;
        advance(it, rec[i].length);
        for (j = ; j < rec[i].width; j++) {
            s.erase(itt, it);
            itt = it;
            advance(it, rec[i].l);
            advance(itt, rec[i].l - rec[i].length);
        }
        itt = it = s.begin();
        for (; it != s.end(); it++) {
            xor1 ^= *it;
        }
        cout << xor1 << endl;
        xor1 = ;
        s.clear();
    }
    return ;
}

題目舉例6：求斐波那契數列的前n項值（這個n上界是22400）

代碼舉例：

#include <iostream>
#include<stdio.h>
#include<vector>
#include<string>
#include<algorithm>
#include<string.h>
#include<queue>
using namespace std;

struct Bigintegar {
    static const int BASE = ;
    static const int WIDTH = ;
    vector<int>s;

    Bigintegar(long long num = ) { *this = num; }
    Bigintegar operator = (long long num) {
        s.clear();
        do {
            s.push_back(num%BASE);
            num /= BASE;
        } while (num > );
        return *this;
    }
    Bigintegar operator=(const string& str) {
        s.clear();
        int x, len = (str.length() - ) / 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;
    }
    Bigintegar operator + (const Bigintegar& b)const {
        Bigintegar c;
        c.s.clear();
        for (int i = , g = ;; i++) {
            if (g ==  && i >= s.size() && i >= 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;
    }
};
ostream& operator<<(ostream &out, const Bigintegar& 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++) out << buf[j];
    }
    return out;
}

Bigintegar fibMinus1 = ;
Bigintegar fibMinus2 = ;
Bigintegar fibN;

Bigintegar step(int n)
{
    for (int i = ; i < n; i++)
    {
        fibN = fibMinus1 + fibMinus2;
        fibMinus1 = fibMinus2;
        fibMinus2 = fibN;
    }
    return fibN;
}

int main()
{
    int s, kase;
    int t, i = , k;
    cin >> t;
    while (i<t) {
        cin >> k;
        i++;

    }
    i = ;

}

爬格子呀--IEEE極限程式設計大賽留念

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