Codeforces round #265 (Div. 2)

Codeforces round #265 (Div. 2)

Question Link

A: Just change the number and compare the differences.

B: Two or more consecutive 0 values are treated as one operation. The 0 at the beginning and the 0 at the end can be ignored.

C: greedy structure from the end. Because it is ensured that the first part is a back-to-text, it is only possible that the Back-to-text may be 2 or 3 after modification. In this way, the complexity of judgment is very small.

D: brute-force enumeration, and then judge

E: calculate the number of digits and corresponding numbers corresponding to each number and calculate the answer once.

Code:

A:

#include <cstdio>#include <cstring>int n, num[105], sb[105];char s[105];int main() {    scanf("%d", &n);    scanf("%s", s);    for (int i = 0; i < n; i++) {        num[i] = s[i] - '0';        sb[i] = num[i];    }    num[0]++;    for (int i = 0; i < n; i++) {        if (num[i] == 2) {            num[i] = 0;            num[i + 1]++;        }    }    int ans = 0;    for (int i = 0; i < n; i++)        if (sb[i] != num[i])            ans++;    printf("%d\n", ans);    return 0;}

B:

#include <cstdio>#include <cstring>const int N = 1005;int n, num[N];int solve() {int ans = 0;int s = 0, e = n - 1;while (num[s] == 0 && s < n) s++;while (num[e] == 0 && e >= 0) e--;for (int i = s ; i <= e; i++) {if (num[i] == num[i - 1] && num[i] == 0)continue;ans++;}return ans;}int main() {scanf("%d", &n);for (int i = 0; i < n; i++)scanf("%d", &num[i]);printf("%d\n", solve());return 0;}

C:

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int N = 1005;int n, p;char str[N];bool solve(int u) {if (u < 0) return false;str[u]++;if (str[u] - 'a' == p) {if (solve(u - 1)) {str[u] = 'a' - 1;return solve(u);}} else {if (u - 1 >= 0 && str[u] == str[u - 1]) return solve(u);if (u - 2 >= 0 && str[u] == str[u - 2]) return solve(u);return true;}}int main() {scanf("%d%d", &n, &p);scanf("%s", str);if (solve(n - 1)) printf("%s\n", str);else printf("NO\n");return 0;}

D:

#include <cstdio>#include <cstring>#include <cmath>#include <algorithm>#include <iostream>#include <set>using namespace std;struct Point {int v[3];void read() {for (int i = 0; i < 3; i++)scanf("%d", &v[i]);}bool operator == (const Point& c) const {return (v[0] == c.v[0] && v[1] == c.v[1] && v[2] == c.v[2]);}} p[10];typedef long long ll;ll dis(Point a, Point b) {ll dx = a.v[0] - b.v[0];ll dy = a.v[1] - b.v[1];ll dz = a.v[2] - b.v[2];return dx * dx + dy * dy + dz * dz;}ll d[10];bool judge() {for (int i = 1; i < 8; i++)for (int j = 0; j < i; j++) {if (p[i] == p[j]) return false;}d[0] = 0;for (int i = 1; i < 8; i++)d[i] = dis(p[0], p[i]);sort(d, d + 8);ll a = d[1], b = d[4], c = d[7];if (a != d[2] || a != d[3] || d[2] != d[3]) return false;if (b != d[5] || b != d[6] || d[5] != d[6]) return false;if (2 * a != b) return false;if (a + b != c) return false;return true;}bool dfs(int u) {if (u == 8) {if (judge())return true;return false;}sort(p[u].v, p[u].v + 3);do {if (dfs(u + 1)) return true;} while (next_permutation(p[u].v, p[u].v + 3));return false;}int main() {for (int i = 0; i < 8; i++)p[i].read();if (!dfs(0)) printf("NO\n");else {printf("YES\n");for (int i = 0; i < 8; i++)printf("%d %d %d\n", p[i].v[0], p[i].v[1], p[i].v[2]);}return 0;}

E:

#include <cstdio>#include <cstring>#include <cmath>#include <vector>using namespace std;typedef long long ll;const ll MOD = 1000000007;const int N = 100005;char str[N], sss[N];int n;ll pow_mod(ll x, ll k) {    ll ans = 1;    while (k) {        if (k&1) ans = (ans * x) % MOD;        x = x * x % MOD;        k >>= 1;    }    return ans;}ll v[15], l[15];ll idx(char c) {    return c - '0';}struct State {    ll u;    vector<ll> v;    void init(char *str) {        int len = strlen(str);        u = idx(str[0]);        v.clear();        for (int i = 3; i < len; i++)            v.push_back(idx(str[i]));    }} s[N];int main() {    scanf("%s", str);    scanf("%d", &n);    for (int i = 0; i < n; i++) {        scanf("%s", sss);        s[i].init(sss);    }    for (int i = 0; i < 10; i++) {        v[i] = i;        l[i] = 1;    }    for (int i = n - 1; i >= 0; i--) {        ll lt = 0, vt = 0;        for (int j = 0; j < s[i].v.size(); j++) {            ll nu = s[i].v[j];            vt = (vt * pow_mod(10, l[nu]) % MOD + v[nu]) % MOD;            lt = (lt + l[nu]) % (MOD - 1);        }        v[s[i].u] = vt;        l[s[i].u] = lt;    }    ll ans = 0;    for (int i = 0; i < strlen(str); i++) {        ll nu = idx(str[i]);        ans = (ans * pow_mod(10, l[nu]) % MOD + v[nu]) % MOD;    }    printf("%lld\n", ans);    return 0;}

Codeforces round #265 (Div. 2)