「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[];
     inline void clear()
     {
         memset(num, ,  << );
     }
     inline Data operator + (const Data &that) const
     {
         R Data ret; memcpy(ret.num, num,  << );
         for (R int i = ; i <= ; ++i) ret.num[i] += that.num[i];
         return ret;
     }
     inline void operator += (const Data &that)
     {
         for (R int i = ; i <= ; ++i) num[i] += that.num[i];
     }
     inline Data operator - (const Data &that) const
     {
         R Data ret; memcpy(ret.num, num,  << );
         for (R int i = ; i <= ; ++i) ret.num[i] -= that.num[i];
         return ret;
     }
     inline void operator *= (const int &that)
     {
         for (R int i = ; i <= ; ++i) num[i] *= that;
     }
 } ;
 inline Data calc2(R int N)
 {
     R ll tmp; R Data ret; ret.clear();
     for (tmp = ; tmp -  <= N; tmp *= )
         for (R int i = ; i <= ; ++i)
             ret.num[i] += tmp / ;
     tmp /= ;
     for (R int i = ; i < (N / tmp); ++i) ret.num[i] += tmp;
     ret.num[N / tmp] += N % tmp + ;
 //    printf("calc2(%d) = \n", N);
 //    for (R int i = 1; i <= 9; ++i) printf("%lld\n", ret.num[i]);
     return ret;
 }
 inline Data calc(R int N)
 {
     R Data ret; ret.clear();
     for (R int i = , j; i <= N; i = j + )
     {
         j = N / (N / i);
         R Data tmp = calc2(j) - calc2(i - );
         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 - );
     for (R int i = ; i <= ; ++i)
         printf("%lld\n", ans.num[i]);
     return ;
 }

数码

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 << ], *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 = ; i < step; ++i) printf("%c", st[i]); puts("");
         exit();
     }
     if (ins[x])
     {
         puts("Infinity!");
         exit();
     }
     ins[x] = ;
     if (x + a[x] >  && x + a[x] <= n)
     {
         st[step] = 'a';
         dfs(x + a[x], step + );
     }
     if (x + b[x] >  && x + b[x] <= n)
     {
         st[step] = 'b';
         dfs(x + b[x], step + );
     }
 }
 int main()
 {
     scanf("%d", &n);
     for (R int i = ; i <= n; ++i) scanf("%d", a + i), i + a[i] >  && i + a[i] <= n ? link(i + a[i], i),  : ;
     for (R int i = ; i <= n; ++i) scanf("%d", b + i), i + b[i] >  && i + b[i] <= n ? link(i + b[i], i),  : ;
     R int head = , tail = ; arv[q[] = n] = ;
     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] = ;
     }
     if (!arv[]) {puts("No solution!"); return ;}
     dfs(, );
     return ;
 }

跳格子

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 = ; scanf("%d", &n);
     for (R int i = ; i <= n; ++i)
     {
         char opt[]; scanf("%s", opt);
         if (opt[] == 'I')
         {
             R int x; scanf("%d", &x);
             p[i] = (Opt) {, x};
         }
         if (opt[] == 'O')
         {
             R int x; scanf("%d", &x);
             p[i] = (Opt) {, x};
         }
         if (opt[] == '?') p[i] = (Opt) {, };
     }
     for (R int i = ; i <= n; ++i)
     {
         if (p[i].type == ) 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 = ; i < v[].size(); ++i) q.push(v[][i]);
     for (R int i = ; i <= n; ++i)
     {
         for (R int j = ; j < v[i].size(); ++j) q.push(v[i][j]);
         if (p[i].type ==  && !q.empty())
         {
             R int top = q.top(); q.pop();
             if (top < i) return !printf("%d\n", top);
         }
     }
     if (q.empty()) puts("-1");
     else printf("%d\n", q.top());
     return ;
 }

优惠劵