Educational Codeforces Round 89 (Rated for Div. 2)

A. Shovels and Swords

思路

題意非常簡(jiǎn)單，就是得到最多的物品嘛，我們假定$a,b$中$a$是最小的一個(gè)，分兩種情況。

如果$2?a<=b$，那么我們只需要購(gòu)買(mǎi)花費(fèi)是$1,2$的東西即可，也就是最后能購(gòu)買(mǎi)得到$a$件物品。

否則的話，我們一定是先讓數(shù)量更多的去減$2$，用數(shù)量更少的去減$1$，直到兩個(gè)物品的數(shù)量相等，再通過(guò)$1,2$，$2,1$的順序去交換執(zhí)行，總結(jié)一下最后的答案就是$(n+m)/3$。

代碼

#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

就是一個(gè)區(qū)間有重合判斷并集的問(wèn)題，如果我們給定的區(qū)間$[l,r]$在原本的區(qū)間外也就是$r<L||l>R$，否則的話我們就更新$L,R$的最大最小值

代碼

#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)) {//比賽時(shí)判斷條件寫(xiě)的比較繁瑣。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

開(kāi)兩個(gè)數(shù)組，$num1[i]$記錄的是步數(shù)為$i$的時(shí)候的位置上的$1$的個(gè)數(shù)，$num0[i]$記錄的是步數(shù)為$i$的時(shí)候的位置上$0$的個(gè)數(shù)，因?yàn)檎w的步數(shù)就是在$[1,n+m?1]$之間，所以我們可以通過(guò)對(duì)每一步全變?yōu)?span id="ozvdkddzhkzd" class="katex--inline">$0$或者全變?yōu)?span id="ozvdkddzhkzd" class="katex--inline">$1$中挑選一個(gè)最小值，作為我們的花費(fèi)，然后累加花費(fèi)就是答案。

代碼

#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) {//好像這個(gè)if語(yǔ)句并沒(méi)有什么用，不知道比賽的時(shí)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

先引入一個(gè)定理$gcd(a,b)=1=gcd(a+b,a?b)$，所以這道題就簡(jiǎn)簡(jiǎn)單單就可以水過(guò)了，但是我的賽況卻不是如此，，，，，

那我們來(lái)證明一下這個(gè)定理的正確性吧：

假設(shè)有KaTeX parse error: Undefined control sequence: \and at position 15: gcd(x, y) = 1 \?a?n?d? ?gcd(x, z) = 1，所以$gcd(x,y?z)=1$

$gcd(a,b)=1?>gcd(a+b,b)=1=gcd(a,a+b)?>gcd(a+b,a?b)=1$

代碼

#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; }

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