「美团 CodeM 资格赛」试题泛做

LibreOJ真是吼啊！

数码

推个式子，把枚举因数转为枚举倍数。然后就发现它是根号分段的。然后每一段算一下就好了。

#include <cstdio>
#include <cstring>

#define R register
typedef long long ll;
struct Data {
    ll num[10];
    inline void clear()
    {
        memset(num, 0, 10 << 3);
    }
    inline Data operator + (const Data &that) const
    {
        R Data ret; memcpy(ret.num, num, 10 << 3);
        for (R int i = 1; i <= 9; ++i) ret.num[i] += that.num[i];
        return ret;
    }
    inline void operator += (const Data &that)
    {
        for (R int i = 1; i <= 9; ++i) num[i] += that.num[i];
    }
    inline Data operator - (const Data &that) const
    {
        R Data ret; memcpy(ret.num, num, 10 << 3);
        for (R int i = 1; i <= 9; ++i) ret.num[i] -= that.num[i];
        return ret;
    }
    inline void operator *= (const int &that)
    {
        for (R int i = 1; i <= 9; ++i) num[i] *= that;
    }
} ;
inline Data calc2(R int N)
{
    R ll tmp; R Data ret; ret.clear();
    for (tmp = 10; tmp - 1 <= N; tmp *= 10)
        for (R int i = 1; i <= 9; ++i)
            ret.num[i] += tmp / 10;
    tmp /= 10;
    for (R int i = 1; i < (N / tmp); ++i) ret.num[i] += tmp;
    ret.num[N / tmp] += N % tmp + 1;
//    printf("calc2(%d) = 
", N);
//    for (R int i = 1; i <= 9; ++i) printf("%lld
", ret.num[i]);
    return ret;
}
inline Data calc(R int N)
{
    R Data ret; ret.clear();
    for (R int i = 1, j; i <= N; i = j + 1)
    {
        j = N / (N / i);
        R Data tmp = calc2(j) - calc2(i - 1);
        tmp *= N / i;
        ret += tmp;
    }
    return ret;
}
int main()
{
    R int l, r; scanf("%d%d", &l, &r);
    R Data ans = calc(r) - calc(l - 1);
    for (R int i = 1; i <= 9; ++i)
        printf("%lld
", ans.num[i]);
    return 0;
}

跳格子

预处理出每个点能不能到终点。然后直接暴搜就好了。

#include <cstdio>
#include <cstdlib>

#define R register
#define maxn 100010
struct Edge {
    Edge *next;
    int to;
} *last[maxn], e[maxn << 1], *ecnt = e;
inline void link(R int a, R int b)
{
    *++ecnt = (Edge) {last[a], b}; last[a] = ecnt;
}
int q[maxn], n, a[maxn], b[maxn];
bool arv[maxn], ins[maxn];
char st[maxn];
void dfs(R int x, R int step)
{
    if (!arv[x]) return ;
    if (x == n)
    {
        for (R int i = 1; i < step; ++i) printf("%c", st[i]); puts("");
        exit(0);
    }
    if (ins[x])
    {
        puts("Infinity!");
        exit(0);
    }
    ins[x] = 1;
    if (x + a[x] > 0 && x + a[x] <= n)
    {
        st[step] = 'a';
        dfs(x + a[x], step + 1);
    }
    if (x + b[x] > 0 && x + b[x] <= n)
    {
        st[step] = 'b';
        dfs(x + b[x], step + 1);
    }
}
int main()
{
    scanf("%d", &n);
    for (R int i = 1; i <= n; ++i) scanf("%d", a + i), i + a[i] > 0 && i + a[i] <= n ? link(i + a[i], i), 1 : 0;
    for (R int i = 1; i <= n; ++i) scanf("%d", b + i), i + b[i] > 0 && i + b[i] <= n ? link(i + b[i], i), 1 : 0;
    R int head = 0, tail = 1; arv[q[1] = n] = 1;
    while (head < tail)
    {
        R int now = q[++head];
        for (R Edge *iter = last[now]; iter; iter = iter -> next)
            if (!arv[iter -> to]) arv[q[++tail] = iter -> to] = 1;
    }
    if (!arv[1]) {puts("No solution!"); return 0;}
    dfs(1, 1);
    return 0;
}

优惠券

一开始傻逼了，以为只要前缀就好了，后来才发现是区间。。。对于每个不满足的条件的左/右括号扔进一个数据结构里，然后每次遇到问号的时候，去消右端点最近的一个括号。然后这个数据结构用堆就够啦~

#include <cstdio>
#include <vector>
#include <queue>

#define R register
#define maxn 500010
int last[maxn], lastt[maxn];
struct Opt {int type, x;} p[maxn];
std::vector<int> v[maxn];
std::priority_queue<int, std::vector<int>, std::greater<int> > q;
int main()
{
    R int n, num = 0; scanf("%d", &n);
    for (R int i = 1; i <= n; ++i)
    {
        char opt[3]; scanf("%s", opt);
        if (opt[0] == 'I')
        {
            R int x; scanf("%d", &x);
            p[i] = (Opt) {1, x};
        }
        if (opt[0] == 'O')
        {
            R int x; scanf("%d", &x);
            p[i] = (Opt) {0, x};
        }
        if (opt[0] == '?') p[i] = (Opt) {2, 0};
    }
    for (R int i = 1; i <= n; ++i)
    {
        if (p[i].type == 2) continue;
        if (lastt[p[i].x] == p[i].type)
        {
            v[last[p[i].x]].push_back(i);
        }
        last[p[i].x] = i;
        lastt[p[i].x] = p[i].type;
    }
    for (R int i = 0; i < v[0].size(); ++i) q.push(v[0][i]);
    for (R int i = 1; i <= n; ++i)
    {
        for (R int j = 0; j < v[i].size(); ++j) q.push(v[i][j]);
        if (p[i].type == 2 && !q.empty())
        {
            R int top = q.top(); q.pop();
            if (top < i) return !printf("%d
", top);
        }
    }
    if (q.empty()) puts("-1");
    else printf("%d
", q.top());
    return 0;
}