Codeforces_803
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A. 填k个1,使矩阵主对角线对称,相同情况选择上面1数量多的。
#include<bits/stdc++.h> using namespace std; int n,k,a[105][105] = {0}; int main() { ios::sync_with_stdio(false); cin >> n >> k; if(k > n*n) { cout << - 1 << endl; return 0; } for(int i = 1;i <= n;i++) { for(int j = 1;j <= n;j++) { if(a[i][j]) continue; if(k >= 1 && i == j) { k--; a[i][j] = 1; } else if(k >= 2) { k -= 2; a[i][j] = 1; a[j][i] = 1; } } } if(k != 0) { cout << -1 << endl; return 0; } for(int i = 1;i <= n;i++) { for(int j = 1;j <= n;j++) cout << a[i][j] << " "; cout << endl; } return 0; }
B.两个方向各跑一次,更新最邻近的0位置就可以了。
#include<bits/stdc++.h> using namespace std; int n,a[200005],l[200005],r[200005]; int main() { ios::sync_with_stdio(false); cin >> n; for(int i = 1;i <= n;i++) cin >> a[i]; int last = -1e9; for(int i = 1;i <= n;i++) { if(a[i] == 0) last = i; l[i] = i-last; } last = 1e9; for(int i = n;i >= 1;i--) { if(a[i] == 0) last = i; r[i] = last-i; } for(int i = 1;i <= n;i++) cout << min(l[i],r[i]) << " "; cout << endl; return 0; }
C.因为和为n,所以可以是某个最小的初始小序列的某倍数为n,达到gcd最大,枚举n的每一个因子即可。
#include<bits/stdc++.h> using namespace std; long long n,k; int main() { ios::sync_with_stdio(fals#include<bits/stdc++.h> using namespace std; long long n,k; int main() { ios::sync_with_stdio(false); cin >> n >> k; if(log(n*2) < log(k)+log(k+1)-1e-9) { cout << -1 << endl; return 0; } long long t = k*(k+1)/2,maxx = 1; for(long long i = 1;i*i <= n;i++) { if(n%i) continue; { if(i*t <= n) maxx = max(maxx,i); if(t <= i) maxx = max(maxx,n/i); } } for(long long i = 1;i < k;i++) cout << i*maxx << " "; cout << n-maxx*k*(k-1)/2 << endl; return 0; } e); cin >> n >> k; if(log(n*2) > log(k)+log(k+1)-1e-9) { cout << -1 << endl; return 0; } long long t = k*(k+1)/2,maxx = 1; for(long long i = 1;i*i <= n;i++) { if(n%i) continue; { if(i*t <= n) maxx = max(maxx,i); if(n/i*t <= n) maxx = max(maxx,n/i); } } for(int i = 1;i <= n;i++) cout << i*maxx << " "; cout << n-maxx*k*(k-1)/2 << endl; return 0; }
D.直接划分成每个最小单词的字符个数,然后二分答案。
#include<bits/stdc++.h> using namespace std; int n,cnt = 1,a[1000005] = {0}; string s; bool ok(int x) { int cntt = 0,now = 0; for(int i = 1;i <= cnt;i++) { if(a[i] > x) return 0; now += a[i]; if(now > x) { cntt++; now = a[i]; } } if(now > 0) cntt++; if(cntt <= n) return 1; return 0; } int main() { cin >> n; getchar(); getline(cin,s); for(int i = 0;i < s.length();i++) { a[cnt]++; if(s[i] == ‘ ‘ || s[i] == ‘-‘) cnt++; } int l = 1,r = 1e6; while(l < r) { int mid = (l+r)/2; if(ok(mid)) r = mid; else l = mid+1; } cout << l << endl; return 0; }
E.dp[i][j]表示到i轮状态为j的可能,j可以用赢的量表示,因为会有负值,可以设置一个偏移量,因为要记录路径,直接把dp[i][j]用来记录上一状态,然后逆向输出路径就可以了。
#include<bits/stdc++.h> using namespace std; int n,k,a[1005][2005]; string s; int main() { ios::sync_with_stdio(false); cin >> n >> k >> s; memset(a,-1,sizeof(a)); a[0][n] = 0; for(int i = 0;i < n;i++) { for(int j = n-k+1;j <= n+k-1;j++) { if(a[i][j] == -1) continue; if(s[i] == ‘?‘ || s[i] == ‘W‘) a[i+1][j+1] = j; if(s[i] == ‘?‘ || s[i] == ‘L‘) a[i+1][j-1] = j; if(s[i] == ‘?‘ || s[i] == ‘D‘) a[i+1][j] = j; } } if(a[n][n-k] == -1 && a[n][n+k] == -1) { cout << "NO" << endl; return 0; } int now = n-k; if(a[n][n-k] == -1) now = n+k; for(int i = n;i >= 1;i--) { if(a[i][now] == now-1) s[i-1] = ‘W‘; else if(a[i][now] == now+1) s[i-1] = ‘L‘; else s[i-1] = ‘D‘; now = a[i][now]; } cout << s << endl; return 0; }
F.gcd为1的序列数较难求得,我们可以求gcd不为1的序列数,预处理统计每个数的约数,求得所有约数的个数,其中,k个数中的选择情况有2^k-1种,容斥原理求得答案。
#include<bits/stdc++.h> #define MOD 1000000007 using namespace std; int n,cnt[100005] = {0},ans[100005],two[100005]; int main() { ios::sync_with_stdio(false); cin >> n; while(n--) { int x; cin >> x; for(int i = 1;i*i <= x;i++) { if(x%i == 0) { cnt[i]++; if(i*i != x) cnt[x/i]++; } } } two[0] = 1; for(int i = 1;i <= 100000;i++) two[i] = two[i-1]*2%MOD; for(int i = 1;i <= 100000;i++) ans[i] = two[cnt[i]]-1; for(int i = 100000;i >= 1;i--) { for(int j = i*2;j <= 100000;j += i) ans[i] = (ans[i]-ans[j]+MOD)%MOD; } cout << ans[1] << endl; return 0; }
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