#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 pii pair<int, int>
#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
const ll mod = 1e9 + 7;
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
void solve() {
int n;
cin >> n;
vector<pii> a(n);
int cnt = 0;
for (int i = 0 ;i < n ;i ++) {
int t;
cin >> t;
cnt += t;
a[i] = mp(t, i);
}
sort(all(a));
int s = n;
vector<int> ans;
bool f= 0;
for (int s = (cnt / n) * n; s ; s -= n) {
int t = s;
bool f = 0;
ans.clear();
for (int i = n - 1; i >= 0; i --) {
if (s - a[i].fi >= 0) {
s -= a[i].fi;
ans.pb(a[i].se + 1);
}
if (s == 0) {
f = 1;
break;
}
}
if (f) {
sort(all(ans));
cout << ans.size() << endl;
for (auto t : ans) cout << t << " ";
return;
}
else s = t;
}
puts("-1");
}
signed main () {
int t;
t =1;
//cin >> t;
while (t --) solve();
return 0;
}
```#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 pii pair<int, int>
#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
const ll mod = 1e9 + 7;
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
void solve() {
int n;
int a[1000005];
memset(a, 0, sizeof a);
cin >> n;
for (int i= 1; i<= n ;i ++) {
cin >> a[i];
a[i] =(a[i - 1] + a[i]) % n;
}
map<int, int> st;
st[0] = 0;
for (int i = 1; i <= n ;i ++) {
a[i] %= n;
if (st.count(a[i])) {
cout << i - st[a[i]] << endl;
for (int j = st[a[i]] + 1; j <= i; j ++) cout << j << " ";
puts("");
return;
}
st[a[i]] =i;
}
puts("-1");
}
signed main () {
int t;
t =1;
//cin >> t;
while (t --) solve();
return 0;
}
<p>
通过从大到小进行选择。对于每一种N，我们判断一下是否能够得到0，如果能就输出现在这个答案。否则最终输出-1
解法2
我们可以利用前缀和前缀和有n+1个结果，取模只有n个结果，所以必然有两个相同.这两个相同的数说构成一个周期.选择这个区间内的数就是n的倍数
</p>
