National training Day1

National training Day1 solution to a problem

T1 divided

Meaning of the questions:

There \ (n-\) number \ (a_1, a_2, ..., a_n \) has a number m \ (B_1, B_2, ..., B_n \)

令\(a = a_1\times a_2\,\times ... \times \,a_n\)

令\(b = b_1\times\,b_2 \times\,...\,\times\,b_n\)

Analyzing \ (A \) whether \ (B \) multiple

Input: \ (n-, m \)

Output: \ (Yes / No \)

Practice: is a prime factorization of a simple

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int n, m;
int a[1010], b[1010];
int p[3][20]={};
void divid(int op, int x)
{
    if(x == 1) return;
    if(x == 2) p[op][1]++;
    if(x == 3) p[op][2]++;
    if(x == 4) p[op][1] += 2;
    if(x == 5) p[op][3]++;
    if(x == 6) {p[op][1]++; p[op][2]++;}
    return;
}
int main()
{
    freopen("divide.in","r",stdin);
    freopen("divide.out","w",stdout);
    scanf("%d%d", &n, &m);
    for(int i = 1; i <= n; i++)
    {
        scanf("%d", &a[i]);
        divid(1, a[i]);
    }
    for(int i = 1; i <= m; i++)
    {
        scanf("%d", &b[i]);
        divid(2, b[i]);
    }
    for(int i = 1; i <= 3; i++)
    {
        if(p[1][i] >= p[2][i])continue;
        else
        {
            printf("No\n"); return 0;
        }
    }
    printf("Yes\n");
    return 0;
}
/*
样例：
输入：2 3
     6 6
     1 3 4
输出：Yes
*/

T2 graph

Meaning of the questions:

To generate n now points a directed graph. Requirements are met:

1. If \ (a-> b \) side, there are (b-> a \) \ edge

2. If \ (a-> b \) sides and \ (b-> c \) side, there are \ (a-> c \) edge

3. At least one point no loopback

Find a few programs. May be larger as a result, the results of \ (m \) modulo

Input: \ (n-, m \)

Output: The answer

practice:

This question is the meaning of problems in addition to relatively pit outside, it does not have much difficulty. Note that, according to the meaning of the questions, if a point where the communication block size is greater than or equal to 2, the point must have a self-loop.

My approach is: Let \ (f [i] [0 ] \) represents \ (I \) points freely combined, and each point there are several solutions of loopback

\ (f [i] [1 ] \) represents \ (I \) points freely combined with at least \ (1 \) point of the program number without loopback

The method I transfer equation somewhat unique, I was watching \ (1 \) the number of points where the communication block of the dot were transferred to note that the transfer is to be discussed \ (1 \) No. point is isolated as a point If isolated, there are \ (1 \) dot from the ring, and two cases not looping, need to be discussed separately, the state transition equation is as follows:

\ (f [i] [1 ] + = f [i - 1] [1] \, \, \, \, 1 \) dots and isolated \ (1 \) exists dot loopback

\ (f [i] [1 ] + = f [i - 1] [1] + f [i - 1] [0] \, \, \, \, 1 \) dots isolated and \ (1 \) No. loopback point is not present, then the other point may loopback, loopback may not

\ (f [i] [1 ] + = \ sum_ {j = 2} ^ {i} f [i - j] [1] \ times C_ {i-1} ^ {j-1} \, \, \ , \, \) is calculated \ (1 \) communication block size number of point as \ (J \) number of scheme

\ (f [i] [0 ] + = f [i-1] [0] \, \, \, \, 1 \) can only be isolated from the loop when the number of points

\ (f [i] [0 ] + = \ sum_ {j = 2} ^ {i} f [i - j] [0] \ times C_ {i-1} ^ {j-1} \, \, \ , \, \) is calculated \ (1 \) communication block size number of point as \ (J \) number of scheme

Pascal's Triangle combination with a number on it.

Teacher practice is to "first \ (1 \) points" to "first \ (i \) points", in fact, the wording is the same, but the idea is different.

#include <cstdio>
#include <algorithm>
#include <cstring>
using namespace std;
#define maxn 2010
#define ll long long
int n, mol;
ll f[maxn][2];//  1:符合题意 0:不符合题意 
ll C[maxn][maxn];
int main()
{
    freopen("graph.in", "r", stdin);
    freopen("graph.out", "w", stdout);
    scanf("%d%d", &n, &mol);
    for(int i = 0; i <= n; i++) C[i][0] = 1;
    for(int i = 1; i <= n; i++)
    {
        for(int j = 1; j <= i; j++)
        {
            C[i][j] = (C[i - 1][j] + C[i - 1][j - 1]) % mol;
        }
    }
/*  for(int i = 0; i <= n; i++)
    {
        for(int j = 0; j <= i; j++)
        printf("C[%d][%d] = %lld ", i, j, C[i][j]);
        printf("\n");
    }
*/
    f[1][1] = 1; f[1][0] = 1;
    for(int i = 2; i <= n; i++)
    {
        f[i][1] = (f[i][1] + f[i - 1][1] + f[i - 1][0] + f[i - 1][1]) % mol;
        f[i][0] = (f[i][0] + f[i - 1][0]) % mol;
        for(int j = 2; j < i; j++)
        {
//          f[i][1] = (f[i][1] + ( ( ( (C[i - 1][j - 1] * S[j - 1][j - 1]) % mol ) * ( (f[i - j][0] + f[i - j][1]) % mol) ) % mol)) % mol;                                               
            f[i][1] = (f[i][1] + (C[i - 1][j - 1] * f[i - j][1]) % mol) % mol;
            f[i][0] = (f[i][0] + (C[i - 1][j - 1] * f[i - j][0]) % mol) % mol;
        }
        f[i][0] = (f[i][0] + 1) % mol;
    }
//  for(int i = 1; i <= n; i++)
//  printf("f[%d][0] = %lld f[%d][1] = %lld\n", i, f[i][0], i, f[i][1]);
    printf("%lld\n", f[n][1]);
    return 0;
}             
/*
样例：
输入：2 5
输出：3
*/

FIG T3 generation