最大公约数.nd

#include<bits/stdc++.h>
using namespace std;

#define rep(i,l,r) for(int i=(l);i<=(r);i++)
#define per(i,l,r) for(int i=(l);i>=(r);i--)
#define ll long long
#define mset(s,t) memset(s,t,sizeof(t))
#define mcpy(s,t) memcpy(s,t,sizeof(t))
#define fi first
#define se second
#define pb push_back
#define all(x) (x).begin(),(x).end()
#define SZ(x) ((int)(x).size())
#define mp make_pair

typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef vector<int> vi;   
typedef vector<ll> Vll;               
typedef vector<pair<int, int> > vpii;
typedef vector<pair<ll, ll> > vpll;                        

const ll mod = 1e9 + 7;
//const ll mod = 998244353;
const double pi  = acos(-1.0);

inline ll qmi (ll a, ll b) {
    ll ans = 1;
    while (b) {
        if (b & 1) ans = ans * a%mod;
        a = a * a %mod;
        b >>= 1;
    }
    return ans;
}
inline int read () {
    int x = 0, f = 0;
    char ch = getchar();
    while (!isdigit(ch)) f |= (ch=='-'),ch= getchar();
    while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
    return f?-x:x;
}
template<typename T> void print(T x) {
    if (x < 0) putchar('-'), x = -x;
    if (x >= 10) print(x/10);
    putchar(x % 10 + '0');
}
inline ll sub (ll a, ll b) {
    return ((a - b ) %mod + mod) %mod;
}
inline ll add (ll a, ll b) {
    return (a + b) %mod;
}
inline ll inv (ll a) {
    return qmi(a, mod - 2);
}
#define int long long
const int N = 1e5 + 10;
int ans[N];
int a[N], pre[N];
int n;
void cal (int x) {
    map<int,int>cnt;
    int tot = 0;
    for (int i = 1; i <= n ;i ++) {
        cnt[pre[i] % x] ++;
        tot = max (tot, cnt[pre[i] % x]);
    }
    ans[tot] = max (ans[tot], x);
}
void solve() {
    cin >> n;
    for (int i = 1; i <= n; i ++)
        cin >> a[i];
    for (int i =1; i <= n; i ++)
        pre[i] = pre[ i -1] + a[i];
    int sq = sqrt(pre[n]);
    for (int i = 1; i <= sq; i ++) {
        if (pre[n] % i == 0) {
            cal(i);
            cal(pre[n] / i);
        }
    }
    for (int i = n; i >= 1; i --)
        ans[i] =max(ans[i], ans[i + 1]);
    for (int i =1; i <= n; i ++)
        cout << ans[i]<<endl;
}
signed main () {
    // ios::sync_with_stdio(0),cin.tie(0), cout.tie(0);
    int t;
    t =1;
    //cin >> t;
    while (t --) solve();
    return 0;
}

题意:给定一个环，换上有数，求划分成i段的最漂亮的值

问题结构:转化形式
为什么还要求一遍后缀最大值。因为并不能保证所有的i都被遍历到。对于ans[i]可以用ans[i + 1]的组合出来。并且所以ans[i] = max (ans[i], ans[i  +1]);