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;
}