多校 9

Arithmetic Sequence

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 0    Accepted Submission(s): 0

Problem Description

A sequence  b1,b2,⋯,bn  are called  (d1,d2) -arithmetic sequence if and only if there exist  i(1≤i≤n)  such that for every  j(1≤j<i),bj+1=bj+d1  and for every  j(i≤j<n),bj+1=bj+d2 .

Teacher Mai has a sequence  a1,a2,⋯,an . He wants to know how many intervals  [l,r](1≤l≤r≤n)  there are that  al,al+1,⋯,ar  are  (d1,d2) -arithmetic sequence.

 

Input

There are multiple test cases.

For each test case, the first line contains three numbers  n,d1,d2(1≤n≤105,|d1|,|d2|≤1000) , the next line contains  n  integers  a1,a2,⋯,an(|ai|≤109) .

 

Output

For each test case, print the answer.

 

Sample Input

  

   5 2 -2
0 2 0 -2 0
5 2 3
2 3 3 3 3
  

 

Sample Output

  

   12
5
  

 

Statistic |  Submit |  Clarifications |  Back

#include <cstdio>
#include <iostream>
#include <cstring>

using namespace std;
#define N 100000 + 5
#define LL long long

int a[N];
int D1[N], D2[N];
int d1, d2;
int n;

int main()
{
    while(~scanf("%d%d%d", &n, &d1, &d2))
    {
        for(int i = 1; i <= n; i++)
        scanf("%d", a + i);

        D1[1] = D2[1] = 1;
        int cn1 = 1, cn2 = 1;
        for(int i = 2; i <= n; i++)
        {
            if(a[i] == a[i - 1] + d1)
            cn1++;
            else cn1 = 1;
            D1[i] = cn1;

            if(a[i] == a[i - 1] + d2)
            cn2++;
            else
            cn2= 1;
            D2[i] = cn2;
        }

//        for(int i = 1; i <= n; i++)
//        cout<<D1[i]<<" ";
//        cout<<endl;
//
//        for(int i = 1; i <= n; i++)
//        cout<<D2[i]<<" ";
//        cout<<endl;

        LL ans = 0;
        for(int i = 1; i <= n; i++)
        {
            if(D1[i] >= 2)
            {
                ans += D1[i];
            }
            else if(D2[i] >= 2)
            {
                int t = i - D2[i] + 1;
                ans += (D1[t] +  D2[i] - 1);
                //cout<<i<<": "<<D1[t] + D2[i]<<endl;
            }
            else ans += 1;
        }
        printf("%I64d\n", ans);
    }
}


/*


9 1 -1
1 2 3 4 5 4 3 2 1

*/

Too Simple

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 429    Accepted Submission(s): 83

Problem Description

Rhason Cheung had a simple problem, and asked Teacher Mai for help. But Teacher Mai thought this problem was too simple, sometimes naive. So she ask you for help.

Teacher Mai has  m  functions  f1,f2,⋯,fm:{1,2,⋯,n}→{1,2,⋯,n} (that means for all  x∈{1,2,⋯,n},f(x)∈{1,2,⋯,n} ). But Rhason only knows some of these functions, and others are unknown.

She wants to know how many different function series  f1,f2,⋯,fm  there are that for every  i(1≤i≤n) , f1(f2(⋯fm(i)))=i . Two function series  f1,f2,⋯,fm  and  g1,g2,⋯,gm  are considered different if and only if there exist  i(1≤i≤m),j(1≤j≤n) , fi(j)≠gi(j) .

 

Input

For each test case, the first lines contains two numbers  n,m(1≤n,m≤100) .

The following are  m  lines. In  i -th line, there is one number  −1  or  n  space-separated numbers.

If there is only one number  −1 , the function  fi  is unknown. Otherwise the  j -th number in the  i -th line means  fi(j) .

 

Output

For each test case print the answer modulo  109+7 .

 

Sample Input

  

   3 3
1 2 3
-1
3 2 1
  

 

Sample Output

  

   1

   

    

     Hint
    
The order in the function series is determined. What she can do is to assign the values to the unknown functions.
   
  

 

Statistic |  Submit |  Clarifications |  Back

#include <cstdio>
#include <iostream>
#include <cstring>

using namespace std;
#define N 100 + 5
#define LL long long
const int mod = 1000000000 + 7;

int n, m;
int f[N][N];
int cnt;

int mul_pow(int a, int k)
{
    int res = 1;
    while(k)
    {
        if(k & 1)
        res = ((long long)res * a) % mod;
        a = ((long long)a * a) % mod;
        k >>= 1;
    }
    return res;
}

int fact(int a)
{
    int res = 1;
    for(int i = 1; i <= a; i++)
    res = ((LL)res * i) % mod;
    return res;
}

bool h[N];
int op[N];

int main()
{
    while(~scanf("%d%d", &n, &m))
    {
        memset(op, 0, sizeof op);
        int t;
        cnt = 0;
        for(int i = 1; i <= m; i++)
        {
            scanf("%d", &t);
            if(t == -1) {
                op[i] = 1;
                cnt++;
            }
            else
            {
                f[i][1] = t;
                for(int j = 2; j <= n; j++)
                scanf("%d", &f[i][j]);
            }
        }

        int flag = 0;

        for(int i = 1; i <= m; i++)
        {
            memset(h, false, sizeof h);
            if(op[i] == 0)
            for(int j = 1; j <= n; j++)
            {
                t = f[i][j];
                if(h[t] || t > n || t <= 0)
                {
                    //cout<<i<<" "<<j<<endl;
                    flag = 1;
                    break;
                }
                h[t] = true;
            }
        }
        if(flag)
        {
            printf("0\n");
            continue;
        }

        flag  = 0;
        if(cnt == 0)
        {
            for(int i = 1; i <= n; i++)
            {
                t = i;
                for(int j = m; j >= 1; j--)
                t = f[j][t];
                if(t != i)
                {
                    flag = 1;
                    break;
                }
            }
            if(flag) printf("0\n");
            else printf("1\n");
        }
        else
        {
            t = fact(n);
            int ans = mul_pow(t, cnt - 1);
            printf("%d\n", ans);
        }
    }
    return 0;
}