Educational Codeforces Round 32

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#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>
#include <bitset>

using namespace std;
#define LL long long
#define pb push_back
#define mk make_pair
#define fi first
#define se second
#define lson (rt << 1)
#define rson ((rt << 1) | 1)
#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 LL INF = 1e18 + 10;
int num[qq];

int main(){
	int n;	scanf("%d", &n);
	int ans = 0;
	for(int i = 1; i <= n; ++i) {
		scanf("%d", num + i);
	}
	for(int i = 2; i < n; ++i) {
		if(num[i] > num[i - 1] && num[i] > num[i + 1])	ans++;
		else if(num[i] < num[i - 1] && num[i] < num[i + 1])	ans++;
	}
	printf("%d\n", ans);
	return 0;
}

很明显要求L和R数目相同并且最多，U和D也是一样，所以统计一下数目然后求最大值即可

#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>
#include <bitset>

using namespace std;
#define LL long long
#define pb push_back
#define mk make_pair
#define fi first
#define se second
#define lson (rt << 1)
#define rson ((rt << 1) | 1)
#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 LL INF = 1e18 + 10;
int num[4];
char st[105];

int main(){
	int n;	scanf("%d", &n);
	scanf("%s", st);
	int len = strlen(st);
	for(int i = 0; i < len; ++i) {
		if(st[i] == 'L')	num[0]++;
		else if(st[i] == 'R')	num[1]++;
		else if(st[i] == 'U')	num[2]++;
		else	num[3]++;
	}
	printf("%d\n", min(num[0], num[1]) * 2 + min(num[2], num[3]) * 2);
	return 0;
}

二分答案判断可行性

#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>
#include <bitset>

using namespace std;
#define LL long long
#define pb push_back
#define mk make_pair
#define fi first
#define se second
#define lson (rt << 1)
#define rson ((rt << 1) | 1)
#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 LL INF = 1e18 + 10;
char st[qq];
int len;
bool Check(int x) {
	bool vis[26];
	int num[26];
	mst(vis, true), mst(num, 0);
	for(int i = 0; i < x - 1; ++i) {
		num[st[i] - 'a']++;
	}
	for(int i = x - 1; i < len; ++i) {
		num[st[i] - 'a']++;
		for(int j = 0; j < 26; ++j) {
			if(num[j] == 0)	vis[j] = false;
		}
		num[st[i - x + 1] - 'a']--;
	}
	for(int i = 0; i < 26; ++i) {
		if(vis[i])	return true;
	}
	return false;
}

int main(){
	scanf("%s", st);
	len = strlen(st);
	int l = 1, r = len;
	int ans;
	while(l <= r) {
		int mid = (l + r) >> 1;
		if(Check(mid)) {
			ans = mid;
			r = mid - 1;
		} else {
			l = mid + 1;
		}
	}
	printf("%d\n", ans);
	return 0;
}

k很小可以组合 + 错排

#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>
#include <bitset>

using namespace std;
#define LL long long
#define pb push_back
#define mk make_pair
#define fi first
#define se second
#define lson (rt << 1)
#define rson ((rt << 1) | 1)
#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 LL INF = 1e18 + 10;

int main(){
	LL n, k;	scanf("%lld%lld", &n, &k);
	if(k == 1) {
		puts("1");
	} else if(k == 2) {
		printf("%lld\n", n * (n - 1) / 2 + 1);
	} else if(k == 3) {
		LL a = n * (n - 1) / 2, b = n * (n - 1) * (n - 2) / 6LL * 2LL;
		printf("%lld\n", a + b + 1);
	} else if(k == 4) {
		LL a = n * (n - 1) / 2, b = n * (n - 1) * (n - 2) / 6LL * 2LL, c = n * (n - 1) * (n - 2) * (n - 3) / 24LL * 9LL;
		printf("%lld\n", a + b + c + 1);
	}
	return 0;
}

n很小，我们可以考虑分半枚举求出两部分所有选择的结果，然后枚举其中一部分的结果二分找一下另外一部分的最优解

#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>
#include <bitset>

using namespace std;
#define LL long long
#define pb push_back
#define mk make_pair
#define fi first
#define se second
#define lson (rt << 1)
#define rson ((rt << 1) | 1)
#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 LL INF = 1e18 + 10;
LL n, m;
LL a[105], b[105];
vector<int> x, y;
void Solve(int l, int r) {
	for(int i = 0; i < (1 << l); ++i) {
		LL sum = 0;
		for(int j = 0; j < l; ++j) {
			if(i & (1 << j))	sum += a[j];
		}
		x.pb(sum % m);
	}
	for(int i = 0; i < (1 << r); ++i) {
		LL sum = 0;
		for(int j = 0; j < r; ++j) {
			if(i & (1 << j))	sum += b[j];
		}
		y.pb(sum % m);
	}
	sort(y.begin(), y.end());
	LL maxn = 0;
	for(int i = 0; i < x.size(); ++i) {
		int k = lower_bound(y.begin(), y.end(), m - 1 - x[i]) - y.begin();
		if(k > 0 && y[k] > (m - 1 - x[i]))	k--;
		maxn = max(maxn, max((x[i] + y[k]) % m, (x[i] + y[y.size() - 1]) % m));
	}
	printf("%lld\n", maxn);
}

int main(){
	scanf("%lld%lld", &n, &m);
	if(n == 1) {
		scanf("%lld", &a[0]);
		printf("%lld\n", a[0] % m);
	} else {
		int l = n / 2, r = n - l;
		for(int i = 0; i < l; ++i) {
			scanf("%lld", a + i);
		}
		for(int i = 0; i < r; ++i) {
			scanf("%lld", b + i);
		}
		Solve(l, r);
	}
	return 0;
}