2018-div-matrix 题解(打表)

题目链接

题目大意

要你求有多少个满足题目条件的矩阵mod 1e9+7

(a[1][1]=2018;;a[i][j]为a[i-1][j]和a[i][j-1]的因子)

题目思路

dp也就图一乐，真正比赛还得看打表

一直在想dp，其实却是打表找规律

只能说看到答案固定的题目就应该要去想打表

然后发现规律

打表代码

#include<set>
#include<map>
#include<queue>
#include<stack>
#include<cmath>
#include<cstdio>
#include<vector>
#include<string>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<unordered_map>
#define fi first
#define se second
#define debug printf(" I am here
");
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> pii;
const ll INF=0x3f3f3f3f3f3f3f3f;
const int maxn=1e3+5,inf=0x3f3f3f3f,mod=1e9+7;
const double eps=1e-10;
int n,m,a[maxn][maxn],cnt;
int num[]={0,1,2,1009,2018};
void dfs(int x,int y,int n,int m){
    if(x>n||y>m) return ;
    if(x==1&&y==1){
        a[1][1]=2018;
        if(x==n&&y==m){
            cnt++;
        }else if(y==m){
            dfs(x+1,1,n,m);
        }else{
            dfs(x,y+1,n,m);
        }
    }else if(x==1){
        for(int i=1;i<=4;i++){
            if(a[x][y-1]%num[i]!=0) continue;
            a[x][y]=num[i];
            if(x==n&&y==m){
                cnt++;
            }else if(y==m){
                dfs(x+1,1,n,m);
            }else{
                dfs(x,y+1,n,m);
            }
        }
    }else if(y==1){
        for(int i=1;i<=4;i++){
            if(a[x-1][y]%num[i]!=0) continue;
            a[x][y]=num[i];
            if(x==n&&y==m){
                cnt++;
            }else if(y==m){
                dfs(x+1,1,n,m);
            }else{
                dfs(x,y+1,n,m);
            }
        }
    }else{
        for(int i=1;i<=4;i++){
            if(a[x-1][y]%num[i]!=0) continue;
            if(a[x][y-1]%num[i]!=0) continue;
            a[x][y]=num[i];
            if(x==n&&y==m){
                cnt++;
            }else if(y==m){
                dfs(x+1,1,n,m);
            }else{
                dfs(x,y+1,n,m);
            }
        }
    }
}
signed main(){
    for(int i=1;i<=10;i++){
        for(int j=1;j<=10;j++){
            cnt=0;
            dfs(1,1,i,j);
            printf("%d ",cnt);
        }
        cout<<endl;
    }
    return 0;
}

代码

#include<set>
#include<map>
#include<queue>
#include<stack>
#include<cmath>
#include<cstdio>
#include<vector>
#include<string>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<unordered_map>
#define fi first
#define se second
#define debug printf(" I am here
");
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> pii;
const ll INF=0x3f3f3f3f3f3f3f3f;
const int maxn=2e3+5,inf=0x3f3f3f3f,mod=1e9+7;
const double eps=1e-10;
int n,m;
int dp[maxn][maxn];
signed main(){
    dp[1][1]=1;
    for(int i=1;i<=2000;i++){
        for(int j=1;j<=2000;j++){
            if(i==1&&j==1) continue;
            dp[i][j]=(dp[i-1][j]+dp[i][j-1]+1)%mod;
        }
    }
    while(scanf("%d%d",&n,&m)!=-1){
        printf("%d
",1ll*dp[n][m]*dp[n][m]%mod);
    }
    return 0;
}