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BestCoder Round #3 A,B

2024-07-15 15:39:43   225人阅读

A.预处理出来，0(1)输出。

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
有一台机器，并且给你这台机器的工作表，工作表上有n个任务，机器在ti时间执行第i个任务，1秒即可完成1个任务。
有m个询问，每个询问有一个数字q，表示如果在q时间有一个工作表之外的任务请求，请计算何时这个任务才能被执行。
机器总是按照工作表执行，当机器空闲时立即执行工作表之外的任务请求。
 

Input
输入的第一行包含一个整数T， 表示一共有T组测试数据。

对于每组测试数据：
第一行是两个数字n, m，表示工作表里面有n个任务, 有m个询问；
第二行是n个不同的数字t1, t2, t3....tn，表示机器在ti时间执行第i个任务。
接下来m行，每一行有一个数字q，表示在q时间有一个工作表之外的任务请求。

特别提醒：m个询问之间是无关的。

[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
对于每一个询问，请计算并输出该任务何时才能被执行，每个询问输出一行。
 

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.统计左右比m大的个数，然后求和。

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;
}



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