CodeForces 709A Juicer

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Dacc123

最新推荐文章于 2024-10-30 02:03:01 发布

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本文介绍了一道关于橙汁机制作过程中的算法问题。主人公Kolya将按顺序放入不超过限定大小的橙子榨汁，并在累积大小超过限定值时清空废料箱。文章探讨了如何计算在给定橙子大小列表的情况下需要清空废料箱的次数。

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Kolya is going to make fresh orange juice. He has n oranges of sizes a1, a2, ..., an. Kolya will put them in the juicer in the fixed order, starting with orange of size a1, then orange of size a2 and so on. To be put in the juicer the orange must have size not exceeding b, so if Kolya sees an orange that is strictly greater he throws it away and continues with the next one.

The juicer has a special section to collect waste. It overflows if Kolya squeezes oranges of the total size strictly greater than d. When it happens Kolya empties the waste section (even if there are no more oranges) and continues to squeeze the juice. How many times will he have to empty the waste section?

Input

The first line of the input contains three integers n, b and d (1 ≤ n ≤ 100 000, 1 ≤ b ≤ d ≤ 1 000 000) — the number of oranges, the maximum size of the orange that fits in the juicer and the value d, which determines the condition when the waste section should be emptied.

The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 1 000 000) — sizes of the oranges listed in the order Kolya is going to try to put them in the juicer.

Output

Print one integer — the number of times Kolya will have to empty the waste section.

Examples

input
2 7 10
5 6

output
1

input
1 5 10
7

output
0

input
3 10 10
5 7 7

output
1

input
1 1 1
1

output
0






#include <iostream>
#include <string.h>
#include <stdlib.h>
#include <algorithm>
#include <math.h>
#include <stdio.h>
#include <math.h>

using namespace std;
int n,b,d;
int a[100005];
int main()
{
    scanf("%d%d%d",&n,&b,&d);
    for(int i=1;i<=n;i++)
    {
        scanf("%d",&a[i]);
    }
    int ans=0;
    int sum=0;
    for(int i=1;i<=n;i++)
    {
        if(a[i]>b) continue;
        else sum+=a[i];
        if(sum>d)
            ans++,sum=0;
    }
    printf("%d\n",ans);
    return 0;
}