「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。
1 #include <cstdio> 2 #include <cstring> 3 4 #define R register 5 typedef long long ll; 6 struct Data { 7 ll num[10]; 8 inline void clear() 9 { 10 memset(num, 0, 10 << 3); 11 } 12 inline Data operator + (const Data &that) const 13 { 14 R Data ret; memcpy(ret.num, num, 10 << 3); 15 for (R int i = 1; i <= 9; ++i) ret.num[i] += that.num[i]; 16 return ret; 17 } 18 inline void operator += (const Data &that) 19 { 20 for (R int i = 1; i <= 9; ++i) num[i] += that.num[i]; 21 } 22 inline Data operator - (const Data &that) const 23 { 24 R Data ret; memcpy(ret.num, num, 10 << 3); 25 for (R int i = 1; i <= 9; ++i) ret.num[i] -= that.num[i]; 26 return ret; 27 } 28 inline void operator *= (const int &that) 29 { 30 for (R int i = 1; i <= 9; ++i) num[i] *= that; 31 } 32 } ; 33 inline Data calc2(R int N) 34 { 35 R ll tmp; R Data ret; ret.clear(); 36 for (tmp = 10; tmp - 1 <= N; tmp *= 10) 37 for (R int i = 1; i <= 9; ++i) 38 ret.num[i] += tmp / 10; 39 tmp /= 10; 40 for (R int i = 1; i < (N / tmp); ++i) ret.num[i] += tmp; 41 ret.num[N / tmp] += N % tmp + 1; 42 // printf("calc2(%d) = \n", N); 43 // for (R int i = 1; i <= 9; ++i) printf("%lld\n", ret.num[i]); 44 return ret; 45 } 46 inline Data calc(R int N) 47 { 48 R Data ret; ret.clear(); 49 for (R int i = 1, j; i <= N; i = j + 1) 50 { 51 j = N / (N / i); 52 R Data tmp = calc2(j) - calc2(i - 1); 53 tmp *= N / i; 54 ret += tmp; 55 } 56 return ret; 57 } 58 int main() 59 { 60 R int l, r; scanf("%d%d", &l, &r); 61 R Data ans = calc(r) - calc(l - 1); 62 for (R int i = 1; i <= 9; ++i) 63 printf("%lld\n", ans.num[i]); 64 return 0; 65 }
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。
1 #include <cstdio> 2 #include <cstdlib> 3 4 #define R register 5 #define maxn 100010 6 struct Edge { 7 Edge *next; 8 int to; 9 } *last[maxn], e[maxn << 1], *ecnt = e; 10 inline void link(R int a, R int b) 11 { 12 *++ecnt = (Edge) {last[a], b}; last[a] = ecnt; 13 } 14 int q[maxn], n, a[maxn], b[maxn]; 15 bool arv[maxn], ins[maxn]; 16 char st[maxn]; 17 void dfs(R int x, R int step) 18 { 19 if (!arv[x]) return ; 20 if (x == n) 21 { 22 for (R int i = 1; i < step; ++i) printf("%c", st[i]); puts(""); 23 exit(0); 24 } 25 if (ins[x]) 26 { 27 puts("Infinity!"); 28 exit(0); 29 } 30 ins[x] = 1; 31 if (x + a[x] > 0 && x + a[x] <= n) 32 { 33 st[step] = 'a'; 34 dfs(x + a[x], step + 1); 35 } 36 if (x + b[x] > 0 && x + b[x] <= n) 37 { 38 st[step] = 'b'; 39 dfs(x + b[x], step + 1); 40 } 41 } 42 int main() 43 { 44 scanf("%d", &n); 45 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; 46 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; 47 R int head = 0, tail = 1; arv[q[1] = n] = 1; 48 while (head < tail) 49 { 50 R int now = q[++head]; 51 for (R Edge *iter = last[now]; iter; iter = iter -> next) 52 if (!arv[iter -> to]) arv[q[++tail] = iter -> to] = 1; 53 } 54 if (!arv[1]) {puts("No solution!"); return 0;} 55 dfs(1, 1); 56 return 0; 57 }
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~
1 #include <cstdio> 2 #include <vector> 3 #include <queue> 4 5 #define R register 6 #define maxn 500010 7 int last[maxn], lastt[maxn]; 8 struct Opt {int type, x;} p[maxn]; 9 std::vector<int> v[maxn]; 10 std::priority_queue<int, std::vector<int>, std::greater<int> > q; 11 int main() 12 { 13 R int n, num = 0; scanf("%d", &n); 14 for (R int i = 1; i <= n; ++i) 15 { 16 char opt[3]; scanf("%s", opt); 17 if (opt[0] == 'I') 18 { 19 R int x; scanf("%d", &x); 20 p[i] = (Opt) {1, x}; 21 } 22 if (opt[0] == 'O') 23 { 24 R int x; scanf("%d", &x); 25 p[i] = (Opt) {0, x}; 26 } 27 if (opt[0] == '?') p[i] = (Opt) {2, 0}; 28 } 29 for (R int i = 1; i <= n; ++i) 30 { 31 if (p[i].type == 2) continue; 32 if (lastt[p[i].x] == p[i].type) 33 { 34 v[last[p[i].x]].push_back(i); 35 } 36 last[p[i].x] = i; 37 lastt[p[i].x] = p[i].type; 38 } 39 for (R int i = 0; i < v[0].size(); ++i) q.push(v[0][i]); 40 for (R int i = 1; i <= n; ++i) 41 { 42 for (R int j = 0; j < v[i].size(); ++j) q.push(v[i][j]); 43 if (p[i].type == 2 && !q.empty()) 44 { 45 R int top = q.top(); q.pop(); 46 if (top < i) return !printf("%d\n", top); 47 } 48 } 49 if (q.empty()) puts("-1"); 50 else printf("%d\n", q.top()); 51 return 0; 52 }
