Codeforces Round #613 (Div. 2)

Posted 2021-03-16 jiaaaaaaaqi

• Mezo Playing Zoma
• Just Eat It!
• Fadi and LCM
• Dr. Evil Underscores

Mezo Playing Zoma

[ Time Limit: 1 squad Memory Limit: 256 MB ]
可以到达的最左是 (-L个数)，最右是 (R个数)，所以答案就是相减一下。

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/*************************************************************** 
    > File Name        : a.cpp
    > Author           : Jiaaaaaaaqi
    > Created Time     : 2020/1/10 22:04:03
 ***************************************************************/

#include <bits/stdc++.h>
#define  fi         first
#define  se         second
#define  pb         push_back
#define  pii        pair<int, int>
#define  dbg(x)     cout << #x << " = " << (x) << endl
#define  mes(a, b)  memset(a, b, sizeof a)

using namespace std;
typedef long long int ll;
typedef unsigned long long int ull;
const int    maxn = 1e5 + 10;
const ll     mod  = 1e9 + 7;
const ll     INF  = 1e18 + 100;
const int    inf  = 0x3f3f3f3f;
const double pi   = acos(-1.0);
const double eps  = 1e-8;

int n, m;
int cas, tol, T;

char s[maxn];

int main() {
    // freopen("in", "r", stdin);
    scanf("%d", &n);
    scanf("%s", s+1);
    int x = 0, y = 0;
    for(int i=1; i<=n; i++) {
        x -= s[i]=='L';
        y += s[i]=='R';
    }
    printf("%d
", y-x+1);
    return 0;
}

Just Eat It!

[ Time Limit: 1 squad Memory Limit: 256 MB ]
对于第二个人肯定是抛弃左边一侧或者右边一侧的若干个数字，所以只需要判断是否存在前缀或者后缀小于等于0即可。

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/*************************************************************** 
    > File Name        : b.cpp
    > Author           : Jiaaaaaaaqi
    > Created Time     : 2020/1/10 22:07:25
 ***************************************************************/

#include <bits/stdc++.h>
#define  fi         first
#define  se         second
#define  pb         push_back
#define  pii        pair<int, int>
#define  dbg(x)     cout << #x << " = " << (x) << endl
#define  mes(a, b)  memset(a, b, sizeof a)

using namespace std;
typedef long long int ll;
typedef unsigned long long int ull;
const int    maxn = 1e5 + 10;
const ll     mod  = 1e9 + 7;
const ll     INF  = 1e18 + 100;
const int    inf  = 0x3f3f3f3f;
const double pi   = acos(-1.0);
const double eps  = 1e-8;

int n, m;
int cas, tol, T;

int a[maxn];

bool solve() {
    ll ans = 0;
    for(int i=1; i<=n; i++) {
        ans += a[i];
        if(ans <= 0)    return false;
    }
    ans = 0;
    for(int i=n; i>=1; i--) {
        ans += a[i];
        if(ans <= 0)    return false;
    }
    return true;
}

int main() {
    // freopen("in", "r", stdin);
    scanf("%d", &T);
    while(T--) {
        scanf("%d", &n);
        for(int i=1; i<=n; i++) {
            scanf("%d", &a[i]);
        }
        puts(solve() ? "YES" : "NO");
    }
    return 0;
}

Fadi and LCM

[ Time Limit: 1 squad Memory Limit: 256 MB ]
先质因子分解，然后就得到了若干个数把他们分配到两个数字上，由于数字只会有 (11) 个，因为前 (12) 个素数相乘就超过 (1e12) 了，所以可以直接暴力状压或者暴力 (01) 背包。

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/*************************************************************** 
    > File Name        : c.cpp
    > Author           : Jiaaaaaaaqi
    > Created Time     : 2020/1/10 22:13:23
 ***************************************************************/

#include <bits/stdc++.h>
#define  fi         first
#define  se         second
#define  pb         push_back
#define  pii        pair<int, int>
#define  dbg(x)     cout << #x << " = " << (x) << endl
#define  mes(a, b)  memset(a, b, sizeof a)

using namespace std;
typedef long long int ll;
typedef unsigned long long int ull;
const int    maxn = 1e6 + 10;
const ll     mod  = 1e9 + 7;
const ll     INF  = 1e18 + 100;
const int    inf  = 0x3f3f3f3f;
const double pi   = acos(-1.0);
const double eps  = 1e-8;

ll n, m;
int cas, tol, T;

bool dp[maxn];

int main() {
    // freopen("in", "r", stdin);
    scanf("%lld", &n);
    vector<ll> g;
    for(ll i=2; i*i<=n; i++) {
        if(n%i==0) {
            ll res = 1;
            while(n%i==0) {
                res *= i;
                n /= i;
            }
            g.pb(res);
        }
    }
    if(n!=1)    g.pb(n);
    g.pb(1);
    sort(g.begin(), g.end());
    tol = g.size();
    ll ans = 1;
    for(int i=0; i<tol; i++)    ans *= g[i];
    ll n = 0;
    for(; n*n<=ans; n++);
    mes(dp, 0);
    dp[1] = 1;
    for(int i=0; i<tol; i++) {
        for(ll j=n-1; j>0; j--) {
            if(j%g[i]==0)   dp[j] |= dp[j/g[i]];
        }
    }
    for(ll i=n-1; i>0; i--) {
        if(dp[i])   return 0*printf("%lld %lld
", i, ans/i);
    }
    return 0;
}

Dr. Evil Underscores

[ Time Limit: 1 squad Memory Limit: 256 MB ]
先建 (01) 字典树，然后从高位到低位开始贪心，如果字典树某个节点同时存在 (0) 和 (1) 的边，那么不管 (X) 这一位是 (0) 或者 (1)，总会与一个数字 (xor) 出来后，这一位为 (1)，其他情况都可以使这一位为 (0)，一直贪心到根即可。

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/*************************************************************** 
    > File Name        : d.cpp
    > Author           : Jiaaaaaaaqi
    > Created Time     : 2020/1/10 22:53:46
 ***************************************************************/

#include <bits/stdc++.h>
#define  fi         first
#define  se         second
#define  pb         push_back
#define  pii        pair<int, int>
#define  dbg(x)     cout << #x << " = " << (x) << endl
#define  mes(a, b)  memset(a, b, sizeof a)

using namespace std;
typedef long long int ll;
typedef unsigned long long int ull;
const int    maxn = 1e5 + 10;
const ll     mod  = 1e9 + 7;
const ll     INF  = 1e18 + 100;
const int    inf  = 0x3f3f3f3f;
const double pi   = acos(-1.0);
const double eps  = 1e-8;

int n, m;
int cas, tol, T;

int a[maxn];
int node[maxn*30][2];

void insert(int x) {
    int root = 0;
    for(int i=30; i>=1; i--) {
        int k = x&(1ll<<(i-1)) ? 1 : 0;
        if(node[root][k]==0) {
            mes(node[++tol], 0);
            node[root][k] = tol;
        }
        root = node[root][k];
    }
}

ll dfs(int root, int deep) {
    if(deep == 0)   return 0;
    ll ans = 0;
    if(node[root][0] && node[root][1])  
        ans += (1ll<<(deep-1));
    ll tmp = INF;
    if(node[root][0])   tmp = min(tmp, dfs(node[root][0], deep-1));
    if(node[root][1])   tmp = min(tmp, dfs(node[root][1], deep-1));
    return tmp+ans;
}

int main() {
    // freopen("in", "r", stdin);
    tol = 0;
    scanf("%d", &n);
    for(int i=1; i<=n; i++) {
        scanf("%d", &a[i]);
        insert(a[i]);
    }
    ll ans = dfs(0, 30);
    printf("%lld
", ans);
    return 0;
}