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;
}