B. Vanya and Food Processor【转】
B. Vanya and Food Processortime limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard outputVanya 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.
InputThe 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.
OutputPrint a single integer — the number of seconds required to smash all the potatoes following the process described in the problem statement.
Examplesinput5 6 3
5 4 3 2 1output5input5 6 3
5 5 5 5 5output10input5 6 3
1 2 1 1 1output2NoteConsider 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.
题解代码:
1 #include <iostream> 2 #include <stdio.h> 3 using namespace std; 4 typedef long long ll; 5 ll i,n,h,ans,x,cur_h,k; 6 int main() 7 { 8 cin >> n >> h >> k; 9 ans = 0; 10 cur_h = 0; 11 for (i = 0; i < n; i++) 12 { 13 scanf("%I64d", &x); 14 if (cur_h + x <= h) 15 cur_h += x; 16 else 17 ans++, cur_h = x; 18 ans += cur_h/k; 19 cur_h %= k; 20 } 21 ans += cur_h/k; 22 cur_h %= k; 23 ans += (cur_h>0); 24 cout << ans << endl; 25 return 0; 26 }