Codeforces Round #355 (Div. 2) B. Vanya and Food Processor(贪心)
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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:
- 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.
- 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.
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.
Print a single integer — the number of seconds required to smash all the potatoes following the process described in the problem statement.
5 6 35 4 3 2 1
5
5 6 35 5 5 5 5
10
5 6 31 2 1 1 1
2
Consider the first sample.
- First Vanya puts the piece of potato of height 5 into processor. At the end of the second there is only amount of height 2 remaining inside.
- Now Vanya puts the piece of potato of height 4. At the end of the second there is amount of height 3 remaining.
- Vanya puts the piece of height 3 inside and again there are only 3 centimeters remaining at the end of this second.
- Vanya finally puts the pieces of height 2 and 1 inside. At the end of the second the height of potato in the processor is equal to 3.
- During this second processor finally smashes all the remaining potato and the process finishes.
In the second sample, Vanya puts the piece of height 5 inside and waits for 2 seconds while it is completely smashed. Then he repeats the same process for 4 other pieces. The total time is equal to 2·5 = 10 seconds.
In the third sample, Vanya simply puts all the potato inside the processor and waits 2 seconds.
题解:有一个榨汁机,有n个potato,一个一个地放进去,榨汁机可以一次性榨掉的高度为h,每秒钟可以榨k米,问榨完最少需要多少时间。
贪心呗,从最小的开始榨,然后就是数学了。。
AC代码;
#include<bits/stdc++.h>using namespace std;typedef long long LL;LL a[100010]={0};int main(){LL n,h,k;while(cin>>n>>h>>k){for(int i=0;i<n;i++){scanf("%I64d",&a[i]);}LL sum=0,ans=0;for(int i=0;i<n;i++){if(sum+a[i]<=h) sum+=a[i];else{sum=a[i];ans++;}LL t=sum/k;ans+=t;sum-=t*k;}if(sum)ans++;printf("%I64d\n",ans);}return 0;}
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