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 -20 2 0 -2 05 2 32 3 3 3 3

 

Sample Output

125

 

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#include <cstdio>#include <iostream>#include <cstring>using namespace std;#define N 100000 + 5#define LL long longint 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 -11 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 31 2 3-13 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.

 

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#include <cstdio>#include <iostream>#include <cstring>using namespace std;#define N 100 + 5#define LL long longconst 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;}