A Star

#include <cstdio>
using namespace std;

int n;

int main()
{
    scanf("%d", &n);
    printf("%d\n", 100 - n % 100);
    return 0;
}

B uNrEaDaBlE sTrInG

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <cctype>
using namespace std;

#define IOS ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)

string str;

bool check()
{
    for(int i = 0; i < (int)str.size(); ++ i)
        if(i % 2 && islower(str[i])) return 0;
        else if(i % 2 == 0 && isupper(str[i])) return 0;
    return 1;
}

int main()
{
    IOS;
    cin >> str;
    if(check()) puts("Yes");
    else puts("No");
    return 0;
}

C Kaprekar Number

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

typedef long long LL;
#define IOS ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)

LL n, k;

LL f(LL x)
{
    vector<int> g1, g2;
    while(x) g1.push_back(x % 10), g2.push_back(x % 10), x /= 10;
    sort(g1.begin(), g1.end(), greater<int>());
    sort(g2.begin(), g2.end());
    LL res = 0;
    for(int i = 0; i < (int)g1.size(); ++ i)
        res = res * 10 + g1[i] - g2[i];
    return res;
}

int main()
{
    IOS;
    cin >> n >> k;
    LL a0 = n;
    for(int i = 1; i <= k; ++ i) a0 = f(a0);
    cout << a0 << endl;
    return 0;
}

D Base n

特判 \(X\) 只有一位的情况，无论基底是几，得到的数都是原数，判断原数和 \(M\) 的大小关系即可

对于其他情况，当基底越大，得到的值也就越大，具有单调性，可以二分基底

计算以 \(mid\) 为基底的数与 \(M\) 的大小关系时，直接计算答案会爆 \(long \; long\), 注意在过程中判断

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

typedef unsigned long long LL;
#define IOS ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)

LL m;
string s;

int find()
{
    int res = 0;
    for(auto x : s) res = max(res, x - '0');
    return res;
}

bool check(LL mid)
{
    LL res = 0;
    for(auto x : s)
    {
        if(res > m / mid) return 0;
        res = res * mid + x - '0';
        if(res > m) return 0;
    }
    return 1;
}

int main()
{
    IOS;
    cin >> s >> m;
    int d = find();
    if(s.size() == 1)
    {
        if((LL)d <= m) cout << "1" << endl;
        else cout << "0" << endl;
        return 0;
    }
    LL l = d + 1, res = 0, r = m;
    while(l <= r)
    {
        LL mid = (l + r) >> 1;
        if(check(mid))
        {
            res = mid;
            l = mid + 1;
        }
        else r = mid - 1;
    }
    if(res) cout << res - d << endl;
    else cout << "0" << endl;
    return 0;
}

E Train

从 \(u\) 到达 \(v\) 的时间为 \(\left\lceil\ \dfrac{d[u]}{K_i}\right\rceil * K_i + T_i\)， 建双向边跑 \(dijkstra\) 即可

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
using namespace std;

typedef long long LL;
const int M = 2e5 + 10;
const int N = 1e5 + 10;
const LL INF = 1e18;

int n, m, S, T;
LL d[N];
struct Edge
{
    int to, nxt, T, K;
}line[M];
int fist[N], idx;
bool st[N];
struct zt
{
    int from;
    LL d;
};

bool operator < (zt a, zt b)
{
    return a.d > b.d;
}

void add(int x, int y, int z, int w)
{
    line[idx] = {y, fist[x], z, w};
    fist[x] = idx ++;
}

void heap_dijkstra()
{
    for(int i = 1; i <= n; ++ i) d[i] = INF;
    priority_queue<zt> q;
    q.push({S, 0}), d[S] = 0;
    while(!q.empty())
    {
        zt u = q.top(); q.pop();
        if(st[u.from]) continue;
        st[u.from] = 1;
        for(int i = fist[u.from]; ~i; i = line[i].nxt)
        {
            int v = line[i].to;
            LL tim = (d[u.from] + line[i].K - 1) / line[i].K * line[i].K + line[i].T;
            if(d[v] > tim)
            {
                d[v] = tim;
                q.push({v, d[v]});
            }
        }
    }
}

int main()
{
    scanf("%d%d%d%d", &n, &m, &S, &T);
    memset(fist, -1, sizeof fist);
    for(int i = 1; i <= m; ++ i)
    {
        int a, b, c, d;
        scanf("%d%d%d%d", &a, &b, &c, &d);
        add(a, b, c, d);
        add(b, a, c, d);
    }
    heap_dijkstra();
    if(d[T] < INF) printf("%lld\n", d[T]);
    else puts("-1");
    return 0;
}

F Potion

令 \(\sum A_i\) 为从 \(n\) 个元素中选出 \(p\) 个元素的权值和

目标是使在满足\(p | (X - \sum A_i)\) 条件下最小化 \(\dfrac{X - \sum A_i}{p}\)

故对于每个 \(p\) ，求出在 \(\sum A_i\) mod \(p\) == \(X\) mod \(p\) 的条件下 \(\sum A_i\) 的最大值即可

问题转化为求 \(i\) 个数中取出 \(j\) 个且和的余数为 \(k\) 的最大和

令\(f[i][j][k]\) 为前 \(i\) 个数中选择 \(j\) 个且和的余数为 \(k\) 的最大和

\(f[i][j][k] = max(f[i - 1][j][k], f[i - 1][j - 1][(k - a_i) \% p] + a_i)\)

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
using namespace std;

#define IOS ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)
typedef long long LL;
const int N = 100 + 20;
const LL INF = 1e18;

LL f[N][N][N], m;
int n, a[N];

int main()
{
    IOS;
    cin >> n >> m;
    for(int i = 1; i <= n; ++ i) cin >> a[i];
    LL ans = INF;
    for(int p = 1; p <= n; ++ p)
    {
        memset(f, -0x3f, sizeof f);
        f[0][0][0] = 0;
        for(int i = 1; i <= n; ++ i)
        {
            f[i][0][0] = 0;
            for(int j = 1; j <= p; ++ j)
                for(int k = 0; k < p; ++ k)
                    f[i][j][k] = max(f[i - 1][j][k], f[i - 1][j - 1][((k - a[i]) % p + p) % p] + a[i]);
        }
        if(f[n][p][m % p] >= 0) ans = min(ans, (m - f[n][p][m % p]) / p);
    }
    cout << ans << endl;
    return 0;
}