AT_abc200_d [ABC200D] Happy Birthday! 2 Solution

AT_abc200_d [ABC200D] Happy Birthday! 2 Solution

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train of thought

We can prove that as long as $N\ge 8$ , then there must be a solution.

The proof is as follows:

• 8 8 The subsequences that can be composed of$8$$2^8-1=255$ species. (Each element can be selected or not selected, remove all unselected cases)

• According to the drawer principle, we put $255$ seed sequences according to their division by$If\; the\; remainder\; of\; 200$ is put into drawers respectively, then at least two subsequences are in one drawer, that is, there must be legal$A$ and$B$。

当 $N<At\; 8$ o'clock, we violently enumerate all possibilities;

This $N\ge 8$ , we violently enumerate any8 of them 8All the possibilities composed of 8 elements can find the solution$.$

the code

#include <bits/stdc++.h>
#define lowbit(x) x & (-x)
#define endl "\n"
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
namespace fastIO {
    
    
	inline int read() {
    
    
		register int x = 0, f = 1;
		register char c = getchar();
		while (c < '0' || c > '9') {
    
    
			if(c == '-') f = -1;
			c = getchar();
		}
		while (c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
		return x * f;
	}
	inline void write(int x) {
    
    
		if(x < 0) putchar('-'), x = -x;
		if(x > 9) write(x / 10);
		putchar(x % 10 + '0');
		return;
	}
}
inline void print(const vector<int>& v) {
    
    
	printf("%llu", v.size());
	for(int x: v) {
    
    
		printf(" %d", x + 1);
	}
	putchar('\n');
}
using namespace fastIO;
int a[15];
vector<int> bkt[200];
int main() {
    
    
	//freopen(".in","r",stdin);
	//freopen(".out","w",stdout);
	int n;
	n = read();
	if(n > 8) {
    
    
		n = 8;	
	}
	for(int i = 0; i < n; i ++) {
    
    
		scanf("%d", a + i);
	}
	int lim = 1 << n;
	for(int st = 0; st < lim; st ++) {
    
    
		int s = 0;
		vector<int> pos;
		for(int i = 0; i < n; i ++) {
    
    
			if(st >> i & 1) {
    
    
				s = (s + a[i]) % 200;
				pos.push_back(i);
			}
		}				
		if(!bkt[s].empty()) {
    
    
			puts("Yes");
			print(bkt[s]);
			print(pos);
			return 0;
		}
		else bkt[s] = pos;
	}
	puts("No");
	return 0;
}