code force 893D Credit Card
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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 0, then ai bourles are withdrawn.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.
Example
5 10
-1 5 0 -5 3
0
3 4
-10 0 20
-1
5 10
-5 0 10 -11 0
2
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:看了很多人的代码,终于明白了,题意在上面,其实意思就是为了让自己每次去查账的时候资金数不为负,所以要不定期的去存储钱,但是现在问最少次数
所以参考了别人的思路之后就是说设置两个变量去记录最大值和最小值,当去检查资金的时候假如最大值都小于0说明你可以在查资金的前一天去存储一次,如果最小值小于0,此时最小值应该重新置为0,而资金就一次存入最大值,减少后面的存储次数. 在不查询资金的时候那么最大值与最小值都加上这天的资金数,并且如果当最小值都大于限制的最大资金的话就可以直接输出-1了,并且要对最大值一直进行更行.#include <iostream>#include <cstdio>#include <stack>using namespace std;int n,d,a[100000+20];int main(){ scanf("%d%d",&n,&d); for(int i=0;i<n;i++) scanf("%d",&a[i]); int minn=0,maxx=0;//当前账户的最小值和最大值 int ans = 0; for(int i=0;i<n;i++) { if(a[i]==0) { //如果最小值小于零,则将最小值重新置为零 if(minn<0)minn=0; //如果最大值小于零,则将最大值置为最大额度,次数加一 //表示第i天早上去了一次银行,并且将账户存到最大额度 if(maxx<0){maxx=d;ans++;} } else { //每次都要更新当前账户的最值 minn+=a[i]; maxx+=a[i]; if(minn>d)//如果最小值大于最大额度输出-1 { printf("-1\n"); return 0; } if(maxx>d)//如果最大值大于最大额度则将最大值置为最大额度 { maxx=d; } } } printf("%d\n",ans); return 0;}
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