Zoj monthly, January 2015 (B, E, G, H)

Question B:

First, the existing card is processed, and then the DFS is run. The DFS has a large pruning function, that is, if the current position is smaller than the Lexicographic Order, you do not need to continue to release it later, you can directly use the combined mathematics for computation. If it is greater than, you don't have to consider it. If it is equal to, you can continue to search later. In this way, the time is acceptable as long as you search on a path with the same Lexicographic Order.

Question E: It can be simulated. There is no solution.

Question G: first, all the numbers are GCD. If it is not 1, there is no solution. If it is 1, the answer is constructed as long as the first number is used, you can perform a GCD for each other number. The first number is 1, and then the first number and the subsequent number are used as GCD. The total number is 1 and N-2 times.

Question H: there is a big pitfall in this question. As long as the string is satisfied, it can be a sub-string.-I feel that the question is not so clear, therefore, you only need to construct an existing string into a dictionary tree, and then DP [x] [y] [u] indicates that the current position is at the position of X and Y, in the dictionary tree, the minimum step tree required for the u state is followed by a short circuit. When you encounter a word node, you can record the minimum value of the answer.

Code:

B:

#include <cstdio>#include <cstring>typedef long long ll;const int MOD = 1000000007;char str[105];int st[52], have[14], stn, hn, C[105][105];int cal(int tot) {    int ans = 1;    for (int i = 1; i <= 13; i++) {if (have[i]) {    ans = (int)((ll)ans * C[tot][have[i]] % MOD);    tot -= have[i];}    }    return ans;}int dfs(int u) {    int ans = 0;    if (u == stn) return 0;    if (hn - u == 0) return 1;    for (int i = 1; i <= st[u]; i++) {if (!have[i]) continue;have[i]--;if (i == st[u]) {    ans = (ans + dfs(u + 1)) % MOD;}else if (i < st[u]) {    ans = (ans + cal(hn - u - 1)) % MOD;}have[i]++;    }    return ans;}int main() {    for (int i = 0; i < 105; i++) {C[i][0] = C[i][i] = 1;for (int j = 1; j < i; j++) {    C[i][j] = (C[i - 1][j] + C[i - 1][j - 1]) % MOD;}    }    while (gets(str) != NULL) {stn = 0;hn = 0;int len = strlen(str);memset(have, 0, sizeof(have));for (int i = 0; i < len; i++) {    if (str[i] == 'A') st[stn++] = 1;    else if (str[i] == '1') {st[stn++] = 10;i++;    }    else if (str[i] >= '2' && str[i] <= '9')st[stn++] = str[i] - '2' + 2;    else if (str[i] == 'J') st[stn++] = 11;    else if (str[i] == 'Q') st[stn++] = 12;    else if (str[i] == 'K') st[stn++] = 13;    have[st[stn - 1]]++;}for (int i = 1; i <= 13; i++) {    have[i] = 4 - have[i];    hn += have[i];}printf("%d\n", dfs(0));    }    return 0;}

E:

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int N = 15;const int INF = 0x3f3f3f3f;int t, n, a[N];void gao() {    while (1) {int Min = INF, Max = -INF;int Minv, Maxv;for (int i = 0; i < n; i++) {    if (Min > a[i]) {Min = a[i];Minv = i;    }    if (Max < a[i]) {Max = a[i];Maxv = i;    }}if (Min == Max) break;a[Minv] = a[Maxv] = Max - Min;    }}int main() {    scanf("%d", &t);    while (t--) {scanf("%d", &n);for (int i = 0; i < n; i++)    scanf("%d", &a[i]);gao();printf("%d\n", a[0]);    }    return 0;}

G:

#include <cstdio>#include <cstring>const int N = 100005;int gcd(int a, int b) {    if (!b) return a;    return gcd(b, a % b);}int n, a;int main() {    int cas = 0;    while (~scanf("%d", &n)) {int s = 0;for (int i = 0; i < n; i++) {    scanf("%d", &a);    s = gcd(s, a);}printf("Case %d: ", ++cas);if (s != 1) {    printf("-1\n");} else {    printf("%d\n", 2 * n - 2);    for (int i = 2; i <= n; i++)printf("1 %d\n", i);    for (int i = 2; i <= n; i++)printf("1 %d\n", i);}printf("\n");    }    return 0;}

H:

#include <cstdio>#include <cstring>#include <algorithm>#include <queue>using namespace std;const int N = 25;const int MAXN = 20005;const int INF = 0x3f3f3f3f;int t, n, m;char str[N][N], s[105];int ch[MAXN][26], sz, dp[N][N][MAXN];bool val[MAXN];void init() {    sz = 0;    memset(val, 0, sizeof(val));    memset(ch[0], 0, sizeof(ch[0]));}int idx(char c) {    return c - 'A';}void insert(char *str) {    int len = strlen(str);    int u = 0;    for (int i = 0; i < len; i++) {int c = idx(str[i]);if (!ch[u][c]) {    ch[u][c] = ++sz;    memset(ch[sz], 0, sizeof(ch[sz]));}u = ch[u][c];    }    val[u] = true;}struct State {    int x, y, u, val;    State() {}    State(int x, int y, int u, int val) {this->x = x;this->y = y;this->u = u;this->val = val;    }    bool operator < (const State& c) const {return val > c.val;    }};const int d[4][2] = {0, 1, 1, 0, 0, -1, -1, 0};int main() {    scanf("%d", &t);    while (t--) {priority_queue<State> Q;scanf("%d%d", &n, &m);memset(dp, INF, sizeof(dp));for (int i = 0; i < n; i++) {    scanf("%s", str[i]);    for (int j = 0; j < m; j++) {if (str[i][j] == '@') {    Q.push(State(i, j, 0, 0));    dp[i][j][0] = 0;}    }}int q;int ans = INF;scanf("%d", &q);init();while (q--) {    scanf("%s", s);    insert(s);}while (!Q.empty()) {    State now = Q.top();    Q.pop();    if (val[now.u])ans = min(ans, now.val);    if (now.val >= ans) continue;    for (int i = 0; i < 4; i++) {int x = now.x + d[i][0];int y = now.y + d[i][1];if (x < 0 || x >= n || y < 0 || y >= m || str[x][y] == '#') continue;if (dp[x][y][0] > now.val + 1) {    dp[x][y][0] = now.val + 1;    Q.push(State(x, y, 0, now.val + 1));}if (str[x][y] >= 'A' && str[x][y] <= 'Z') {    int u = now.u;    int c = idx(str[x][y]);    while (1) {u = ch[u][c];if (u == 0) break;if (dp[x][y][u] > now.val + 1) {    dp[x][y][u] = now.val + 1;    Q.push(State(x, y, u, now.val + 1));}    }} else {    int u = now.u;    if (dp[x][y][u] > now.val + 1) {dp[x][y][u] = now.val + 1;Q.push(State(x, y, u, now.val + 1));    }}    }}if (ans == INF) ans = -1;printf("%d\n", ans);    }    return 0;}

Zoj monthly, January 2015 (B, E, G, H)