codeforces--C，codeforces267c

codeforces--C，codeforces267c
C - Anya and GhostsTime Limit:2000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64uSubmit Status

 

Appoint description: System Crawler  (2015-01-28)

Description

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 w_{i 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 w_{i (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.

Sample Input

Input

1 8 310

Output

3

Input

2 10 15 8

Output

1

Input

1 1 310

Output

-1

Hint

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.

 

題目大意：給出n個鬼訪問的時間，和每個蠟燭可以持續的時間，每次訪問最少需要的點燃的蠟燭數。問一共最少需要幾個蠟燭。

將每個蠟燭的點燃的時間盡量靠近訪問的時間，所以，統計第i個訪問時點燃的蠟燭，如果少於r，那麼要訪問的時間開始，向前遍曆，找出最近的可以點燃蠟燭的時刻，這樣可以保證使用的蠟燭最少。

注意

1、可以在子夜之前點蠟燭，所以要將時間右移。

2、蠟燭持續的時間t和點燃的時間j的關係為，j時刻點蠟燭，j+1到j+t時刻蠟燭亮著。

 

#include <cstdio>#include <cstring>#include <algorithm>using namespace std ;int a[400] , k[2000] , flag[2000] ;int main(){    int m , t , r , num , i , j , ans ;    while( scanf("%d %d %d", &m, &t, &r) != EOF )    {        memset(k,0,sizeof(k)) ;        memset(flag,0,sizeof(flag)) ;        num = 0 ; ans = 0 ;        for(i = 0 ; i < m ; i++)        {            scanf("%d", &a[i] ) ;            a[i] += 1000 ;            //flag[ a[i] ] = 1 ;        }        for(i = 0 ; i < m ; i++)        {            if( num < r )            {                for(j = a[i]-1 ; j >= a[i]-t ; j--)                {                    if( flag[j] == 0 )                    {                        flag[j] = 1 ;                        k[j+1] += 1 ;                        k[j+1+t] += -1 ;                        num++ ;                        ans++ ;                    }                    if( num == r )                        break ;                }                if( num < r )                    break ;            }            for(j = a[i]+1 ; i < m-1 && j <= a[i+1] ; j++)                num += k[j] ;        }        if( num < r )            printf("-1\n") ;        else            printf("%d\n", ans) ;    }    return 0 ;}