多校 9

时间：2015-08-18 19:33:25 阅读：147 评论：0 收藏：0 [点我收藏+]

标签：acm hdu

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

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

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;
}

多校 9

标签：acm hdu

原文地址：http://blog.csdn.net/dojintian/article/details/47754535

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