40分求调

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

const int N = 1e5 + 5, M = 105;

int n, ans;
int a[N], dp[M];

int gcd(int m, int n) {
	if (n) return gcd(n, m % n);
	return m;
}

bool pd(int *a) {
	int g = a[1];
	for (int i = 2; i <= n; i ++) {
		g = gcd(g, a[i]);
		if (g == 1) {
			return false;
		}
	}
	return g > 1;
}

int main() {
	scanf("%d", &n);
	for (int i = 1; i <= n; i ++) scanf("%d", &a[i]);

	if (pd(a)) {
		printf("INF");
		return 0;
	}

	dp[1] = 1;
	for (int i = 1; i <= n; i ++) 
		for (int j = a[i]; j <= M; j ++) 
			dp[j] = max(dp[j], dp[j - a[i]]);
	
	for (int i = 1; i <= M; i ++) 
		if (!dp[i]) ans ++;
	
	printf("%d\n", ans);
	return 0;
}