Educational Codeforces Round 26

A

#include <cstdio>#include <cstring>#include <cmath>#include <cstdlib>#include <ctime>#include <iostream>#include <algorithm>#include <sstream>#include <string>#include <vector>#include <queue>#include <stack>#include <map>#include <set>#include <utility>using namespace std;#define LL long long#define pb push_back#define mk make_pair#define mst(a, b)memset(a, b, sizeof a)#define REP(i, x, n)for(int i = x; i <= n; ++i)const int MOD = 1e9 + 7;const int qq = 200 + 10;char st[qq];int main(){int n;scanf("%d", &n);getchar();gets(st);int maxn = 0;string x = "";int cnt = 0;for(int i = 0; i < n; ++i) {if(isalpha(st[i])) {if(isupper(st[i]))cnt++;x += st[i];} else {maxn = max(maxn, cnt);cnt = 0;}}maxn = max(maxn, cnt);printf("%d\n", maxn);return 0;}

B

题意：整个图是三条条纹状，每个颜色是一条条纹。

思路：分别找出R、B、G三种颜色的左上角和右下角，如果成条纹状那么整个区域肯定得是同一种颜色，并且大小得相同，三条条纹大小得相等

#include <cstdio>#include <cstring>#include <cmath>#include <cstdlib>#include <ctime>#include <iostream>#include <algorithm>#include <sstream>#include <string>#include <vector>#include <queue>#include <stack>#include <map>#include <set>#include <utility>using namespace std;#define LL long long#define pb push_back#define mk make_pair#define mst(a, b)memset(a, b, sizeof a)#define REP(i, x, n)for(int i = x; i <= n; ++i)const int MOD = 1e9 + 7;const int qq = 105 + 10;char st[qq][qq];int n, m;int width[qq], heigh[qq];bool f = true;void Check(char ch) {int x1, y1, x2, y2;x1 = y1 = 1e9;x2 = y2 = -1e9;for(int i = 1; i <= n; ++i) {for(int j = 1; j <= m; ++j) {if(st[i][j] == ch) {x1 = min(x1, i);y1 = min(y1, j);x2 = max(x2, i);y2 = max(y2, j);}}}for(int i = x1; i <= x2; ++i) {for(int j = y1; j <= y2; ++j) {if(st[i][j] != ch) {f = false;}}}if(ch == 'R') {width[1] = x2 - x1;heigh[1] = y2 - y1;} else if(ch == 'B') {width[2] = x2 - x1;heigh[2] = y2 - y1;} else {width[3] = x2 - x1;heigh[3] = y2 - y1;}}int main(){scanf("%d%d", &n, &m);for(int i = 1; i <= n; ++i) {scanf("%s", st[i] + 1);}Check('R'), Check('B'), Check('G');if(!f) {puts("NO");return 0;}if(width[1] == width[2] && width[2] == width[3] && heigh[1] == heigh[2] && heigh[2] == heigh[3]) {puts("YES");} else {puts("NO");}return 0;}

C

题意：给出n个矩形的左上角和右下角，给出一个a * b的矩形，问你再在n个中选两个放入a * b的矩形要求被占据的面积最大

思路：枚举，然后判断是否能放下

#include <cstdio>#include <cstring>#include <cmath>#include <cstdlib>#include <ctime>#include <iostream>#include <algorithm>#include <sstream>#include <string>#include <vector>#include <queue>#include <stack>#include <map>#include <set>#include <utility>using namespace std;#define LL long long#define pb push_back#define mk make_pair#define pill pair<int, int> #define mst(a, b)memset(a, b, sizeof a)#define REP(i, x, n)for(int i = x; i <= n; ++i)const int MOD = 1e9 + 7;const int qq = 105 + 10;pill node[qq];bool Check(int a, int b, int x1, int y1, int x2, int y2) {if(x1 <= a && y1 <= b && (a - x1) >= x2 && b >= y2)return true;if(x1 <= a && y1 <= b && x2 <= a && (b - y1) >= y2)return true;if(x1 <= a && y1 <= b && y2 <= a && (b - y1) >= x2)return true;if(x1 <= a && y1 <= b && x2 <= b && (a - x1) >= y2)return true;if(y1 <= a && x1 <= b && (b - x1) >= x2 && y2 <= a)return true;if(y1 <= a && x1 <= b && x2 <= b && (a - y1) >= y2)return true;if(y1 <= a && x1 <= b && y2 <= b && (a - y1) >= x2)return true;if(y1 <= a && x1 <= b && y2 <= (b - x1) && x2 <= a)return true;return false;}int main(){int n, a, b;scanf("%d%d%d", &n, &a, &b);if(a > b)swap(a, b);for(int i = 0; i < n; ++i) {int x, y;scanf("%d%d", &x, &y);if(x > y)swap(x, y);node[i] = mk(x, y);//if(node[i].first > node[i].second)swap(node[i].first, node[i].second);}int maxn = 0;for(int i = 0; i < n; ++i) {for(int j = i + 1; j < n; ++j) {if(a * b < node[i].first * node[i].second + node[j].first * node[j].second)continue;if(Check(a, b, node[i].first, node[i].second, node[j].first, node[j].second)) {maxn = max(maxn, node[i].first * node[i].second + node[j].first * node[j].second);}}}printf("%d\n", maxn);return 0;}

D

题意：n个数中选k个使得最后乘积含0最多

思路：很显然这题不是简单的贪心，如果一个数是5^15 ，一个数是2^20，那么他们乘积后由15个0，先预处理出每个数含因子2和5的个数，然后dp[i][j] = k，表示选i个因子5个数为j的情况下含最多的5是k

#include <cstdio>#include <cstring>#include <cmath>#include <cstdlib>#include <ctime>#include <iostream>#include <algorithm>#include <sstream>#include <string>#include <vector>#include <queue>#include <stack>#include <map>#include <set>#include <utility>using namespace std;#define LL long long#define pb push_back#define mk make_pair#define mst(a, b)memset(a, b, sizeof a)#define REP(i, x, n)for(int i = x; i <= n; ++i)const int MOD = 1e9 + 7;const int qq = 5500 + 10;int n, k;struct Node {int two;int five;}node[qq];int Cal(LL x, int id) {int a = 0;while(x % 2 == 0) {a++;x /= 2;}int b = 0;while(x % 5 == 0) {b++;x /= 5;}node[id].two = a;node[id].five = b;}int dp[205][qq];int main(){scanf("%d%d", &n, &k);int maxnX, maxnY;maxnY = maxnX = 0;for(int i = 0; i < n; ++i) {LL x;scanf("%lld", &x);Cal(x, i);maxnX += node[i].two;maxnY += node[i].five;}mst(dp, -1);dp[0][0] = 0;for(int i = 0; i < n; ++i) {for(int j = k; j >= 1; --j) {for(int t = maxnY; t >= 0; --t) {if(t - node[i].five < 0)break;if(dp[j - 1][t - node[i].five] == -1)continue;dp[j][t] = max(dp[j][t], dp[j - 1][t - node[i].five] + node[i].two);}}}int ans = 0;for(int i = 0; i <= maxnY; ++i) {if(dp[k][i] == -1)continue;ans = max(ans, min(i, dp[k][i]));}printf("%d\n", ans);return 0;}

E

参考：传送门

#include <cstdio>#include <cstring>#include <cmath>#include <sstream>#include <iostream>#include <algorithm>#include <string>#include <stack>#include <queue>#include <vector>#include <map>#include <set>#include <utility>using namespace std;#define LL long long#define pb push_back#define mk make_pair#define pill pair<int, int>#define ft first#define sd second#define mst(a, b)memset(a, b, sizeof a)#define REP(i, x, n)for(int i = x; i <= n; ++i)const int qq = 1e5 + 10;const int MOD = 1e9 + 7;LL x, y;LL Gcd(LL a, LL b) {return b == 0 ? a : Gcd(b, a % b);}int main(){scanf("%lld%lld", &x, &y);LL g = Gcd(x, y);x /= g, y /= g;vector<LL> a;for(LL i = 2; i * i <= x; ++i) {while(x % i == 0) {x /= i;a.pb(i);}}if(x > 1)a.pb(x);LL ans = 0;while(y > 0) {LL minx = y;for(LL i : a) {minx = min(minx, y % i);}ans += minx;y -= minx;vector<LL> b;for(LL i : a) {if(y % i == 0) {y /= i;} else {b.pb(i);}}a.swap(b);}printf("%lld\n", ans); return 0;}