CodeForce edu round 53 Div 2. D:Berland Fair

D. Berland Fair

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

XXI Berland Annual Fair is coming really soon! Traditionally fair consists of $\mathrm{nn booths,\; arranged\; in\; a\; circle.\; The\; booths\; are\; numbered 11through nn clockwise\; with nn being\; adjacent\; to 11.\; The ii-th\; booths\; sells\; some\; candies\; for\; the\; price\; of aiai burles\; per\; item.\; Each\; booth\; has\; an\; unlimited\; supply\; of\; candies.}$

Polycarp has decided to spend at most $\mathrm{TT burles\; at\; the\; fair.\; However,\; he\; has\; some\; plan\; in\; mind\; for\; his\; path\; across\; the\; booths:}$

• at first, he visits booth number $\mathrm{11;}$
• if he has enough burles to buy exactly one candy from the current booth, then he buys it immediately; 
• then he proceeds to the next booth in the clockwise order (regardless of if he bought a candy or not). 

Polycarp's money is finite, thus the process will end once he can no longer buy candy at any booth.

Calculate the number of candies Polycarp will buy.

Input

The first line contains two integers $\mathrm{nn and TT (1\le n\le 2\cdot 1051\le n\le 2\cdot 105, 1\le T\le 10181\le T\le 1018)\; —\; the\; number\; of\; booths\; at\; the\; fair\; and\; the\; initial\; amount\; of\; burles\; Polycarp\; has.}$

The second line contains $\mathrm{nn integers a1,a2,\dots ,ana1,a2,\dots ,an (1\le ai\le 1091\le ai\le 109)\; —\; the\; price\; of\; the\; single\; candy\; at\; booth\; number ii.}$

Output

Print a single integer — the total number of candies Polycarp will buy.

Examples

input

Copy

3 38
5 2 5

output

Copy

10

input

Copy

5 21
2 4 100 2 6

output

Copy

6

Note

Let's consider the first example. Here are Polycarp's moves until he runs out of money:

1. Booth $\mathrm{11,\; buys\; candy\; for 55, T=33T=33;}$
2. Booth $\mathrm{22,\; buys\; candy\; for 22, T=31T=31;}$
3. Booth $\mathrm{33,\; buys\; candy\; for 55, T=26T=26;}$
4. Booth $\mathrm{11,\; buys\; candy\; for 55, T=21T=21;}$
5. Booth $\mathrm{22,\; buys\; candy\; for 22, T=19T=19;}$
6. Booth $\mathrm{33,\; buys\; candy\; for 55, T=14T=14;}$
7. Booth $\mathrm{11,\; buys\; candy\; for 55, T=9T=9;}$
8. Booth $\mathrm{22,\; buys\; candy\; for 22, T=7T=7;}$
9. Booth $\mathrm{33,\; buys\; candy\; for 55, T=2T=2;}$
10. Booth $\mathrm{11,\; buys\; no\; candy,\; not\; enough\; money;}$
11. Booth $\mathrm{22,\; buys\; candy\; for 22, T=0T=0.}$

No candy can be bought later. The total number of candies bought is $1010.$

In the second example he has $\mathrm{11 burle\; left\; at\; the\; end\; of\; his\; path,\; no\; candy\; can\; be\; bought\; with\; this\; amount.}$

之前用线段树做的。超时了很难受。因为没有有效的处理无效的点。之后借鉴了网上的答案。改成链表去做。链表有些不熟悉了。比如删除结点那里都搞错了。

#include <iostream>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <cstdio>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <stack>
#define ll long long
//#define local

using namespace std;

const int MOD = 1e9+7;
const int inf = 0x3f3f3f3f;
const double PI = acos(-1.0);
const int maxn = 2e5+10;

ll a[maxn];
ll t;
int n;
int nex[maxn];
int pre[maxn];

int main() {
#ifdef local
    if(freopen("/Users/Andrew/Desktop/data.txt", "r", stdin) == NULL) printf("can't open this file!
");
#endif
    
    scanf("%d%lld", &n , &t);
    int num = 1;
    for (int i = 0; i < n; ++i) {
        ll tmp;
        scanf("%lld", &tmp);
        if (tmp > t) continue;
        a[num++] = tmp;
    }
    n = num-1;
    if (n == 0) {
        printf("0
");
        return 0;
    }
    for (int i = 0; i < n; ++i) {
        nex[i] = i+1;
    }
    nex[n] = -1;
    for (int i = 1; i <= n; ++i)
        pre[i] = i-1;
    ll cnt = 0;
    while (nex[0] != -1) {
        ll sum = 0;
        int num = 0;
        ll rem = t;
        for (int i = nex[0]; i != -1; i = nex[i]) {
            if (a[i] <= rem) {
                sum += a[i];
                num++;
                rem -= a[i];
            } else {
                nex[pre[i]] = nex[i];
                pre[nex[i]] = pre[i];
            }
        }
        if (sum == 0) break;
        cnt += t/sum*num;
        t %= sum;
    }
    printf("%lld
", cnt);
#ifdef local
    fclose(stdin);
#endif
    return 0;
}