codeforces 677B B. Vanya and Food Processor(模拟)

题目链接：

B. Vanya and Food Processor

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Vanya smashes potato in a vertical food processor. At each moment of time the height of the potato in the processor doesn't exceed hand the processor smashes k centimeters of potato each second. If there are less than k centimeters remaining, than during this second processor smashes all the remaining potato.

Vanya has n pieces of potato, the height of the i-th piece is equal to ai. He puts them in the food processor one by one starting from the piece number 1 and finishing with piece number n. Formally, each second the following happens:

1. If there is at least one piece of potato remaining, Vanya puts them in the processor one by one, until there is not enough space for the next piece.
2. Processor smashes k centimeters of potato (or just everything that is inside).

Provided the information about the parameter of the food processor and the size of each potato in a row, compute how long will it take for all the potato to become smashed.

 

Input

 

The first line of the input contains integers n, h and k (1 ≤ n ≤ 100 000, 1 ≤ k ≤ h ≤ 109) — the number of pieces of potato, the height of the food processor and the amount of potato being smashed each second, respectively.

The second line contains n integers ai (1 ≤ ai ≤ h) — the heights of the pieces.

 

Output

 

Print a single integer — the number of seconds required to smash all the potatoes following the process described in the problem statement.

 

Examples

 

input

5 6 3
5 4 3 2 1

output

5

input

5 6 3
5 5 5 5 5

output

10

input

5 6 3
1 2 1 1 1

output

2

题意：

有这么多高为a[i]的土豆,每次最多放h高度的土豆,超过了就不能放进去了,每秒削k高度的,问这些得用多长时间;

思路：

模拟削土豆的过程算一下时间就好了;

AC代码：

#include <bits/stdc++.h>
/*#include <vector>
#include <iostream>
#include <queue>
#include <cmath>
#include <map>
#include <cstring>
#include <algorithm>
#include <cstdio>
*/
using namespace std;
#define Riep(n) for(int i=1;i<=n;i++)
#define Riop(n) for(int i=0;i<n;i++)
#define Rjep(n) for(int j=1;j<=n;j++)
#define Rjop(n) for(int j=0;j<n;j++)
#define mst(ss,b) memset(ss,b,sizeof(ss));
typedef long long LL;
template<class T> void read(T&num) {
    char CH; bool F=false;
    for(CH=getchar();CH<'0'||CH>'9';F= CH=='-',CH=getchar());
    for(num=0;CH>='0'&&CH<='9';num=num*10+CH-'0',CH=getchar());
    F && (num=-num);
}
int stk[70], tp;
template<class T> inline void print(T p) {
    if(!p) { puts("0"); return; }
    while(p) stk[++ tp] = p%10, p/=10;
    while(tp) putchar(stk[tp--] + '0');
    putchar('
');
}

const LL mod=1e9+7;
const double PI=acos(-1.0);
const LL inf=1e10;
const int N=1e5+15;

int n,h,k;
int a[N];
int main()
{
    read(n);read(h);read(k);
    Riep(n)read(a[i]);
    LL ans=0,sum=0;
    Riep(n)
    {
        if(sum+a[i]>h)
        {
            if(sum%k==0)ans=ans+sum/k;
            else ans=ans+sum/k+1;
            ans=ans+a[i]/k;
            sum=a[i]%k;
        }
        else
        {
            sum=sum+a[i];
            ans=ans+sum/k;
            sum=sum%k;
        }
    }
    if(sum>0)ans++;
print(ans);
    return 0;
}