思维太僵硬了，我从余数入手，嫩是记录不了每个数要操作多少次。但是如果考虑每个数的贡献，即操作多少次能使得满足条件，就好写了，实际上也是暴力。

#include
     
      #define ll long long#define P pair
      
       #define pb push_back#define lson root << 1#define INF (int)2e9 + 7#define maxn (int)2e5 + 7#define rson root << 1 | 1#define LINF (unsigned long long int)1e18#define mem(arry, in) memset(arry, in, sizeof(arry))using namespace std;int n, m, t;int a[maxn], pre[maxn], sum[maxn];int Find(int x){    return (sum[x] < t ? x : pre[x] = Find(pre[x]));}int main(){    ios::sync_with_stdio(false);    cin.tie(NULL);    cout.tie(NULL);        cin >> n >> m;    for(int i = 0; i < m; i++) pre[i] = (i + 1) % m;    t = n / m;    ll ans = 0;    for(int i = 1; i <= n; i++) {        int tp = 0;        cin >> tp;        int x = Find(tp % m);        sum[x]++;        a[i] = (x - tp % m + m) % m + tp;        ans += a[i] - tp;    }    cout << ans << endl;    for(int i = 1; i <= n; i++) cout << a[i] << " ";    cout << endl;    return 0;}
      
     

题解：先从起点搜索一遍，对不能到达的点加一条从首都到这个点的边（加边操作只能是思维上的，不能真的addedge(st, i)，因为后面还有删除边的操作），并标记，假如后加入的边能访问到先前的点，就删除原先对应加的边（说明允许犯错）。

#include
     
      #define ll long long#define P pair
      
       #define pb push_back#define lson root << 1#define INF (int)2e9 + 7#define maxn (int)5e3 + 7#define rson root << 1 | 1#define LINF (unsigned long long int)1e18#define mem(arry, in) memset(arry, in, sizeof(arry))using namespace std;int n, m, tot, st;int head[maxn];bool use[maxn], connect[maxn], link[maxn];struct node{ int to, next; } g[maxn << 1];void Inite(){    tot = 0;    mem(head, -1);}void addedge(int u, int v){    g[tot].to = v;    g[tot].next = head[u];    head[u] = tot++;}void DFS(int u){    use[u] = connect[u] = 1;    //connect数组表示从首都能到这个点    for(int i = head[u]; i != -1; i = g[i].next){        int v = g[i].to;        if(!use[v]){            link[v] = 0;   //删除标记            DFS(v);        }    }}int main(){    cin >> n >> m >> st;    Inite();    for(int i = 1; i <= m; i++){        int u, v;        cin >> u >> v;        addedge(u, v);    }    DFS(st);    mem(use, 0);    for(int i = 1; i <= n; i++){        if(!connect[i]){            link[i] = 1;            DFS(i);            mem(use, 0);        }    }    int res = 0;    for(int i = 1; i <= n; i++) if(link[i]) res++;    cout << res << "\n";    return 0;}
      
     

题解：dp[ i ][ j ]：同一爱好的 i 个人分同样的 j 件物品得到的最大价值。

#include
     
      #define ll long long#define P pair
      
       #define pb push_back#define lson root << 1#define INF (int)2e9 + 7#define maxn (int)1e5 + 7#define rson root << 1 | 1#define LINF (unsigned long long int)1e18#define mem(arry, in) memset(arry, in, sizeof(arry))using namespace std;int n, k;int cntf[maxn], cnta[maxn], h[15], dp[505][5005];int main(){    cin >> n >> k;    int m = n * k;    int tp;    for(int i = 1; i <= m; i++) cin >> tp, cnta[tp]++;    for(int i = 1; i <= n; i++) cin >> tp, cntf[tp]++;    for(int i = 1; i <= k; i++) cin >> h[i];    for(int i = 1; i <= n; i++) {        for(int j = 1; j <= m; j++){            for(int p = 1; p <= min(j, k); p++) dp[i][j] = max(dp[i][j], dp[i - 1][j - p] + h[p]);        }    }    ll ans = 0;    for(int i = 0; i < maxn; i++) if(cntf[i]) ans += (ll)dp[cntf[i]][cnta[i]];    cout << ans << "\n";    return 0;}