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

题目链接

A：把数字变换后比较一下几个不一样即可

B：连续2个以上0当作一次操作，开头的0和结尾的0可以忽略

C：贪心从末尾去构造，由于保证一开始是回文，所以保证修改后出现回文只可能为长度2或3的，这样的话判断复杂度就很小了

D：暴力枚举情况，然后判断

E：把操作逆过来处理出每个数字对应的位数和相应数字，然后在for一遍计算答案即可

代码：

A：

#include #include 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;}

#include #include 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;}

#include #include #include 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;}

#include #include #include #include #include #include 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;}

#include #include #include #include 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 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;}