BestCoder Round #3 A,B

畫船聽雨發表於2014-08-04

A.預處理出來，0(1)輸出。

Task schedule

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 387    Accepted Submission(s): 193


Problem Description
有一臺機器，並且給你這臺機器的工作表，工作表上有n個任務，機器在ti時間執行第i個任務，1秒即可完成1個任務。
有m個詢問，每個詢問有一個數字q，表示如果在q時間有一個工作表之外的任務請求，請計算何時這個任務才能被執行。
機器總是按照工作表執行，當機器空閒時立即執行工作表之外的任務請求。
 

Input
輸入的第一行包含一個整數T， 表示一共有T組測試資料。

對於每組測試資料：
第一行是兩個數字n, m，表示工作表裡面有n個任務, 有m個詢問；
第二行是n個不同的數字t1, t2, t3....tn，表示機器在ti時間執行第i個任務。
接下來m行，每一行有一個數字q，表示在q時間有一個工作表之外的任務請求。

特別提醒：m個詢問之間是無關的。

[Technical Specification]
1. T <= 50
2. 1 <= n, m <= 10^5
3. 1 <= ti <= 2*10^5, 1 <= i <= n
4. 1 <= q <= 2*10^5
 

Output
對於每一個詢問，請計算並輸出該任務何時才能被執行，每個詢問輸出一行。
 

Sample Input
1
5 5
1 2 3 5 6
1
2
3
4
5





 



Sample Output




4
4
4
4
7





 



Source



BestCoder Round #3



 


#include <algorithm>
#include <iostream>
#include <stdlib.h>
#include <string.h>
#include <iomanip>
#include <stdio.h>
#include <string>
#include <queue>
#include <cmath>
#include <stack>
#include <map>
#include <set>
#define eps 1e-10
///#define LL __int64
#define LL long long
///#define INF 0x7ffffff
#define INF 0x3f3f3f3f
#define PI 3.1415926535898
#define zero(x) ((fabs(x)<eps)?0:x)

const int maxn = 201000;

using namespace std;

int vis[maxn];
int num[maxn];

int main()
{
    int T;
    cin >>T;
    while(T--)
    {
        int n, m;
        cin >>n>>m;
        memset(vis, 0, sizeof(vis));
        int x;
        for(int i = 1; i <= n; i++)
        {
            scanf("%d",&x);
            vis[x] = 1;
        }
        int now;
        for(int i = maxn-1; i >= 1; i--)
        {
            if(!vis[i])
            {
                now = i;
                num[i] = now;
                continue;
            }
            num[i] = now;
        }
        while(m--)
        {
            scanf("%d",&x);
            printf("%d\n",num[x]);
        }
    }
    return 0;
}


B.統計左右比m大的個數，然後求和。






BestCoder Sequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 258    Accepted Submission(s): 108



Problem Description



Mr Potato is a coder.

Mr Potato is the BestCoder.


One night, an amazing sequence appeared in his dream. Length of this sequence is odd, the median number is M, and he named this sequence as Bestcoder Sequence.


As the best coder, Mr potato has strong curiosity, he wonder the number of consecutive sub-sequences which are bestcoder sequences in a given permutation of 1 ~ N.



 



Input



Input contains multiple test cases. 

For each test case, there is a pair of integers N and M in the first line, and an permutation of 1 ~ N in the second line.

[Technical Specification]

1. 1 <= N <= 40000

2. 1 <= M <= N



 



Output



For each case, you should output the number of consecutive sub-sequences which are the Bestcoder Sequences. 



 



Sample Input




1 1
1
5 3
4 5 3 2 1





 



Sample Output




1
3

Hint

For the second case, {3},{5,3,2},{4,5,3,2,1} are Bestcoder Sequence.

 





 



Source



BestCoder Round #3


#include <algorithm>
#include <iostream>
#include <stdlib.h>
#include <string.h>
#include <iomanip>
#include <stdio.h>
#include <string>
#include <queue>
#include <cmath>
#include <stack>
#include <map>
#include <set>
#define eps 1e-10
#define LL __int64
///#define LL long long
///#define INF 0x7ffffff
#define INF 0x3f3f3f3f
#define PI 3.1415926535898
#define zero(x) ((fabs(x)<eps)?0:x)

const int maxn = 101000;

using namespace std;

int num[maxn];
LL sum1[maxn];
LL sum2[maxn];
int main()
{
    int n, m;
    while(cin >>n>>m)
    {
        memset(sum1, 0, sizeof(sum1));
        memset(sum2, 0, sizeof(sum2));
        int s;
        for(int i = 1; i <= n; i++)
        {
            scanf("%d",&num[i]);
            if(num[i] == m) s = i;
        }
        int cnt = 0;
        int pp = 40000;
        sum1[pp] = 1;
        for(int i = s-1; i >= 1; i--)
        {
            if(num[i] > m) cnt++;
            else if(num[i] < m) cnt--;
            sum1[cnt+pp]++;
        }
        cnt = 0;
        LL ans = 0;
        ans += sum1[pp];
        for(int i = s+1; i <= n; i++)
        {
            if(num[i] > m) cnt++;
            else if(num[i] < m)cnt--;
            sum2[cnt+pp]++;
            ans += sum1[pp-cnt];
        }
        cout<<ans<<endl;
    }
    return 0;
}









相關文章