9.16 腾讯笔试编程题ac代码

思路: 求lcm(1, 2, …, n, …, m) = lcm(n + 1, …, m)
根据算数基本定理

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
// #include 
#include 

using namespace std;

#define LL long long
#define pb push_back
#define mk make_pair
#define pill pair
#define fi first
#define se second
#define mst(a, b) memset(a, b, sizeof a)
#define lson (rt << 1)
#define rson ((rt << 1) | 1)

const int qq = 1e6 + 10;
const int INF = 1e9 + 10;
const int MOD = 1e9 + 7;

LL gcd(LL a, LL b) {
    return b == 0 ? a : gcd(b, a % b);
}
LL lcm(LL a, LL b) {
    return a / gcd(a, b) * b;
}
int n;

int vis[qq];


int main() {
    #ifdef ONLINE_JUDGE
    #else
        freopen("in.txt", "r", stdin);
    #endif
    scanf("%d", &n);
    if (n == 1) {
        printf("2\n");
        return 0;
    }
    mst(vis, 0);
    for (int i = 2; i <= n; ++i) {
        int x = i;
        for (int j = 2; j * j <= x; ++j) {
            if (x % j == 0) {
                int cnt = 1;
                while (x % j == 0) {
                    cnt *= j;
                    x /= j;
                }
                vis[j] = max(vis[j], cnt);
            }
        }
        if (x > 1) {
            vis[x] = max(vis[x], x);
        }
    }
    int ans = -INF;
    for (int i = 2; i <= n; ++i) {
        if (!vis[i])    continue;
        int x = vis[i];
        for (int j = 2; ; j++) {
            if (x * j > n) {
                x = x * j;
                break;
            }
        }
        ans = max(ans, x);
    }
    printf("%d\n", ans);
    return 0;
}

思路: dfs每一个点走一遍

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
// #include 
#include 

using namespace std;

#define LL long long
#define pb push_back
#define mk make_pair
#define pill pair
#define fi first
#define se second
#define mst(a, b) memset(a, b, sizeof a)
#define lson (rt << 1)
#define rson ((rt << 1) | 1)

const int qq = 1e3 + 10;
const int INF = 1e9 + 10;
const int MOD = 1e9 + 7;
vector<int> vt[qq];
int n, m;
bool vis[qq];
int to[qq], from[qq];
void dfs(int u) {
    for (int i = 0; i < vt[u].size(); ++i) {
        int v = vt[u][i];
        if (!vis[v]) {
            vis[v] = true;
            from[v]++;
            dfs(v);
        }
    }
}
mapbool> mp;

int main() {
    #ifdef ONLINE_JUDGE
    #else
        freopen("in.txt", "r", stdin);
    #endif
    mst(to, 0);
    mst(from, 0);
    scanf("%d%d", &n, &m);
    for (int i = 0, a, b; i < m; ++i) {
        scanf("%d%d", &a, &b);
        if (a == b) continue;
        if (mp.find(mk(a, b)) == mp.end()) {
            vt[a].pb(b);
            mp[mk(a, b)] = true;
        }
    }
    for (int i = 1; i <= n; ++i) {
        mst(vis, false);
        vis[i] = true;
        dfs(i);
        int cnt = 0;
        for (int j = 1; j <= n; ++j) {
            if (vis[j]) cnt++;
        }
        to[i] = cnt - 1;
    }
    int cnt = 0;
    for (int i = 1; i <= n; ++i) {
        if (to[i] < from[i]) {
            cnt++;
        }
    }
    printf("%d\n", cnt);
    return 0;
}

思路: 这题怎么做都可以
A%B, 2A%B, 3A%B …. 因为B <= 100 , 根据抽屉原理, 再加上A的倍数关系, 这里一定会有循环节, 剩下的就是求一下背包, 看看成组合哪些数

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
// #include 
#include 

using namespace std;

#define LL long long
#define pb push_back
#define mk make_pair
#define pill pair
#define fi first
#define se second
#define mst(a, b) memset(a, b, sizeof a)
#define lson (rt << 1)
#define rson ((rt << 1) | 1)

const int qq = 1e5 + 10;
const int INF = 1e9 + 10;
const int MOD = 1e9 + 7;
map<int, bool> mp;
vector<int> vt;
vector<int> dp;
bool vis[105];

int main() {
    #ifdef ONLINE_JUDGE
    #else
        freopen("in.txt", "r", stdin);
    #endif
    int t;  scanf("%d", &t);
    int a, b, c;
    while (t--) {
        mp.clear();
        dp.clear();
        vt.clear();
        scanf("%d%d%d", &a, &b, &c);
        for (int i = 1; ; ++i) {
            if (mp.find((a * i) % b) != mp.end()) {
                break;
            }
            mp[(a * i) % b] = true;
            vt.pb((a * i) % b);
        }
        dp.pb(0);
        mst(vis, false);
        bool f = false;
        for (int i = 0; i < vt.size() && !f; ++i) {
            int len = dp.size();
            for (int j = 0; j < len && !f; ++j) {
                int tmp = (dp[j] + vt[i]) % b;
                if (vis[tmp])   continue;
                vis[tmp] = true;
                dp.pb(tmp);
                if (tmp == c)   f = true;
            }
        }
        printf("%s\n", f ? "YES" : "NO");
    }
    return 0;
}