解析:
维护一个sum(一圈下来的和),循环内,每次先把可以整圈的部分算了(除法取模)
不能整圈(sum>m)的情况,显然某个前缀的sum已经>m了,所以我们二分这个前缀(动态前缀和使用树状数组维护),将第一个大于m的前缀的那个位置删掉。
详情看代码:
/*
* Author : Jk_Chen
* Date : 2020-04-02-20.14.15
*/
#include<bits/stdc++.h>
using namespace std;
#define LL long long
#define rep(i,a,b) for(int i=(int)(a);i<=(int)(b);i++)
#define per(i,a,b) for(int i=(int)(a);i>=(int)(b);i--)
#define mmm(a,b) memset(a,b,sizeof(a))
#define pb push_back
#define pill pair<int, int>
#define fi first
#define se second
void test(){cerr<<"\n";}
template<typename T,typename... Args>void test(T x,Args... args){cerr<<x<<" ";test(args...);}
const LL mod=1e9+7;
const int maxn=1e5+9;
const int inf=0x3f3f3f3f;
LL rd(){ LL ans=0; char last=' ',ch=getchar();
while(!(ch>='0' && ch<='9'))last=ch,ch=getchar();
while(ch>='0' && ch<='9')ans=ans*10+ch-'0',ch=getchar();
if(last=='-')ans=-ans; return ans;
}
#define rd rd()
/*_________________________________________________________begin*/
int n;
LL a[maxn];
LL tr[maxn];
void update(int pos,LL val){
while(pos<=n){
tr[pos]+=val;
pos+=pos&-pos;
}
}
LL query(int pos){
LL res=0;
while(pos>0){
res+=tr[pos];
pos-=pos&-pos;
}
return res;
}
int main(){
n=rd;
LL m=rd;
LL sum=0,res=n;
rep(i,1,n){
a[i]=rd;
sum+=a[i];
update(i,a[i]);
}
LL ans=0;
while(res>0&&m){
LL lun=m/sum;
m=m%sum;
ans+=lun*res;
int l=1,r=n,A=-1;
while(l<=r){
int mid=(l+r)>>1;
if(query(mid)>m)A=mid,r=mid-1;
else l=mid+1;
}
update(A,-a[A]);
res--;
sum-=a[A];
continue;
}
printf("%lld\n",ans);
return 0;
}
/*_________________________________________________________end*/