Educational Codeforces Round 20

A

按题意来即可

#include <bits/stdc++.h>using namespace std;#define LL long long#define pb push_back#define mst(a, b)memset(a, b, sizeof a)#define REP(i, x, n)for(int i = x; i < n; ++i)const int qq = 1e3 + 10;int gra[qq][qq];int main(){int n, k;scanf("%d%d", &n, &k);for(int i = 1; i <= n; ++i){for(int j = 1; j <= n; ++j){if(gra[i][j] == 1)continue;if(i == j){if(k >= 1)gra[i][j] = 1, k--;}else{if(k >= 2)gra[i][j] = gra[j][i] = 1, k -= 2;}}}if(k){puts("-1");return 0;}for(int i = 1; i <= n; ++i){for(int j = 1; j <= n; ++j)printf("%d ", gra[i][j]);puts("");}return 0;}

B

每个非0距离0的最短距离一定是它左边的第一个0或者它右边的第一个0，计算这两个位置的0即可

#include <bits/stdc++.h>using namespace std;#define LL long long#define pb push_back#define mst(a, b)memset(a, b, sizeof a)#define REP(i, x, n)for(int i = x; i < n; ++i)const int qq = 2e5 + 10;int num[qq], dist[qq];int main(){int n;scanf("%d", &n);for(int i = 1; i <= n; ++i){scanf("%d", num + i);}int pos = -1;for(int i = 1; i <= n; ++i){if(num[i] == 0){if(pos == -1){for(int j = i - 1; j >= 1; --j)dist[j] = i - j;pos = i;}else{for(int j = i - 1; j > pos; --j)dist[j] = min(dist[j], i - j);pos = i;}continue;}if(pos == -1)continue;dist[i] = i - pos;}for(int i = 1; i <= n; ++i)printf("%d ", dist[i]);puts("");return 0;}

C

题意：给出一个n，一个k，要求在1到n中选k个数，使得和为n，并且这k个数的gcd要最大。

思路：假设选出来的数  a1 + a2 + ... + ak = n， 那么令 g = gcd(a1, a2, ... , ak)， 则 b1 + b2 + ... + bk = n/g， 很显然这个g是n的因子，那么我们枚举n的因子判断可行性即可，注意这里n的最大值是1e10， 那么当k大于1e6的时候其实就是不可能的，因为算最小的情况 从1加到1e6是大于1e10的，并且k > n的时候也不可能，排除这两个情况就可以直接做了。

#include <bits/stdc++.h>using namespace std;#define LL long long#define pb push_back#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 qq = 1e5 + 10;int main(){LL n, k;scanf("%lld%lld", &n, &k);if(k >= 1e6 || k > n){puts("-1");return 0;}LL f = (1 + k) * k / 2;LL ans = 0;for(LL i = 1; i * i <= n; ++i){if(n % i == 0){if(i >= f){ans = n / i;break;}else if(n / i >= f){ans = i;}}}if(!ans){puts("-1");return 0;}LL sum = 0;for(int i = 1; i < k; ++i)printf("%lld ", i * ans), sum += i * ans;printf("%lld\n", n - sum);return 0;}

E

题意：给出一个n一个k，然后给出一个长度为n的字符串，字符只有四种，L,W,D,?，分别代表输，赢，平，任意，现在要求在最后一个回合的时要求输赢数的绝对值等于k，并且前面任意一个回合结束以后输赢的绝对值不能等于k，现在问你把？填完，求一种符合要求的解。

思路：dp[i][j]代表第i回合，赢 - 输数为j的字符串是否存在，因为有负数，所以可以把整体向右平移一个n，然后就状态转移就好。

#include <bits/stdc++.h>using namespace std;#define LL long long#define pb push_back#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 qq = 2e3 + 10;int dp[qq][qq], pre[qq][qq];char st[qq];int n, k;int main(){scanf("%d%d", &n, &k);scanf("%s", st + 1);if(k > n){puts("NO");return 0;}dp[0][n] = 1;for(int i = 1; i <= n; ++i){for(int j = 0; j <= n << 1; ++j){if((j == n - k or j == n + k) and i != n)continue;if(st[i] == 'W' and j - 1 >= 0 and dp[i - 1][j - 1])dp[i][j] = 1, pre[i][j] = -1;if(st[i] == 'D' and dp[i - 1][j])dp[i][j] = 1, pre[i][j] = 0;if(st[i] == 'L' and j + 1 <= (n << 1) and dp[i - 1][j + 1])dp[i][j] = 1, pre[i][j] = 1;if(st[i] == '?'){if(dp[i - 1][j - 1] and j - 1 >= 0)dp[i][j] = 1, pre[i][j] = -1;if(dp[i - 1][j])dp[i][j] = 1, pre[i][j] = 0;if(dp[i - 1][j + 1] and j + 1 <= (n << 1))dp[i][j] = 1, pre[i][j] = 1;}}}//printf("%d %d\n", dp[n][n + k], dp[n][n - k]);if(!dp[n][n - k] and !dp[n][n + k]){puts("NO");return 0;}int a = n, b;if(dp[n][n + k])b = n + k;elseb = n - k;while(a > 0){if(pre[a][b] == -1)st[a] = 'W';else if(pre[a][b] == 1)st[a] = 'L';elsest[a] = 'D';b = b + pre[a][b];a = a - 1;}printf("%s\n", st + 1);return 0;}

F

题意：给出n个数，要求求出gcd = 1的子序列的数量

思路：我们可以处理出gcd = i * k （k = 1， 2， 3 ......） 的个数，如果我们要求出gcd = i的数量 就减去gcd等于i的倍数即可得到

#include <bits/stdc++.h>using namespace std;#define LL long long#define pb push_back#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 ans[qq], p[qq];LL num[qq];int main(){int n;scanf("%d", &n);p[0] = 1;for(int i = 1; i <= n; ++i)p[i] = (p[i - 1] * 2LL) % MOD;int maxn = 0;for(int i = 1; i <= n; ++i){int x;scanf("%d", &x);maxn = max(maxn, x);num[x]++;}for(int i = 1; i<= maxn; ++i){for(int j = i + i; j <= maxn; j += i)num[i] += num[j];}for(int i = maxn; i > 0; --i){ans[i] = p[num[i]] - 1;for(int j = i + i; j <= maxn; j += i)ans[i] = (ans[i] - ans[j] + MOD) % MOD;}printf("%lld\n", ans[1]);return 0;}