2 seconds

256 megabytes

standard input

standard output

Recenlty Luba got a credit card and started to use it. Let's consider n consecutive days Luba uses the card.

She starts with 0 money on her account.

In the evening of i-th day a transaction ai occurs. If ai > 0, then ai bourles are deposited to Luba's account. If ai < 0, then ai bourles are withdrawn. And if ai = 0, then the amount of money on Luba's account is checked.

In the morning of any of n days Luba can go to the bank and deposit any positive integer amount of burles to her account. But there is a limitation: the amount of money on the account can never exceed d.

It can happen that the amount of money goes greater than d by some transaction in the evening. In this case answer will be «-1».

Luba must not exceed this limit, and also she wants that every day her account is checked (the days when ai = 0) the amount of money on her account is non-negative. It takes a lot of time to go to the bank, so Luba wants to know the minimum number of days she needs to deposit some money to her account (if it is possible to meet all the requirements). Help her!

The first line contains two integers n, d (1 ≤ n ≤ 105, 1 ≤ d ≤ 109) —the number of days and the money limitation.

The second line contains n integer numbers a1, a2, ... an ( - 104 ≤ ai ≤ 104), where ai represents the transaction in i-th day.

Print -1 if Luba cannot deposit the money to her account in such a way that the requirements are met. Otherwise print the minimum number of days Luba has to deposit money.

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 const int inf=1e5+10;
 4 int n,a[inf],s[inf],tot,d,ans,hs[inf];
 5 struct ghb{
 6     int seat,ma,ans,Ma;
 7 }p[inf];
 8 int main()
 9 {
10 //    freopen("1.txt","r",stdin);
11     scanf("%d%d",&n,&d);
12     for(int i=1;i<=n;i++)scanf("%d",&a[i]);
13     int now=0,Max=-0x7fffffff;
14     for(int i=1;i<=n;i++){
15         now+=a[i];
16         Max=max(Max,now);
17         if(a[i]==0){
18             p[++tot].ans=now;
19             p[tot].seat=i;
20             hs[i]=tot;
21             p[tot].Ma=Max;
22             Max=-0x7fffffff;
23             now=0;
24         }
25     }
26     int tem1=0,tem2=-0x7fffffff;
27     for(int i=p[tot].seat;i<=n;i++){
28         tem1+=a[i];
29         tem2=max(tem2,tem1);
30     }
31     p[tot].ma=min(d-tem2,d);
32     if(p[tot].ma<0){
33         printf("-1\n");
34         return 0;
35     }
36     for(int i=tot-1;i>=1;i--){
37         p[i].ma=min(d-p[i+1].Ma,p[i+1].ma-p[i+1].ans);
38         p[i].ma=min(p[i].ma,d);
39         if(p[i].ma<0){
40             printf("-1\n");
41             return 0;
42         }
43     }
44     now=0;
45     for(int i=1;i<=n;i++){
46         now+=a[i];
47         if(now>d){
48             printf("-1\n");
49             return 0;
50         }
51         if(a[i]==0){
52             if(now<0){
53                 now=p[hs[i]].ma;
54                 ans++;
55             }
56         }
57     }
58     printf("%d\n",ans);
59     return 0;
60 }