Educational Codeforces Round 28

最新推荐文章于 2021-01-22 23:56:41 发布

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最新推荐文章于 2021-01-22 23:56:41 发布

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题意：可以删除一些数，但不能改变其位置，并且1的右边不能有1，问最后最多剩下多少数

思路：很显然如果存在一个1那么右边不可能存在0，所以对1做一个后缀，从前向后考虑，每碰到一个1就判断不删掉这个1的最大值

#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 = 1e5 + 10;
const int INF = 1e9 + 10;
int num[105];
int suf[105];

int main(){
	int n;	scanf("%d", &n);
	for(int i = 1; i <= n; ++i) {
		scanf("%d", num + i);
	}
	for(int i = n; i >= 1; --i) {
		suf[i] = suf[i + 1] + (num[i] == 1 ? 1 : 0);
	}
	int maxn = 0;
	int cnt = 0;
	for(int i = 1; i <= n; ++i) {
		if(num[i] == 1) {
			maxn = max(maxn, cnt + suf[i]);
		} else {
			cnt++;
			maxn = max(maxn, cnt);
		}
	}
	printf("%d\n", maxn);
	return 0;
}

题意：n个任务，每个任务有k个子任务，给出k个子任务完成需要的时间ti，你可以任意做子任务，每完成一个子任务得到1点，你每做一个任务的所有子任务会额外获得1点

问你在时间M内最多可以获得多少点

思路：直接枚举完成任务的个数，然后贪心即可

#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 = 1e5 + 10;
const int INF = 1e9 + 10;
LL n, k, M;
LL t[qq];
LL Completly(LL num) {
	LL cnt = 0;
	for(int i = 1; i <= k; ++i) {
		cnt += t[i];
	}
	cnt = cnt * num;
	LL tmp = M - cnt;
	if(tmp < 0)	return 0;
	LL ans = k * num + num;
	for(int i = 1; i <= k; ++i) {
		for(int j = num + 1; j <= n; ++j) {
			if(tmp - t[i] >= 0) {
				tmp -= t[i];
				ans++;
				if(i == k)	ans++;
			}
		}
	}
	return ans;
}

int main(){
	scanf("%lld%lld%lld", &n, &k, &M);
	for(int i = 1; i <= k; ++i) {
		scanf("%lld", t + i);
	}
	sort(t + 1, t + 1 + k);
	LL maxn = 0;
	for(int i = 0; i <= n; ++i) {
		maxn = max(maxn, Completly((LL)i));
	}
	printf("%d\n", maxn);
	return 0;
}

题意：求出［0，d0） - ［d0，d1） + ［d1，d2）- ［d2，n）的最大值，其中d0 <= d1 <= d2，输出d0， d1， d2

思路：先枚举d1和d2，dp2[i]代表d1 == i 时［d1，d2）- ［d2，n）的最大值，并且d2的位置用di[i]记录，然后枚举d0，d1即可

#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 = 1e6 + 10;
const int INF = 1e9 + 10;
int num[qq], n;
LL pre[qq];
LL dp2[qq], id[qq];

int main(){
	scanf("%d", &n);
	for(int i = 1; i <= n; ++i) {
		scanf("%lld", num + i);
		pre[i] = num[i];
	}
	for(int i = 1; i <= n + 1; ++i) {
		pre[i] += pre[i - 1];
	}
	for(int i = n + 1; i >= 1; --i) {
		dp2[i] = -(pre[n + 1] - pre[i - 1]);
		id[i] = i;
		for(int j = i; j <= n + 1; ++j) {
			if(pre[j - 1] - pre[i - 1] - (pre[n + 1] - pre[j - 1]) > dp2[i]) {
				dp2[i] = pre[j - 1] - pre[i - 1] - (pre[n + 1] - pre[j - 1]);
				id[i] = j;
			}
		}
	}
	int a, b, c;
	LL maxn = -1e18;
	for(int i = 1; i <= n + 1; ++i) {
		for(int j = i; j <= n + 1; ++j) {
			if(pre[i - 1] - (pre[j - 1] - pre[i - 1]) + dp2[j] > maxn) {
				maxn = pre[i - 1] - (pre[j - 1] - pre[i - 1]) + dp2[j];
				a = i, b = j, c = id[j];
			}
		}
	}
	printf("%d %d %d\n", a - 1, b - 1, c - 1);
	return 0;
}

题意：n × m的矩阵，然后给出一个k，和q，然后给出q，接下来q行给出x，y，t，表示点x，y在时间t坏掉了，如果存在一个k × k的矩阵都换掉了，则整个矩阵坏掉了，求最小时间

思路：二分时间，然后用矩阵前缀和判断即可

#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 = 500 + 10;
const int INF = 1e9 + 10;
int n, m, k, q;
int x[qq * qq], y[qq * qq], t[qq * qq];
int dp[qq][qq];
bool Judge(int tm) {
	for(int i = 0; i <= n; ++i) {
		for(int j = 0; j <= m; ++j) {
			dp[i][j] = 0;
		}
	}
	for(int i = 0; i < q; ++i) {
		if(t[i] <= tm) {
			dp[x[i]][y[i]] = 1;
		}
	}
	for(int i = 1; i <= n; ++i) {
		for(int j = 1; j <= m; ++j) {
			dp[i][j] += dp[i - 1][j] + dp[i][j - 1] - dp[i - 1][j - 1];
		}
	}
	for(int i = 1; i + k - 1 <= n; ++i) {
		for(int j = 1; j + k - 1 <= m; ++j) {
			if(dp[i + k - 1][j + k - 1] - dp[i - 1][j + k - 1] - dp[i + k - 1][j - 1] + dp[i - 1][j - 1] == k * k) {
				return true;
			}
		}
	}
	return false;
}

int main(){
	scanf("%d%d%d%d", &n, &m, &k, &q);
	for(int i = 0; i < q; ++i) {
		scanf("%d%d%d", x + i, y + i, t + i);
	}
	int l = 0, r = 1e9 + 1;
	int ans = -1;
	while(l <= r) {
		int mid = (l + r) >> 1;
		if(Judge(mid)) {
			r = mid - 1;
			ans = mid;
		} else {
			l = mid + 1;
		}
	}
	printf("%d\n", ans);
	return 0;
}