Codeforces_803

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[][] = {};

int main()
{
    ios::sync_with_stdio(false);
    cin >> n >> k;
    if(k > n*n)
    {
        cout << -  << endl;
        return ;
    }
    for(int i = ;i <= n;i++)
    {
        for(int j = ;j <= n;j++)
        {
            if(a[i][j])   continue;
            if(k >=  && i == j)
            {
                k--;
                a[i][j] = ;
            }
            else if(k >= )
            {
                k -= ;
                a[i][j] = ;
                a[j][i] = ;
            }
        }
    }
    if(k != )
    {
        cout << - << endl;
        return ;
    }
    for(int i = ;i <= n;i++)
    {
        for(int j = ;j <= n;j++)   cout << a[i][j] << " ";
        cout << endl;
    }
    return ;
}

---

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[],l[],r[];

int main()
{
    ios::sync_with_stdio(false);
    cin >> n;
    for(int i = ;i <= n;i++)   cin >> a[i];
    int last = -1e9;
    for(int i = ;i <= n;i++)
    {
        if(a[i] == )   last = i;
        l[i] = i-last;
    }
    last = 1e9;
    for(int i = n;i >= ;i--)
    {
        if(a[i] == )   last = i;
        r[i] = last-i;
    }
    for(int i = ;i <= n;i++)   cout << min(l[i],r[i]) << " ";
    cout << endl;
    return ;
}

---

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*) < log(k)+log(k+)-1e-)
    {
        cout << - << endl;
        return ;
    }
    long long t = k*(k+)/,maxx = ;
    for(long long i = ;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 = ;i < k;i++)   cout << i*maxx << " ";
    cout << n-maxx*k*(k-)/ << endl;
    return ;
}
e);
    cin >> n >> k;
    if(log(n*) > log(k)+log(k+)-1e-)
    {
        cout << - << endl;
        return ;
    }
    long long t = k*(k+)/,maxx = ;
    for(long long i = ;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 = ;i <= n;i++)   cout << i*maxx << " ";
    cout << n-maxx*k*(k-)/ << endl;
    return ;
}

---

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 = ,a[] = {};
string s;

bool ok(int x)
{
    int cntt = ,now = ;
    for(int i = ;i <= cnt;i++)
    {
        if(a[i] > x)  return ;
        now += a[i];
        if(now > x)
        {
            cntt++;
            now = a[i];
        }
    }
    if(now > ) cntt++;
    if(cntt <= n)   return ;
    return ;
}
int main()
{
    cin >> n;
    getchar();
    getline(cin,s);
    for(int i = ;i < s.length();i++)
    {
        a[cnt]++;
        if(s[i] == ' ' || s[i] == '-')    cnt++;
    }
    int l = ,r = 1e6;
    while(l < r)
    {
        int mid = (l+r)/;
        if(ok(mid)) r = mid;
        else    l = mid+;
    }
    cout << l << endl;
    return ;
}

---

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[][];
string s;
int main()
{
    ios::sync_with_stdio(false);
    cin >> n >> k >> s;
    memset(a,-,sizeof(a));
    a[][n] = ;
    for(int i = ;i < n;i++)
    {
        for(int j = n-k+;j <= n+k-;j++)
        {
            if(a[i][j] == -)   continue;
            if(s[i] == '?' || s[i] == 'W')  a[i+][j+] = j;
            if(s[i] == '?' || s[i] == 'L')  a[i+][j-] = j;
            if(s[i] == '?' || s[i] == 'D')  a[i+][j] = j;
        }
    }
    if(a[n][n-k] == - && a[n][n+k] == -)
    {
        cout << "NO" << endl;
        return ;
    }
    int now = n-k;
    if(a[n][n-k] == -)   now = n+k;
    for(int i = n;i >= ;i--)
    {
        if(a[i][now] == now-)  s[i-] = 'W';
        else if(a[i][now] == now+) s[i-] = 'L';
        else    s[i-] = 'D';
        now = a[i][now];
    }
    cout << s << endl;
    return ;
}

---

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[] = {},ans[],two[];

int main()
{
    ios::sync_with_stdio(false);
    cin >> n;
    while(n--)
    {
        int x;
        cin >> x;
        for(int i = ;i*i <= x;i++)
        {
            if(x%i == )
            {
                cnt[i]++;
                if(i*i != x)    cnt[x/i]++;
            }
        }
    }
    two[] = ;
    for(int i = ;i <= ;i++)  two[i] = two[i-]*%MOD;
    for(int i = ;i <= ;i++)  ans[i] = two[cnt[i]]-;
    for(int i = ;i >= ;i--)
    {
        for(int j = i*;j <= ;j += i) ans[i] = (ans[i]-ans[j]+MOD)%MOD;
    }
    cout << ans[] << endl;
    return ;
}