【CodeForces】508C - Anya and Ghosts(贪心 & 模拟)
2016-08-15 12:07
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C. Anya and Ghosts
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
Anya loves to watch horror movies. In the best traditions of horror, she will be visited by m ghosts tonight. Anya has lots of candles
prepared for the visits, each candle can produce light for exactly t seconds. It takes the girl one second to light one candle. More
formally, Anya can spend one second to light one candle, then this candle burns for exactly t seconds and then goes out and can no longer
be used.
For each of the m ghosts Anya knows the time at which it comes: the i-th
visit will happen wi seconds
after midnight, all wi's
are distinct. Each visit lasts exactly one second.
What is the minimum number of candles Anya should use so that during each visit, at least r candles are burning? Anya can start to light
a candle at any time that is integer number of seconds from midnight, possibly, at the time before midnight. That means, she can start to light a candle integer number of seconds before midnight or integer
number of seconds after a midnight, or in other words in any integer moment of time.
Input
The first line contains three integers m, t, r (1 ≤ m, t, r ≤ 300),
representing the number of ghosts to visit Anya, the duration of a candle's burning and the minimum number of candles that should burn during each visit.
The next line contains m space-separated numbers wi (1 ≤ i ≤ m, 1 ≤ wi ≤ 300),
the i-th of them repesents at what second after the midnight the i-th
ghost will come. All wi's
are distinct, they follow in the strictly increasing order.
Output
If it is possible to make at least r candles burn during each visit, then print the minimum number of candles that Anya needs to light
for that.
If that is impossible, print - 1.
Examples
input
output
input
output
input
output
Note
Anya can start lighting a candle in the same second with ghost visit. But this candle isn't counted as burning at this visit.
It takes exactly one second to light up a candle and only after that second this candle is considered burning; it means that if Anya starts lighting candle at moment x, candle is buring from second x + 1 to second x + t inclusively.
In the first sample test three candles are enough. For example, Anya can start lighting them at the 3-rd, 5-th
and 7-th seconds after the midnight.
In the second sample test one candle is enough. For example, Anya can start lighting it one second before the midnight.
In the third sample test the answer is - 1, since during each second at most one candle can burn but Anya needs three candles to light up the
room at the moment when the ghost comes.
模拟+贪心就行了。
首先来一个特判,如果需要的蜡烛数大于蜡烛的燃烧时间则肯定不行。
然后从鬼拜访的前一秒开始点蜡烛,一直往前知道满足条件,用一个数组表示当前秒蜡烛亮着的数量。
代码如下:
#include <stdio.h>
#include <cstring>
#include <algorithm>
using namespace std;
#define CLR(a,b) memset(a,b,sizeof(a))
#define INF 0x3f3f3f3f
#define LL long long
int main()
{
int n,l,m,ans;
int num[311];
int g[311];
scanf ("%d %d %d",&n,&l,&m);
for (int i = 1 ; i <= n ; i++)
scanf ("%d",&num[i]);
if (m > l)
{
printf ("-1\n");
return 0;
}
CLR(g,0);
ans = 0;
for (int i = 1 ; i <= n ; i++)
{
int st = num[i];
while (g[num[i]] < m)
{
for (int j = 0 ; j < l && j+st <= num
; j++)
if (st + j >= 1) //从负数的时间点开始点蜡烛(这个居然还可以提前)
g[st+j]++;
st--;
}
ans += num[i] - st;
}
printf ("%d\n",ans);
return 0;
}
C. Anya and Ghosts
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
Anya loves to watch horror movies. In the best traditions of horror, she will be visited by m ghosts tonight. Anya has lots of candles
prepared for the visits, each candle can produce light for exactly t seconds. It takes the girl one second to light one candle. More
formally, Anya can spend one second to light one candle, then this candle burns for exactly t seconds and then goes out and can no longer
be used.
For each of the m ghosts Anya knows the time at which it comes: the i-th
visit will happen wi seconds
after midnight, all wi's
are distinct. Each visit lasts exactly one second.
What is the minimum number of candles Anya should use so that during each visit, at least r candles are burning? Anya can start to light
a candle at any time that is integer number of seconds from midnight, possibly, at the time before midnight. That means, she can start to light a candle integer number of seconds before midnight or integer
number of seconds after a midnight, or in other words in any integer moment of time.
Input
The first line contains three integers m, t, r (1 ≤ m, t, r ≤ 300),
representing the number of ghosts to visit Anya, the duration of a candle's burning and the minimum number of candles that should burn during each visit.
The next line contains m space-separated numbers wi (1 ≤ i ≤ m, 1 ≤ wi ≤ 300),
the i-th of them repesents at what second after the midnight the i-th
ghost will come. All wi's
are distinct, they follow in the strictly increasing order.
Output
If it is possible to make at least r candles burn during each visit, then print the minimum number of candles that Anya needs to light
for that.
If that is impossible, print - 1.
Examples
input
1 8 3 10
output
3
input
2 10 1 5 8
output
1
input
1 1 310
output
-1
Note
Anya can start lighting a candle in the same second with ghost visit. But this candle isn't counted as burning at this visit.
It takes exactly one second to light up a candle and only after that second this candle is considered burning; it means that if Anya starts lighting candle at moment x, candle is buring from second x + 1 to second x + t inclusively.
In the first sample test three candles are enough. For example, Anya can start lighting them at the 3-rd, 5-th
and 7-th seconds after the midnight.
In the second sample test one candle is enough. For example, Anya can start lighting it one second before the midnight.
In the third sample test the answer is - 1, since during each second at most one candle can burn but Anya needs three candles to light up the
room at the moment when the ghost comes.
模拟+贪心就行了。
首先来一个特判,如果需要的蜡烛数大于蜡烛的燃烧时间则肯定不行。
然后从鬼拜访的前一秒开始点蜡烛,一直往前知道满足条件,用一个数组表示当前秒蜡烛亮着的数量。
代码如下:
#include <stdio.h>
#include <cstring>
#include <algorithm>
using namespace std;
#define CLR(a,b) memset(a,b,sizeof(a))
#define INF 0x3f3f3f3f
#define LL long long
int main()
{
int n,l,m,ans;
int num[311];
int g[311];
scanf ("%d %d %d",&n,&l,&m);
for (int i = 1 ; i <= n ; i++)
scanf ("%d",&num[i]);
if (m > l)
{
printf ("-1\n");
return 0;
}
CLR(g,0);
ans = 0;
for (int i = 1 ; i <= n ; i++)
{
int st = num[i];
while (g[num[i]] < m)
{
for (int j = 0 ; j < l && j+st <= num
; j++)
if (st + j >= 1) //从负数的时间点开始点蜡烛(这个居然还可以提前)
g[st+j]++;
st--;
}
ans += num[i] - st;
}
printf ("%d\n",ans);
return 0;
}
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