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A.预处理出来，0(1)输出。

Task schedule

1
5 5
1 2 3 5 6
1
2
3
4
5

4
4
4
4
7

#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;
}

BestCoder Sequence

1 1
1
5 3
4 5 3 2 1

1
3

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

#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;
}

BestCoder Round #3 A,B,布布扣,bubuko.com

BestCoder Round #3 A,B

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原文地址：http://blog.csdn.net/xu12110501127/article/details/38365127