Educational Codeforces Round 33 893D - Credit Card

最新推荐文章于 2020-08-06 11:10:19 发布

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最新推荐文章于 2020-08-06 11:10:19 发布

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本文介绍了一个信用卡收支管理问题，主人公Luba需要确保在一系列交易中，账户余额既不超过上限也不在检查时为负值。文章详细阐述了解决方案，包括如何通过适时地存款来满足这些条件，并给出具体的实现代码。

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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 0, then ai bourles are withdrawn.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!

Input

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.

Output

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.

Examples

input
5 10
-1 5 0 -5 3

output
0

input
3 4
-10 0 20

output
-1

input
5 10
-5 0 10 -11 0

output
2

题意：信用卡n天交易记录，第i天可能赚钱，可能赔钱，可以查账，赚钱后，帐户总余额不能超过上限，超过就

不符合要求输出-1，查账的时候，不希望余额是负值，如果哪天赔钱后，可以去银行存任意数量的钱，把余额变成大

于等于0.希望查账的日子都不会查到负数。

思路：开始找不到哪里是可以决策的，以为按照步骤模拟就可以了，但是错在了test20。后来想到了，在每次存钱的

时候可以存多点，这样赔钱数比较少的情况下，下次查账也是正值，不用去银行。所以每天都有一个区间记录最大

余额，与最小余额。

查账时如果最大余额都是负数，说明之前再怎么存更多的钱都会出现负值，那就需要去银行存钱，这之后，余额区间

就需要更新，最大值改为d，最小值至少是0，因为查账不能是负数，如果原来就是大于零的数，则不处理。

不是查账情况，区间端点加上当前值。如果最大值大于d，则改为d，如果最小值值大于d，说明按规则存钱，

最小也会超限，则输出-1。

从区间的变化看上限与下限的差距只会越来越小，而不是越来越大，所以不会出现当前满足，但是之前不满足的

情况。

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
#define INF 0x3f3f3f3f
#define FI first
#define SE second
int mod = 1000000007;
const int N = 100010;
int rec[N];
int main() 
{
    int n, d;
    while (~scanf("%d%d", &n, &d))
    {
        for (int i = 0; i < n; i++)
            scanf("%d", rec + i);
        int top, bottom, ans = 0;
        bool flag = true;
        top = bottom = 0;
        for (int i = 0; i < n; i++)
        {
            if (rec[i] == 0)
            {
                if (top < 0)
                {
                    ans++;
                    top = d;
                }
                bottom = max(0, bottom);
            }else
            {
                top += rec[i];
                bottom += rec[i];

                top = min(top, d);
                if (bottom > d)
                {
                    flag = false;
                    break;
                }
            }
        }
        printf("%d\n", flag ? ans: -1);
    }

    return 0;
}