2022 robocom world robot developer competition - undergraduate group (provincial competition) -- question 1: don't waste gold (finished)

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RC-u1 Don't waste gold

Zhe Zhe is playing a game recently , You can get gold coins by killing monsters —— Here, remember to kill the first i The number of gold coins obtained by monsters is Pi.

However, there is a limit to the number of gold coins allowed in this game , When over , The part exceeding the upper limit will be eaten openly by the system , Zhe zhe can't get it .

In order not to waste gold , Zhe zhe decides , The next monster to be killed can get gold coins, which will cause the number of gold coins he has to exceed the limit , Just spend once , Use up all the gold coins you have .

Now, given the number of gold coins corresponding to a series of monsters that zhe zhe will kill , Please calculate how many times zhe zhe spent .

Input format :
The first line of input is two integers N,M （1≤N≤1e3 ,1≤M≤1e6）, Indicates the number of monsters killed and the maximum number of gold coins allowed by the system .

The next line is separated by spaces N Number Pi（i=1,⋯,N）, It means to kill the first i The number of gold coins a monster can get .
Suppose zhe zhe kills monsters in the order of input , And each Pi All are No more than 1e6 Non-negative integer .

Output format :
Output the number of times philosophers spend in one line .

** sample input :**

10 10
1 2 3 4 1 2 3 5 11 1

sample output :

4

Sample explanation ：
The consumption time point is ： After the fourth monster is killed 、 After the seventh monster is killed 、 After the eighth monster is killed 、 After the ninth monster is killed .

Answer key

The subject is very simple , Simulate with a loop . We are in the process of summing the array , whenever sum Add the value of the current element to exceed m, We will take sum Set up 0, At the same time, the counter is incremented 1

AC Code

#include <bits/stdc++.h>

using namespace std;

const int N = 1010;
int arr[N];

int main()
{
    
    int n, m;
    cin >> n >> m;
    for(int i=1; i<=n; i++) cin >> arr[i];

    int res = 0;
    int sum = 0;
    for(int i=1; i<=n; i++)
    {
    
        if(sum + arr[i] > m)
        {
    
            res++;
            sum = 0;
        }
        sum += arr[i];
    }
    cout << res << endl;
    return 0;
}

The above code can AC, Welcome to discuss and exchange any questions