爬格子呀--IEEE极限编程大赛留念

10.14，坐标：电子科技大学
24h，不间断的编程，感觉还是很爽的。
排名一般，但是这是开始，未来还很远。

题目举例1：
广袤的非洲大草原上，狮子居住在一个个的网格里，他们的势力范围会以曼哈顿路程的的方式给出，求出哪个狮子重叠在其他狮子领地里的次数最少。
代码如下：

#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[1000][1000];

void draw() {
    int i, j;
    for (i = ri - di + 1; i < ri+di; i++) {
        for (j = ci - di + 1; j < ci+di; j++) {
            if (i < 0 || j < 0 || 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) >= 0) sign[row1][ci]++;
    if ((row2 = ri + di) < n) sign[row2][ci]++;
    if ((col1 = ci - di) >= 0) 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 = 0;
    while (i++ < k) {
        cin >> ri >> ci >> di;
        ri--; ci--;
        row.push_back(ri);
        col.push_back(ci);
        draw();
    }
    i = 0;
    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, -1);
    sort(tar->second.begin(), tar->second.end());
    cout << *(tar->second).begin() << " " << tar->first;
    return 0;
}

题目举例2：
有一堆人，每个人都有身高，将人按身高排序，同样身高的放到一行，同时按名字排序，每一行的最后输出这一行人的始编号和末编号，
代码如下：

#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 = 0, height;
    string s;
    while (t < n) {
        cin >> s >> height;
        num.insert(make_pair(height, s));
        t++;
    }
    vector<string>store[1000];
    multimap<int, string>::iterator it = num.begin(), itt = num.begin();
    int k = 0;
    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 = 0, init = 1;
    for (int j = 0; 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
", init, time);
        init = time+1;
    }
    return 0;
}

题目举例3：
有一个n*n的地图，你从（0，0）出发，目的地是（n,n)， 一路上只能像下走或者向右走，每个格子都有血药或者毒药，走上去会按药的数量扣血或加血，计算开始时的应该有的最低血量。
代码如下：

#include<stdio.h>
#include<iostream>
#include<vector>
using namespace std;
int r, c, health;
int rev_div[2][2] = { {-1,0},{0,-1} };
//vector<vector<int>>step;

struct {
    int potion;
    bool state = false;
}unit[1000][1000];

void init() {
    int i = 0, j = 0;
    while (i < r) {
        unit[i][0].potion += unit[i + rev_div[0][0]][0].potion;
        i++;
    }
    while (j < c) {
        unit[0][j].potion += unit[0][j + rev_div[1][1]].potion;
        j++;
    }
}

int main() {
    cin >> r >> c;
    for (int i = 0; i < r; i++) {
        for (int j = 0; j < c; j++) {
            cin >> unit[i][j].potion;
        }
    }
//  struct type position;
    init();
    int mid = 0;
    while (1) {
        int i = 1, j = 1;
        while (i < r) {
            while (j < c) {
                mid = unit[i + rev_div[0][0]][j + rev_div[0][1]].potion > unit[i + rev_div[1][0]][j + rev_div[1][1]].potion ? unit[i + rev_div[0][0]][j + rev_div[0][1]].potion : unit[i + rev_div[1][0]][j + rev_div[1][1]].potion;
                unit[i][j].potion += mid;
                j++;
            }
            j = 1;
            i++;
        }
        health = (unit[r-1][c-1].potion > 1 ? 1 : 1 - unit[r-1][c-1].potion);
        break;
    }
    cout << health;
    return 0;
}

题目举例4：
这道题就是生命游戏的实现，要求左边界与右边界相接，上边界与下边界相接，输出迭代n次后的结果。
代码如下：
1： 啰里啰唆版：

#include<cstdio>
#include<iostream>
using namespace std;
char cell[26][26], new_cell[26][26];
int n, m, c;

int get_neigh(int row,int col) {
    int count=0, i, j;
    //左上
    if (row == 0 && col == 0) {
        if (cell[0][1] == '*') count++;
        if (cell[1][1] == '*') count++;
        if (cell[1][0] == '*') count++;
        if (cell[n-1][0] == '*') count++;
        if (cell[n-1][1] == '*') count++;
        if (cell[n-1][m-1] == '*') count++;
        if (cell[0][m-1] == '*') count++;
        if (cell[1][m-1] == '*') count++;
        return count;

    }
    //左下
    else if (row == n-1 && col == 0) {
        if (cell[n-2][0] == '*') count++;
        if (cell[n-1][1] == '*') count++;
        if (cell[n-2][1] == '*') count++;
        if (cell[n - 1][m-1] == '*') count++;
        if (cell[n - 2][m-1] == '*') count++;
        if (cell[0][0] == '*') count++;
        if (cell[0][1] == '*') count++;
        if (cell[0][m - 1] == '*') count++;
        return count;
    }
    //右上
    else if (row == 0 && col == m - 1) {
        if (cell[0][0] == '*') count++;
        if (cell[1][0] == '*') count++;
        if (cell[0][m - 2] == '*') count++;
        if (cell[1][m - 2] == '*') count++;
        if (cell[1][m - 1] == '*') count++;
        if (cell[n - 1][0] == '*') count++;
        if (cell[n - 1][m - 2] == '*') count++;
        if (cell[n - 1][m - 1] == '*') count++;
        return count;
    }
    //右下
    else if (row == n-1 && col == m-1) {
        if (cell[0][0] == '*') count++;
        if (cell[n-1][m-2] == '*') count++;
        if (cell[n-2][m - 2] == '*') count++;
        if (cell[n-2][m - 1] == '*') count++;
        if (cell[n-1][0] == '*') count++;
        if (cell[n - 2][0] == '*') count++;
        if (cell[0][m - 2] == '*') count++;
        if (cell[0][m - 1] == '*') count++;
        return count;
    }
    //首行
    else if (row == 0) {
        for (i = row; i <= row + 1; i++) {
            for (j = col - 1; j <= col + 1; j++) {
                if (cell[i][j] == '*')
                    count++;
            }
        }
        for (j = col - 1; j <= col + 1; j++) {
            if (cell[n-1][j] == '*')
                count++;
        }
        if (cell[row][col] == '*')
            count--;
        return count;
    }
    //末行
    else if (row == n - 1) {
        for (i = row-1; i <= row; i++) {
            for (j = col - 1; j <= col + 1; j++) {
                if (cell[i][j] == '*')
                    count++;
            }
        }
        for (j = col - 1; j <= col + 1; j++) {
            if (cell[0][j] == '*')
                count++;
        }
        if (cell[row][col] == '*')
            count--;
        return count;
    }
    //首列
    else if (col == 0) {
        for (i = row-1; i <= row + 1; i++) {
            for (j = col ; j <= col + 1; j++) {
                if (cell[i][j] == '*')
                    count++;
            }
        }
        for (i = row - 1; i <= row + 1; i++) {
            if (cell[i][m-1] == '*')
                count++;
        }
        if (cell[row][col] == '*')
            count--;
        return count;
    }
    //末列
    else if (col == m) {
        for (i = row - 1; i <= row + 1; i++) {
            for (j = col-1; j <= col ; j++) {
                if (cell[i][j] == '*')
                    count++;
            }
        }
        for (i = row - 1; i <= row + 1; i++) {
            if (cell[i][0] == '*')
                count++;
        }
        if (cell[row][col] == '*')
            count--;
        return count;
    }
    //一般
    else {
        for (i = row - 1; i <= row + 1; i++) {
            for (j = col - 1; j <= col + 1; j++) {
                if (cell[i][j] == '*')
                    count++;
            }
        }
        if (cell[row][col] == '*')
            count--;
        return count;
    }
}

int main() {
    cin >> n >> m >> c;
    int i, j;
    for (i = 0; i < n; i++) {
        for (j = 0; j < m; j++) {
            cin >> cell[i][j];
        }
    }
    int kase = 0;
    while (kase < c) {
        for (i = 0; i < n; i++)
        {
            for (j = 0; j < m; j++)
            {
                switch (get_neigh(i, j))
                {
                case 0:
                case 1:
                case 4:
                case 5:
                case 6:
                case 7:
                case 8:
                    new_cell[i][j] = '-';
                    break;
                case 2:
                    new_cell[i][j] = cell[i][j];
                    break;
                case 3:
                    new_cell[i][j] = '*';
                    break;
                }
            }
        }
        kase++;
    }
    cout << endl;
    for (i = 0; i < n; i++) {
        for (j = 0; j < m; j++) {
            cout << cell[i][j];
        }
        cout << endl;
    }
    return 0;
}

代码举例2：
这个最后改的挺多的，其实按照第一个的大思路然后把拿这个的邻居计算思路就行。

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

int n, m, c;
//获取邻居数；
int get_neigh(int row, int col, char (*cell)[26]) {
    int count = 0, i, j;
    for (i = -1; i <= 1; i++) {
        for (j = -1; j <= 1; j++) {
            if (cell[(row + i + n) % n][(col + j + m) % m] == '*')
                count++;
        }
    }
    if (cell[row][col] == '*')
        count--;
    return count;
}

void change(char (*mat1)[26], char (*mat2)[26]) {
    int i, j, num;
    for (i = 0; i < n; i++)
    {
        for (j = 0; j < m; j++)
        {
            num = get_neigh(i, j, mat1);
            if (num == 2)  mat2[i][j] = mat1[i][j];
            else if (num == 3) {
                mat2[i][j] = '*';
            }
            else {
                mat1[i][j] = '-';
            }
        }
    }
}

int main() {
    char cell[26][26], new_cell[26][26] = {};
    cin >> n >> m >> c;
    int i, j;
    for (i = 0; i < n; i++) {
        for (j = 0; j < m; j++) {
            cin >> cell[i][j];
        }
    }
    int kase = 1;
    while (kase <= c) {
        if (kase % 2 != 0)
            change(cell, new_cell);
        else
            change(new_cell, cell);
        kase++;
    }
    for (i = 0; i < n; i++) {
        for (j = 0; j < m; j++) {
            cout << new_cell[i][j];
        }
        cout << endl;
    }
    return 0;
}

题目举例5：有好多个长方形，每个长方形中间都会包一个小的长方形，大长方形已经内部已经一个一个小格的依次编号，题目会给出你小长方形的两个对角线顶点，计算每个长方形在小长方形之外的所有数字的异或和

代码如下：

#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[100];

int main() {
    cin >> t;
    int i = 0, j = 0;
    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 + 1) % rec[i].l;
        a2 = (rec[i].d2 - rec[i].val + 1) % rec[i].l;
        b1 = (rec[i].d1 - rec[i].val + 1) / rec[i].l;
        b2 = (rec[i].d2 - rec[i].val + 1) / 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) + 1;
        rec[i].width = abs(b1 - b2) + 1;
        i++;
    }
    set<int>::iterator it ,itt;
    for (i = 0; i < t; i++) {
        j = 0;
        while (j < rec[i].l*rec[i].h) {
            s.insert(rec[i].val++);
            j++;
        }
        it = s.begin();
        advance(it, rec[i].position-1);
        itt = it;
        advance(it, rec[i].length);
        for (j = 0; 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 = 0;
        s.clear();
    }
    return 0;
}

题目举例6：求斐波那契数列的前n项值（这个n上界是22400）
代码举例：

#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 = 100000000;
    static const int WIDTH = 8;
    vector<int>s;

    Bigintegar(long long num = 0) { *this = num; }
    Bigintegar operator = (long long num) {
        s.clear();
        do {
            s.push_back(num%BASE);
            num /= BASE;
        } while (num > 0);
        return *this;
    }
    Bigintegar operator=(const string& str) {
        s.clear();
        int x, len = (str.length() - 1) / WIDTH + 1;
        for (int i = 0; i < len; i++) {
            int end = str.length() - i*WIDTH;
            int start = max(0, 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 = 0, g = 0;; i++) {
            if (g == 0 && 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() - 2; i >= 0; i--) {
        char buf[20];
        sprintf(buf, "%08d", x.s[i]);
        for (int j = 0; j < strlen(buf); j++) out << buf[j];
    }
    return out;
}

Bigintegar fibMinus1 = 1;
Bigintegar fibMinus2 = 2;
Bigintegar fibN;

Bigintegar step(int n)
{
    for (int i = 2; i < n; i++)
    {
        fibN = fibMinus1 + fibMinus2;
        fibMinus1 = fibMinus2;
        fibMinus2 = fibN;
    }
    return fibN;
}

int main()
{
    int s, kase;
    int t, i = 0, k;
    cin >> t;
    while (i<t) {
        cin >> k;
        i++;

    }
    i = 0;

}