Codeforces Round #265 (Div. 2)_html/css_WEB-ITnose

Codeforces Round #265 (Div. 2)

Question link

A: Just transform the numbers and compare the differences

B: More than 2 consecutive 0s As an operation, the 0 at the beginning and the 0 at the end can be ignored

C: Greedy construction from the end. Since it is guaranteed to be a palindrome at the beginning, it is guaranteed that the length of the palindrome after modification can only be 2 or 3. Yes, in this case the judgment complexity will be very small

D: Violently enumerate the situation, and then judge

E: Reverse the operation to process the number of digits and corresponding digits corresponding to each number , and then calculate the answer in for

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  <br> B: <p></p> <p> </p> <pre name="code" class="sycode">#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 = 0) e--;	for (int i = s ; i  <br> C: <p></p> <p> </p> <pre name="code" class="sycode">#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 && 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;}</algorithm></cstring></cstdio>

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  <br> E: <p></p> <p> </p> <pre name="code" class="sycode">#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 = 0; i--) {        ll lt = 0, vt = 0;        for (int j = 0; j  <br> <br> <p></p> </ll></vector></cmath></cstring></cstdio>