每次先走一圈,计算出可以购买的糖果的总价格 c 和总个数 t,然后 cnt += t*(T/c) T %= c,循环至无糖果可买。
时间复杂度计算如下:
#include<bits/stdc++.h>
using namespace std;
const int maxn = 200000 + 5;
typedef long long ll;
int n,p[maxn];
ll T;
void get(ll &T,ll &c,ll &cnt){
for(int i = 0; i < n;i++){
if(T >= p[i]){
T -= p[i];
c += p[i];
cnt++;
}
}
}
int main()
{
// freopen("data.in","r",stdin);
// freopen("data.out","w",stdout);
scanf("%d%lld",&n,&T);
ll cnt = 0;
int m = 1<<30;
for(int i = 0; i < n; i++){
scanf("%d",&p[i]);
m = min(m,p[i]);
}
while(T >= m){
ll c = 0, t = 0;
get(T,c,t);
cnt += t*(T/c) + t;
T %= c;
}
printf("%lld\n",cnt);
return 0;
}