Codeforces Round #262 (Div. 2 ),

Codeforces Round #262 (Div. 2 ),
Codeforces Round #262 (Div. 2)

A: You can directly simulate the problem.

B: Since the size of s (x) is as large as 1e9, the number and the maximum value are 81. In this way, if we enumerate s (x), we only need to enumerate 1 to 81, then, after calculating the x value, determine whether the x value is correct. If yes, add the answer.

C: two points of height. Then, when you judge it, you need to water it at the wrong position and push it from left to right.

D: constructor. If k> = 5, you can directly put the even, odd, even, and odd numbers. Because the even and even numbers + 1 are different or must be 1, so let's put the four XOR values to the smallest zero, and then k = 1, 2, 4 can be identified. The key lies in how to construct k = 3, in fact, as long as the highest bit of l is found, write down this number as p. If 3 p does not exceed r, it can be constructed. You can select l, l + p, and 3 p to construct 0, if not, select two exclusive or sum. 1 is the best choice.

Code:

A:

#include <cstdio>#include <cstring>#include <cstdlib>int n, m;int main() {int ans = 0;scanf("%d%d", &n, &m);while (n >= m) {ans += n - n % m;int num = n / m;n = n % m + num; } printf("%d\n", ans + n); return 0;}

B:

#include <cstdio>#include <cstring>#include <cstdlib>#include <algorithm>#include <set>#include <iostream>using namespace std;typedef long long ll;ll a, b, c;set<ll> ans;ll out[105], on = 0;ll pow1(ll x, ll k) {ll ans = 1;for (ll i = 0; i < k; i++)ans *= x;return ans;}bool judge(ll num, ll x) {ll sum = 0; while (num) {sum += num % 10;num /= 10; } return sum == x;}int main() {scanf("%lld%lld%lld", &a, &b, &c);for (ll i = 0; i <= 81; i++) { ll tmp = b * pow1(i, a) + c; if (tmp > 0LL && tmp < 1000000000LL && judge(tmp, i)) ans.insert(tmp); } for (set<ll>::iterator it = ans.begin(); it != ans.end(); it++) out[on++] = *it; printf("%lld\n", on);for (ll i = 0; i < on - 1; i++)printf("%lld ", out[i]);if (on)printf("%lld\n", out[on - 1]);return 0;}

C:

#include <cstdio>#include <cstring>#include <cstdlib>#include <algorithm>using namespace std;typedef long long ll;const int N = 100005;const int INF = 1100000000;int n, m, w;int a[N], tmp[N];bool judge(int h) {int cnt = 0;memset(tmp, 0, sizeof(tmp));for (int i = 0; i < n; i++) {if (i) tmp[i] += tmp[i - 1];if (a[i] + tmp[i] < h) {int add = h - tmp[i] - a[i];cnt += add;tmp[i] += add;tmp[min(i + w, n)] -= add;}if (cnt > m) return false;}return true;}int solve() {int l = 0, r = INF;while (l < r) {int mid = (l + r) / 2;if (judge(mid)) l = mid + 1;else r = mid; } return l - 1;}int main() {scanf("%d%d%d", &n, &m, &w); for (int i = 0; i < n; i++)  scanf("%d", &a[i]);printf("%d\n", solve());return 0;}

D:

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;typedef long long ll;ll l, r, ans[10];int k, an = 0;void build() {    if (k == 1) ans[an++] = l;    else if (k == 2) {if (r - l + 1 == 2) {    if (l < (l^r)) ans[an++] = l;    else {ans[an++] = l; ans[an++] = r;}}else {    if (l%2) l++;    ans[an++] = l; ans[an++] = l + 1;}    }    else if (k == 3) {ll p = 1;while (p <= l) p *= 2;p /= 2;if (3 * p <= r) {    ans[an++] = l;    ans[an++] = l + p;    ans[an++] = 3 * p;}else {    if (l%2) l++;    ans[an++] = l;    ans[an++] = l + 1;}    }    else if (k == 4) {if (r - l + 1 > 4) {    if (l % 2) l++;    for (an = 0; an < 4; an++)ans[an] = l + an;}else {    ll Min = l;    ans[an++] = l;    for (int i = 1; i < 16; i++) {ll sum = 0;for (int j = 0; j < 4; j++) {    if (i&(1<<j)) sum ^= (l + j);}if (sum < Min) {    Min = sum;    an = 0;    for (int j = 0; j < 4; j++) {if (i&(1<<j))    ans[an++] = (l + j);    }}    }}    }    else {if (l % 2) l++;for (an = 0; an < 4; an++)    ans[an] = l + an;    }}int main() {    scanf("%lld%lld%d", &l, &r, &k);    build();    ll sum = 0;    for (int i = 0; i < an; i++)sum ^= ans[i];    printf("%lld\n", sum);    printf("%d\n", an);    for (int i = 0; i < an - 1; i++)printf("%lld ", ans[i]);    printf("%lld\n", ans[an - 1]);    return 0;}