题目链接：

题解：直接二分答案然后再判断（a-b）来替代不足的。看代码比较好理解，水题。

#include 
    
     #include 
     
      #include 
      
       using namespace std;typedef long long ll;const int M = 1e5 + 10;ll num[M];ll n , a , b;bool cheak(ll x) {    ll cn = a - b;    ll gl = 0;    for(int i = 0 ; i < n ; i++) {        ll gg = num[i] - b * x;        if(gg <= 0) continue;        else {            gl += gg / cn;            gl += (gg % cn != 0);            if(gl > x) return false;        }    }    if(gl > x) return false;    return true;}int main() {    scanf("%lld%lld%lld" , &n , &a , &b);    for(int i = 0 ; i < n ; i++) {        scanf("%lld" , &num[i]);    }    ll l = 0 , r = 1e9;    ll ans = r;    while(l <= r) {        ll mid = (l + r) >> 1;        if(cheak(mid)) r = mid - 1 , ans = min(mid , ans);        else l = mid + 1;    }    printf("%lld\n" , ans);    return 0;}