第13届景驰-埃森哲杯广东工业大学ACM程序设计大赛

C.平分游戏

D.psd面试

感受：dp还是不会，虽然这是最简单的dp。。。。。

#pragma warning(disable:4996)
#include<cstdio>
#include<string>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;

const int maxn = 1500;

int n, dp[maxn][maxn];
string s;

void LPS() {
    memset(dp, 0, sizeof(dp));
    for (int i = 0; i < n; i++) dp[i][i] = 1;
    for (int i = 1; i < n; i++) {
        int tp = 0;
        for (int j = 0; j + i < n; j++) {
            if (s[j] == s[j + i]) tp = dp[j + 1][j + i - 1] + 2;
            else tp = max(dp[j + 1][j + i], dp[j][j + i - 1]);
            dp[j][j + i] = tp;
        }
    }
    cout << n - dp[0][n - 1] << endl;
}

int main()
{    
    while (cin >> s) {
        n = s.size();
        for (int i = 0; i < n; i++) if (s[i] >= 'A'&& s[i] <= 'Z') s[i] = tolower(s[i]);
        LPS();
    }
    return 0;
}

E.回旋星空

题解：枚举中间的那个点，分别计算其它点到这个点的距离，再排序。那么相同距离的就会排在一起。

感受：想到了枚举中间点，但算错了复杂度。。。。

#pragma warning(disable:4996)
#include<map>
#include<vector>
#include<stack>
#include<queue>
#include<cstdio>
#include<string>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
typedef long long ll;

const int maxn = 1005;

int n, T;

struct node {
    int x, y;
}p[maxn];

/*struct mode {
    int s, e, d;
    bool operator<(const mode& i)const {
        return d < i.d;
    }
}q[maxn];*/

inline int cad(node a, node b) { return (a.x - b.x)*(a.x - b.x) + (a.y - b.y)*(a.y - b.y); }

int main()
{
    cin >> T;
    while (T--) {
        cin >> n;
        for (int i = 1; i <= n; i++) cin >> p[i].x >> p[i].y;

        ll ans = 0;
        for (int i = 1; i <= n; i++) {
            vector<int> d;
            for (int j = 1; j <= n; j++) if (j != i)d.push_back(cad(p[i], p[j]));
            sort(d.begin(), d.end());
            
            int cnt = 1;
            for (int j = 1; j < d.size(); j++) {
                if (d[j] == d[j - 1]) cnt++;
                else {
                    ans += cnt * (cnt - 1);
                    cnt = 1;
                }
            }
            ans += cnt * (cnt - 1);
        }
        if (ans == 0) cout << "WA" << endl;
        else printf("%lld\n", ans);
            
    }
    return 0;
}

F.等式

题解：假设x=n+a,y=n+b;代入方程得：n*n=a*b。所以如果n*n有p个因子，则ans=(p+1)/2;因为n*n的质因子和n的质因子是相同的，所以这里分解n就行了。

感受：网上的大佬就是6啊！不过这题是原题，poj2917。

#pragma warning(disable:4996)
#include<cstdio>
#include<string>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;

int n, T;
int num[64];

int main()
{
    cin >> T;
    while (T--) {
        cin >> n;
        memset(num, 0, sizeof(num));
        int p = 0;
        for (int i = 2; i*i <= n; i++) {
            if (n % i) continue;
            while (n % i == 0) {
                num[p]++;
                n /= i;
            }
            p++;
        }
        if (n != 1) {
            num[p]++;
            p++;
        }
        for (int i = 0; i < p; i++) num[i] *= 2;
        int ans = 1;
        for (int i = 0; i < p; i++) ans *= (num[i] + 1);
        cout << (ans+1)/2 << endl;
    }
    return 0;
}

G.旋转矩阵

题解：LR和RL等同没有旋转，所以旋转到最后等价于只向左旋或只向右旋。

感受：fuckkkkk！if-else结构竟然写挂了，比赛结束后真想找块豆腐撞死。

比赛时写的左旋：

/*左旋*/
void print3() {
    cout << m << " " << n << endl;
    for (int i = m - 1; i >= 0; i--) {
        for (int j = 0; j < n; j++) {
            if (mp[j][i] == '|') mp[j][i] = '-';
            if (mp[j][i] == '-') mp[j][i] = '|';    //竟然没找出错误，orzzzzz!
            cout << mp[j][i];
        }
        cout << endl;
    }
}

最后AC的代码：

#pragma warning(disable:4996)
#include<cstdio>
#include<string>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;

const int maxn = 2000;

int T, n, m;
char mp[100][100];

string s;

void print1(){
    cout << n << " " << m << endl;
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < m; j++) cout << mp[i][j];
        cout << endl;
    }
}
/*右旋*/
void print2() {
    cout << m << " " << n << endl;
    for (int i = 0; i < m; i++) {
        for (int j = n - 1; j >= 0; j--) {
            if (mp[j][i] == '|')  cout << "-";
            else if (mp[j][i] == '-')  cout << "|";
            else cout << mp[j][i];
        }
        cout << endl;
    }
}
/*左旋*/
void print3() {
    cout << m << " " << n << endl;
    for (int i = m - 1; i >= 0; i--) {
        for (int j = 0; j < n; j++) {
            if (mp[j][i] == '|')  cout << "-";
            else if (mp[j][i] == '-') cout << "|";
            else cout << mp[j][i];
        }
        cout << endl;
    }
}
/*左旋两次*/
void print4() {
    cout << n << " " << m << endl;
    for (int i = n - 1; i >= 0; i--) {
        for (int j = m - 1; j >= 0; j--) cout << mp[i][j];
        cout << endl;
    }
}

int main()
{
    cin >> T;
    while (T--) {
        cin >> n >> m;
        for (int i = 0; i < n; i++)
            for (int j = 0; j < m; j++) cin >> mp[i][j];
        cin >> s;

        int l = s.size();
        int p = 0, q = 0;
        for (int i = 0; i < l; i++) {
            if (s[i] == 'L') p++;
            if (s[i] == 'R') q++;
        }

        if (p == q) print1(); 
        else if (p > q) {
            p = (p - q) % 4;
            if (p == 0) print1(); 
            else if (p == 1) print3(); 
            else if (p == 2) print4(); 
            else print2(); 
        }
        else {
            q = (q - p) % 4;
            if (q == 0) print1(); 
            else if (q == 1) print2(); 
            else if (q == 2) print4(); 
            else print3();
        }

        cout << endl;
    }
    return 0;
}

J.强迫症序列

题解：每次只能对n-1个数加一，等价于每次只能对1个数减一。而且每个元素都相等的情况只有一种。

#pragma warning(disable:4996)
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;

const int maxn = 1e5 + 5;

int n;
int a[maxn];

int main()
{    
    int T;
    while (cin >> T) {
        while (T--) {
            cin >> n;
            for (int i = 1; i <= n; i++) scanf("%d", a + i);
            sort(a + 1, a + n + 1);
            int ans = 0;
            for (int i = 1; i <= n; i++) ans += (a[i] - a[1]);
            cout << ans << " " << ans + a[1] << endl;
        }
    }
    return 0;
}

K.密码

题解：找规律。

#pragma warning(disable:4996)
#include<cstdio>
#include<string>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;

const int maxn = 100005;

int n, m;
char s[maxn];

int main()
{
    int T;
    cin >> T;
    while (T--) {
        scanf("%d%s", &n, s);
        m = strlen(s);
        if (n == 1 || n >= m) { printf("%s\n", s); continue; }

        for (int i = 0; i < n; i++) {
            int t = i;
            int p = 0;
            while (t < m) {
                cout << s[t];
                if (i == 0 || i == n - 1) { t += 2 * (n - 1); continue; }
                if (p % 2) t += 2 * i;
                else t += 2 * (n - i - 1);
                p++;
            }
        }
        cout << endl;
    }
    
    return 0;
}

L.用来作弊的药水

题解：分类讨论。

#pragma warning(disable:4996)
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;

const int maxn = 1e5 + 5;

int n;
int a[maxn];

int main()
{    
    
    int T;
    cin >> T;
    while (T--) {
        int x, a, y, b;
        cin >> x >> a >> y >> b;
        bool flag = false;
        while (true) {

            if (x == 1 && y == 1) { flag = true; break; }
            if (x == 1 && y != 1) break;
            if (x != 1 && y == 1) break;

            if (a == b) {
                if (x == y) { flag = true; break; }
                else break;
            }
            else if (a < b) {
                if (x < y || x % y) break;
                else {
                    x = x / y;
                    b = b - a;
                }
            }
            else {
                if (y < x || y % x) break;
                else {
                    y = y / x;
                    a = a - b;
                }
            }
            if (x == y && a != b) break;
        }
        if (flag) cout << "Yes" << endl;
        else cout << "No" << endl;
        
    }
    return 0;
}