Educational Codeforces Round 89 (Rated for Div. 2)

A. Shovels and Swords

思路

题意非常简单，就是得到最多的物品嘛，我们假定a,ba, ba,b中aaa是最小的一个，分两种情况。

如果2∗a<=b2 * a <= b2∗a<=b，那么我们只需要购买花费是1,21, 21,2的东西即可，也就是最后能购买得到aaa件物品。

否则的话，我们一定是先让数量更多的去减222，用数量更少的去减111，直到两个物品的数量相等，再通过1,21, 21,2，2,12, 12,1的顺序去交换执行，总结一下最后的答案就是(n+m)/3(n + m) / 3(n+m)/3。

代码

#include <bits/stdc++.h>
#define mp make_pair
#define pb push_backusing namespace std;typedef pair<int, int> pii;
typedef long long ll;
typedef unsigned long long ull;const double eps = 1e-7;
const double pi = acos(-1.0);
const int inf = 0x3f3f3f3f;inline ll read() {ll f = 1, x = 0;char c = getchar();while(c < '0' || c > '9') {if(c == '-') f = -1;c = getchar();} while(c >= '0' && c <= '9') {x = (x << 1) + (x << 3) + (c ^ 48);c = getchar();}return f * x;
}const int N = 2e5 + 10;int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);int t = read();while(t--) {ll a = read(), b = read();if(a > b)   swap(a, b);if(a * 2 <= b)  printf("%lld\n", a);else    printf("%lld\n", (a + b) / 3);}return 0;
}

B.Shuffle

思路

就是一个区间有重合判断并集的问题，如果我们给定的区间[l,r][l, r][l,r]在原本的区间外也就是r<L∣∣l>Rr < L || l > Rr<L∣∣l>R，否则的话我们就更新L,RL, RL,R的最大最小值

代码

#include <bits/stdc++.h>
#define mp make_pair
#define pb push_backusing namespace std;typedef pair<int, int> pii;
typedef long long ll;
typedef unsigned long long ull;const double eps = 1e-7;
const double pi = acos(-1.0);
const int inf = 0x3f3f3f3f;inline ll read() {ll f = 1, x = 0;char c = getchar();while(c < '0' || c > '9') {if(c == '-') f = -1;c = getchar();} while(c >= '0' && c <= '9') {x = (x << 1) + (x << 3) + (c ^ 48);c = getchar();}return f * x;
}const int N = 2e5 + 10;int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);int t = read();while(t--) {int n = read(), x = read(), m = read();int l = x, r = x;// cout << l << " " << r << endl;for(int i = 1; i <= m; i++) {int a = read(), b = read();if((a <= r && a >= l) || (b >= l && b <= r) || (l >= a && b >= r)) {//比赛时判断条件写的比较繁琐。l = min(l, a);r = max(r, b);}// cout << l << " " << r << endl;}printf("%d\n", r - l + 1);}return 0;
}

C.Palindromic Paths

思路

开两个数组，num1[i]num1[i]num1[i]记录的是步数为iii的时候的位置上的111的个数，num0[i]num0[i]num0[i]记录的是步数为iii的时候的位置上000的个数，因为整体的步数就是在[1,n+m−1][1, n + m - 1][1,n+m−1]之间，所以我们可以通过对每一步全变为000或者全变为111中挑选一个最小值，作为我们的花费，然后累加花费就是答案。

代码

#include <bits/stdc++.h>
#define mp make_pair
#define pb push_backusing namespace std;typedef pair<int, int> pii;
typedef long long ll;
typedef unsigned long long ull;const double eps = 1e-7;
const double pi = acos(-1.0);
const int inf = 0x3f3f3f3f;inline ll read() {ll f = 1, x = 0;char c = getchar();while(c < '0' || c > '9') {if(c == '-') f = -1;c = getchar();} while(c >= '0' && c <= '9') {x = (x << 1) + (x << 3) + (c ^ 48);c = getchar();}return f * x;
}const int N = 2e5 + 10;int a[40][40];int num0[100], num1[100];int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);int t = read();while(t--) {memset(num1, 0, sizeof num1);memset(num0, 0, sizeof num0);int n = read(), m = read();// cout << n << " " << m << endl;for(int i = 1; i <= n; i++)for(int j = 1; j <= m; j++) {a[i][j] = read();if(a[i][j] == 1)    num1[i + j]++;else num0[i + j]++;}if(n == 2 && m == 2) {puts("0");// puts("");continue;}int ans = 0;int l = 2, r = n + m;while(l < r) {//好像这个if语句并没有什么用，不知道比赛的时候怎么想的。if((num1[l] && num0[l]) || (num1[r] && num0[r]) || (num1[l] && num0[r]) || (num0[l] && num1[r]))ans += min(num0[l]+ num0[r], num1[l] + num1[r]);// cout << ans << "\n";l++, r--;}printf("%d\n", ans);// puts("");}return 0;
}

D.Two Divisors

思路

先引入一个定理gcd(a,b)=1=gcd(a+b,a∗b)gcd(a, b) = 1 = gcd(a + b, a * b)gcd(a,b)=1=gcd(a+b,a∗b)，所以这道题就简简单单就可以水过了，但是我的赛况却不是如此，，，，，

那我们来证明一下这个定理的正确性吧：

假设有KaTeX parse error: Undefined control sequence: \and at position 15: gcd(x, y) = 1 \̲a̲n̲d̲ ̲gcd(x, z) = 1，所以gcd(x,y∗z)=1gcd(x, y * z) = 1gcd(x,y∗z)=1

gcd(a,b)=1−>gcd(a+b,b)=1=gcd(a,a+b)−>gcd(a+b,a∗b)=1gcd(a, b) = 1 -> gcd(a + b, b) = 1 = gcd(a, a + b) - > gcd(a + b, a * b) = 1gcd(a,b)=1−>gcd(a+b,b)=1=gcd(a,a+b)−>gcd(a+b,a∗b)=1

代码

#include <bits/stdc++.h>
#define mp make_pair
#define pb push_backusing namespace std;typedef pair<int, int> pii;
typedef long long ll;
typedef unsigned long long ull;const double eps = 1e-7;
const double pi = acos(-1.0);
const int inf = 0x3f3f3f3f;inline ll read() {ll f = 1, x = 0;char c = getchar();while(c < '0' || c > '9') {if(c == '-') f = -1;c = getchar();} while(c >= '0' && c <= '9') {x = (x << 1) + (x << 3) + (c ^ 48);c = getchar();}return f * x;
}const int N1 = 1e7 + 10, N2 = 5e5 + 10;int prime[N1], cnt;
int ans1[N2], ans2[N2], n;
bool st[N1];int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);for(int i = 2; i < N1; i++) {if(!st[i])  prime[++cnt] = i;for(int j = 1; j <= cnt && i * prime[j] < N1; j++) {st[i * prime[j]] = true;if(i % prime[j] == 0)   break;}}n = read();for(int i = 1; i <= n; i++) {int temp = read();ans1[i] = ans2[i] = -1;for(int j = 1; prime[j] * prime[j] <= temp; j++) {int x = 1;if(temp % prime[j] == 0) {while(temp % prime[j] == 0) {temp /= prime[j];x *= prime[j];}if(temp == 1)   break;else {ans1[i] = x;ans2[i] = temp;break;}}}}for(int i = 1; i <= n; i++)printf("%d%c", ans1[i], i == n ? '\n' : ' ');for(int i = 1; i <= n; i++)printf("%d%c", ans2[i], i == n ? '\n' : ' ');return 0;
}