题解 Set

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确实思路题

• 在 \(n\) 个非负整数中找出一个非空子集, 使得它的元素之和能被 \(n\) 整除，并输出方案
  求出前缀和, 前缀和最多只有 \(n\) 个取值, 但是一共有 \(S[0], S[1]..S[N]\) 这 \(n+1\) 个值, 所以一定有某两个 \(S[i], S[j]\) 相等, 找出来再把 \(i\) 到 \(j\) 输出就可以了

Code:

#include <bits/stdc++.h>
using namespace std;
#define INF 0x3f3f3f3f
#define N 1000010
#define ll long long
#define fir first
#define sec second
#define make make_pair
//#define int long long

char buf[1<<21], *p1=buf, *p2=buf;
#define getchar() (p1==p2&&(p2=(p1=buf)+fread(buf, 1, 1<<21, stdin)), p1==p2?EOF:*p1++)
inline int read() {
	int ans=0, f=1; char c=getchar();
	while (!isdigit(c)) {if (c=='-') f=-f; c=getchar();}
	while (isdigit(c)) {ans=(ans<<3)+(ans<<1)+(c^48); c=getchar();}
	return ans*f;
}

int n;

namespace force{
	bool vis[N];
	pair<int, int> back[N];
	int sta[N], top;
	void dfs(int u) {
		if (back[u].fir==-1) {printf("%d ", back[u].sec); return ;}
	}
	void solve() {
		for (int i=1,t; i<=n; ++i) {
			t=read()%n;
			for (int j=n-1; ~j; --j) if (vis[j] && back[j].sec!=i && !vis[(j+t)%n]) {
				vis[(j+t)%n]=1;
				back[(j+t)%n]=make(j, i);
			}
			if (!vis[t%n]) vis[t%n]=1, back[t%n]=make(-1, i);
			if (vis[0]) break;
		}
		// cout<<"pos1"<<endl;
		if (!vis[0]) {puts("-1"); exit(0);}
		int now=0;
		while (1) {
			// printf("%d ", back[now].sec);
			sta[++top]=back[now].sec;
			if (back[now].fir==-1) break;
			now=back[now].fir;
		}
		printf("%d\n", top);
		for (int i=1; i<=top; ++i) printf("%d ", sta[i]);
		printf("\n");
		exit(0);
	}
}

namespace task{
	ll sum[N];
	unordered_map<ll, int> mp;
	void solve() {
		for (int i=0; i<=n; ++i) {
			if (i) sum[i]=(sum[i-1]+read())%n;
			if (mp.find(sum[i])!=mp.end()) {
				printf("%d\n", i-mp[sum[i]]);
				for (int j=mp[sum[i]]+1; j<=i; ++j) printf("%d ", j);
				printf("\n");
				exit(0);
			}
			mp[sum[i]]=i;
		}
	}
}

signed main()
{
	freopen("a.in", "r", stdin);
	freopen("a.out", "w", stdout);

	n=read();
	// cout<<double(sizeof(force::back)+sizeof(force::sta)+sizeof(force::vis))/1024/1024<<endl;
	// force::solve();
	task::solve();
	
	return 0;
}