计蒜客-Arab Collegiate Programming Contest 2015

\(B\)

题意

给定两个有理数 \(\frac{a}{b},\ \frac{c}{d}\)，求他们的 \(gcd\) 和 \(lcm\)

解

考虑求解 \(gcd\)

先将 \(\frac{a}{b},\ \frac{c}{d}\) 化为最简有理数

在保证能被 \(\frac{a}{b},\ \frac{c}{d}\) 整除的情况下，分母必须是最小，分子必须是最大。

分母要最小，即为 \(LCM(b,\ d)\)

所以

\[ gcd = \frac{GCD(a\cdot \frac{LCM(b,\ d)}{b},\ c\cdot \frac{LCM(b,\ d)}{d})}{LCM(b,\ d)} \]

即

\[ gcd = \frac{GCD(a\cdot \frac{d}{GCD(b,\ d)},\ c\cdot \frac{b}{GCD(b,\ d)})}{LCM(b,\ d)} \]

由于

\[ GCD(\frac{d}{GCD(b,\ d)},\ \frac{b}{GCD(b,\ d)}) = 1 \]

所以 \(gcd\) 分子为

\[ GCD(a,\ c) \]

所以

\[ gcd = \frac{GCD(a,\ c)}{LCM(b,\ d)} \]

相应的

\[ lcm = \frac{ac}{bd}\cdot \frac{LCM(b,\ d)}{GCD(a,\ c)} \]

即

\[ lcm = \frac{LCM(a,\ c)}{GCD(b,\ d)} \]

\(code\)

#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int, int>
#define arrayDebug(a, l, r) for(int i = l; i <= r; ++i) printf("%lld%c", a[i], " \n"[i == r])
typedef long long LL;
typedef unsigned long long ULL;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int DX[] = {0, -1, 0, 1, 0, -1, -1, 1, 1};
const int DY[] = {0, 0, 1, 0, -1, -1, 1, 1, -1};
const int MOD = 1e9 + 7;
const int N = 5e4 + 7;
const double PI = acos(-1);
const double EPS = 1e-8;
using namespace std;

int t;
LL a, b, c, d;

LL gcd(LL x, LL y) {return y ? gcd(y, x % y) : x;}

int main()
{
    scanf("%d", &t);
    while(t--) {
        scanf("%lld %lld %lld %lld", &a, &b, &c, &d);
        LL g1 = gcd(a, b);
        LL g2 = gcd(c, d);
        a /= g1, b /= g1;
        c /= g2, d /= g2;
        LL gcdac = gcd(a, c);
        LL lcmbd = b * d / gcd(b, d);
        LL gcdbd = gcd(b, d);
        LL lcmac = a * c / gcd(a, c);
        LL g3 = gcd(gcdac, lcmbd);
        LL g4 = gcd(lcmac, gcdbd);
        printf("%lld/%lld %lld/%lld\n", gcdac / g3, lcmbd / g3, lcmac / g4, gcdbd / g4);
    }
    return 0;
}

下回 \(tmd\) 直接找规律

\(E\)

题意

给定 \(A + B + 1\) 个数，给定 \(A\) 个 \(\&\) 运算符和 \(B\) 个 \(\mid\) 运算符，询问怎样的安排方式，使得从左至右运算的结果最大。

解

考虑分析二进制位。

贪心猜想，必须先连续进行与运算，然后连续进行或运算，这样得到的结果才最大。

一种感性的理解是，即使“或”在前面成功保护 \(1\) ，在后面的“与”中也会丧失这个 \(1\) 。

具体证明不会。

\(code\)

#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int, int>
#define arrayDebug(a, l, r) for(int i = l; i <= r; ++i) printf("%lld%c", a[i], " \n"[i == r])
typedef long long LL;
typedef unsigned long long ULL;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int DX[] = {0, -1, 0, 1, 0, -1, -1, 1, 1};
const int DY[] = {0, 0, 1, 0, -1, -1, 1, 1, -1};
const int MOD = 1e9 + 7;
const int N = 1e5 + 7;
const double PI = acos(-1);
const double EPS = 1e-8;
using namespace std;

int t, A, B;
LL a[N];

int main()
{
    scanf("%d", &t);
    while(t--) {
        scanf("%d %d", &A, &B);
        int n = A + B + 1;
        for(int i = 1; i <= n; ++i)
            scanf("%lld", &a[i]);
        LL ans = a[1];
        for(int i = 2; i <= A + 1; ++i)
            ans &= a[i];
        //cout<<ans<<endl;
        for(int i = A + 2; i <= A + 1 + B; ++i)
            ans |= a[i];
        printf("%lld\n", ans);
    }
    return 0;
}
/*
2
1 1
1 4 5
1
2 2
2 3 11 4 5
*/

不会做就直接找规律

\(K\)

题意

给定一棵 \(n\) 个点的树，定义 \(vulnerable\ road\) 为切断就影响两端点连通性的边，现在允许新建一条路，询问新建一条路之后图上 \(vulnerable\ road\) 的最少数量为？

解

只需要找到一条最长的链，将首尾联通即可，记链上节点共有 \(x\) 个，答案即为 \(n - x\)。

本质上就是求一个树的直径。考虑两次 \(dfs\) ，第一次从任意节点开始找到距离其最远的叶子节点，第二次从该叶子节点开始找到距离其最远的叶子节点。

\(code\)

#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int, int>
#define arrayDebug(a, l, r) for(int i = l; i <= r; ++i) printf("%lld%c", a[i], " \n"[i == r])
typedef long long LL;
typedef unsigned long long ULL;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int DX[] = {0, -1, 0, 1, 0, -1, -1, 1, 1};
const int DY[] = {0, 0, 1, 0, -1, -1, 1, 1, -1};
const int MOD = 1e9 + 7;
const int N = 1e4 + 7;
const double PI = acos(-1);
const double EPS = 1e-8;
using namespace std;

int t, n, vis[N];
vector<int> g[N];

void dfs(int x, int &res, int &root, int step)
{
    if(!vis[x]) {
        vis[x] = 1;
        if(step > res) root = x, res = step;
        for(auto i: g[x])
            dfs(i, res, root, step + 1);
    }
}

void ini(int n)
{
    memset(vis, 0, sizeof(vis));
    for(int i = 1; i <= n; ++i)
        g[i].clear();
}
int main()
{
    scanf("%d", &t);
    while(t--) {
        scanf("%d", &n);
        ini(n);
        for(int i = 1; i < n; ++i) {
            int a, b;
            scanf("%d %d", &a, &b);
            g[a].push_back(b);
            g[b].push_back(a);
        }
        int root = 1, res = 0;
        dfs(root, res, root, 1);
        memset(vis, 0, sizeof(vis));
        dfs(root, res, root, 1);
        printf("%d\n", n - res);
    }
    return 0;
}

模板题，找树的直径就行。