【CodeForces】699C - Vacations(贪心)
2016-07-26 21:31
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C. Vacations
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output
Vasya has n days of vacations! So he decided to improve his IT skills and do sport. Vasya knows the following information about each
of this n days: whether that gym opened and whether a contest was carried out in the Internet on that day. For the i-th
day there are four options:
on this day the gym is closed and the contest is not carried out;
on this day the gym is closed and the contest is carried out;
on this day the gym is open and the contest is not carried out;
on this day the gym is open and the contest is carried out.
On each of days Vasya can either have a rest or write the contest (if it is carried out on this day), or do sport (if the gym is open on this day).
Find the minimum number of days on which Vasya will have a rest (it means, he will not do sport and write the contest at the same time). The only limitation that Vasya has — he does not want to do the
same activity on two consecutive days: it means, he will not do sport on two consecutive days, and write the contest on two consecutive days.
Input
The first line contains a positive integer n (1 ≤ n ≤ 100)
— the number of days of Vasya's vacations.
The second line contains the sequence of integers a1, a2, ..., an (0 ≤ ai ≤ 3)
separated by space, where:
ai equals
0, if on the i-th day of vacations the gym is closed and the contest is not carried out;
ai equals
1, if on the i-th day of vacations the gym is closed, but the contest is carried out;
ai equals
2, if on the i-th day of vacations the gym is open and the contest is not carried out;
ai equals
3, if on the i-th day of vacations the gym is open and the contest is carried out.
Output
Print the minimum possible number of days on which Vasya will have a rest. Remember that Vasya refuses:
to do sport on any two consecutive days,
to write the contest on any two consecutive days.
Examples
input
output
input
output
input
output
Note
In the first test Vasya can write the contest on the day number 1 and do sport on the day number 3. Thus, he will have a rest for only 2 days.
In the second test Vasya should write contests on days number 1, 3, 5 and 7, in other days do sport. Thus, he will not have a rest for a single day.
In the third test Vasya can do sport either on a day number 1 or number 2. He can not do sport in two days, because it will be contrary to the his limitation. Thus, he will have a rest for only one day.
来说好好说一下这道题的策略问题。主要的问题集中在对 3 的处理上,如果碰见两个一样的数(1或2),就把后面的数变为0,方便后面的数进行别的处理(因为前面的数都处理过了,不用再考虑)。
然后是对3的处理,如果前面是1,那么它就等于2,如果前面等于2,那么它就等于1。后面的数不管他,到后面的数的时候如果有重复再进行去重操作。
代码如下:
#include <cstdio>
int main()
{
int op[111];
int n;
while (~scanf ("%d",&n))
{
for (int i = 1 ; i <= n ; i++)
scanf ("%d",&op[i]);
for (int i = 2 ; i <= n ; i++)
{
if (op[i] == 3)
{
if (op[i-1] == 1)
op[i] = 2;
else if (op[i-1] == 2)
op[i] = 1;
}
else if (op[i] == op[i-1] && (op[i] == 1 || op[i] == 2))
op[i] = 0;
}
int ans = 0;
for (int i = 1 ; i <= n ; i++)
if (op[i] == 0)
ans++;
printf ("%d\n",ans);
}
return 0;
}
C. Vacations
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output
Vasya has n days of vacations! So he decided to improve his IT skills and do sport. Vasya knows the following information about each
of this n days: whether that gym opened and whether a contest was carried out in the Internet on that day. For the i-th
day there are four options:
on this day the gym is closed and the contest is not carried out;
on this day the gym is closed and the contest is carried out;
on this day the gym is open and the contest is not carried out;
on this day the gym is open and the contest is carried out.
On each of days Vasya can either have a rest or write the contest (if it is carried out on this day), or do sport (if the gym is open on this day).
Find the minimum number of days on which Vasya will have a rest (it means, he will not do sport and write the contest at the same time). The only limitation that Vasya has — he does not want to do the
same activity on two consecutive days: it means, he will not do sport on two consecutive days, and write the contest on two consecutive days.
Input
The first line contains a positive integer n (1 ≤ n ≤ 100)
— the number of days of Vasya's vacations.
The second line contains the sequence of integers a1, a2, ..., an (0 ≤ ai ≤ 3)
separated by space, where:
ai equals
0, if on the i-th day of vacations the gym is closed and the contest is not carried out;
ai equals
1, if on the i-th day of vacations the gym is closed, but the contest is carried out;
ai equals
2, if on the i-th day of vacations the gym is open and the contest is not carried out;
ai equals
3, if on the i-th day of vacations the gym is open and the contest is carried out.
Output
Print the minimum possible number of days on which Vasya will have a rest. Remember that Vasya refuses:
to do sport on any two consecutive days,
to write the contest on any two consecutive days.
Examples
input
4 1 3 2 0
output
2
input
7 1 3 3 2 1 2 3
output
0
input
22 2
output
1
Note
In the first test Vasya can write the contest on the day number 1 and do sport on the day number 3. Thus, he will have a rest for only 2 days.
In the second test Vasya should write contests on days number 1, 3, 5 and 7, in other days do sport. Thus, he will not have a rest for a single day.
In the third test Vasya can do sport either on a day number 1 or number 2. He can not do sport in two days, because it will be contrary to the his limitation. Thus, he will have a rest for only one day.
来说好好说一下这道题的策略问题。主要的问题集中在对 3 的处理上,如果碰见两个一样的数(1或2),就把后面的数变为0,方便后面的数进行别的处理(因为前面的数都处理过了,不用再考虑)。
然后是对3的处理,如果前面是1,那么它就等于2,如果前面等于2,那么它就等于1。后面的数不管他,到后面的数的时候如果有重复再进行去重操作。
代码如下:
#include <cstdio>
int main()
{
int op[111];
int n;
while (~scanf ("%d",&n))
{
for (int i = 1 ; i <= n ; i++)
scanf ("%d",&op[i]);
for (int i = 2 ; i <= n ; i++)
{
if (op[i] == 3)
{
if (op[i-1] == 1)
op[i] = 2;
else if (op[i-1] == 2)
op[i] = 1;
}
else if (op[i] == op[i-1] && (op[i] == 1 || op[i] == 2))
op[i] = 0;
}
int ans = 0;
for (int i = 1 ; i <= n ; i++)
if (op[i] == 0)
ans++;
printf ("%d\n",ans);
}
return 0;
}
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