Educational Codeforces Round 29

最新推荐文章于 2021-01-15 12:47:58 发布

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最新推荐文章于 2021-01-15 12:47:58 发布

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本文解析了四道编程题目，包括字符串处理、组合优化、循环检测及数组操作回溯等核心算法，通过示例代码详细介绍了每道题目的解决思路。

把末尾的0删除然后匹配

#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 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 = 1e4 + 10;
const int INF = 1e9 + 10;
char st[30];

int main(){
	scanf("%s", st);
	int r = strlen(st) - 1;
	int l = 0;
	while(st[r] == '0')	r--;
	bool f = true;
	while(l <= r) {
		if(st[l] != st[r])	f = false;
		l++, r--;
	}
	if(f)	puts("YES");
	else	puts("NO");
	return 0;
}

枚举两个不参加的人

#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 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[60];

int main(){
	int n;	scanf("%d", &n);
	for(int i = 1; i <= 2 * n; ++i) {
		scanf("%d", num + i);
	}
	sort(num + 1, num + 1 + 2 * n);
	int minx = INF;
	for(int i = 1; i <= 2 * n; ++i) {
		for(int j = i + 1; j <= 2 * n; ++j) {
			int sum = 0, a, b;
			a = b = -1;
			for(int k = 1; k <= 2 * n; ++k) {
				if(k == i || k == j)	continue;
				if(a == -1) {
					a = num[k];
				} else {
					b = num[k];
					sum += abs(a - b);
					a = b = -1;
				}
			}
			minx = min(minx, sum);
		}
	}
	printf("%d\n", minx);
	return 0;
}

题意：k局比赛，第一局Alice出a，Bob出b，之后每一局两人出的数字仅仅只和前一局有关，给出相关矩阵，让你求k局后的最终分数

思路：很显然进行到某个程度之后肯定是有循环节的，所以我们进行比赛直到找到循环节为止，然后处理一下有不循环的部分和循环的部分就好了

#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 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;
LL k, a, b;
int A[4][4], B[4][4];
LL dp[10005][2];
int Check(LL x, LL y) {
	if(x == 1 && y == 3)	return 1;
	if(x == 2 && y == 1)	return 1;
	if(x == 3 && y == 2)	return 1;
	if(x == y)	return 0;
	return 2;
}
map<pill, int> mp;

int main(){
	scanf("%lld%lld%lld", &k, &a, &b);
	for(int i = 1; i <= 3; ++i) {
		for(int j = 1; j <= 3; ++j) {
			scanf("%d", &A[i][j]);
		}
	}
	for(int i = 1; i <= 3; ++i) {
		for(int j = 1; j <= 3; ++j) {
			scanf("%d", &B[i][j]);
		}
	}
	dp[0][0] = dp[0][1] = 0;
	int cnt = 1;
	while(true) {
		int t = Check(a, b);
		dp[cnt][0] = dp[cnt - 1][0];
		dp[cnt][1] = dp[cnt - 1][1];
		if(t == 1) {
			dp[cnt][0]++;
		} else if(t == 2) {
			dp[cnt][1]++;
		}
		if(mp.find(mk(a, b)) == mp.end()) {
			mp[mk(a, b)] = cnt;
			int x, y;
			x = A[a][b], y = B[a][b];
			a = x, b = y;
			cnt++;
		} else {
			break;
		}
	}
	LL x = dp[cnt - 1][0] - dp[mp[mk(a, b)] - 1][0];
	LL y = dp[cnt - 1][1] - dp[mp[mk(a, b)] - 1][1];
	int len = (cnt - 1) - mp[mk(a, b)] + 1;
	if(k <= cnt) {
		printf("%lld %lld\n", dp[k][0], dp[k][1]);
	} else {
		LL ans1, ans2;
		ans1 = dp[mp[mk(a, b)] - 1][0], ans2 = dp[mp[mk(a, b)] - 1][1];
		k -= mp[mk(a, b)] - 1;
		LL tmp = k / len;
		ans1 += x * tmp, ans2 += y * tmp;
		k = k - tmp * len;
		for(int i = mp[mk(a, b)]; k > 0; ++i, --k) {
			ans1 += dp[i][0] - dp[i - 1][0];
			ans2 += dp[i][1] - dp[i - 1][1];
		}
		printf("%lld %lld\n", ans1, ans2);
	}
	return 0;
}

题意：给出数组a，n个数，q次操作，操作1 把区间l r的数向右平移一位，r位置上的数会到l位置上，操作2 反转区间l r，然后出一个数组b，求q次操作后的数组a中索引为b数组的所有数

思路：直接求数组a不好求，考虑到数组b的大小，我们可以将数组b给导回去，这样可以再m × q的复杂度内求出结果

#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 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 = 2e5 + 10;
const int INF = 1e9 + 10;
int n, q, m;
int a[qq], b[105];
int p[qq], l[qq], r[qq];

int main(){
	scanf("%d%d%d", &n, &q, &m);
	for(int i = 1; i <= n; ++i) {
		scanf("%d", a + i);
	}
	for(int i = 1; i <= q; ++i) {
		scanf("%d%d%d", p + i, l + i, r + i);
	}
	for(int i = 1; i <= m; ++i) {
		scanf("%d", b + i);
	}
	for(int i = q; i >= 1; --i) {
		for(int j = 1; j <= m; ++j) {
			if(l[i] <= b[j] && b[j] <= r[i]) {
				if(p[i] == 2) {
					b[j] = r[i] - b[j] + l[i];
				} else {
					if(b[j] == l[i])	b[j] = r[i];
					else	b[j]--;
				}
			}
		}
	}
	for(int i = 1; i <= m; ++i) {
		printf("%d ", a[b[i]]);
	}
	puts("");
	return 0;
}