多校10 1002 HDU 6172 Array Challenge
#include <cstdio>
#include <map>
#include <cstdlib>
#include <queue>
using namespace std;
#define mem(a,b) memset(a,b,sizeof(a))
#define rep(i,l,r) for (int i=l;i<r;++i)
typedef unsigned long long ull;
int dx[4]={1,1,-1,-1},dy[4]={0,1,0,-1};
map<ull,bool>vis;
struct Sta{
int a[6][6],step,x,y;
Sta(){step=x=y=0;}
};
int gujia(Sta s){
int ans=0;
rep(i,0,6)rep(j,0,i+1)
if(s.a[i][j])ans+=abs(s.a[i][j]-i);
return ans;
}
ull haxi(Sta s){
ull ans=0;
rep(i,0,6)rep(j,0,i+1){
ans<<=3;ans|=s.a[i][j];
}
return ans;
}
int bfs(Sta s){
vis.clear();
queue<Sta>q;q.push(s);
while(!q.empty()){
Sta now=q.front();q.pop();
if(gujia(now)==0)return now.step;
rep(i,0,4){
int x=now.x,y=now.y;
int nx=x+dx[i],ny=y+dy[i];
if(nx>=0 && nx<6 && ny>=0 && ny<=nx){
swap(now.a[x][y],now.a[nx][ny]);
now.x=nx,now.y=ny,++now.step;
ull hx=haxi(now);
if(!vis[hx]&&gujia(now)+now.step<21){
q.push(now);
vis[hx]=true;
}
swap(now.a[x][y],now.a[nx][ny]);
now.x-=dx[i],now.y-=dy[i],--now.step;
}
}
}
return -1;
}
int main() {
int t;
scanf("%d",&t);
while(t--){
Sta s;
rep(i,0,6)
rep(j,0,i+1){
scanf("%d",&s.a[i][j]);
if(s.a[i][j]==0)s.x=i,s.y=j;
}
int ans=bfs(s);
if(ans==-1)puts("too difficult");else printf("%d\n",ans);
}
return 0;
}
#include <cstdio>
#include <algorithm>
#include <vector>
using namespace std;
#define rep(i,a,n) for (int i=a;i<n;i++)
#define pb push_back
#define SZ(x) ((int)(x).size())
typedef vector<int> VI;
typedef long long ll;
const ll mod=1000000007;
ll qpow(ll a,ll b) {ll res=1;for(a%=mod;b;b>>=1,a=a*a%mod)if(b&1)res=res*a%mod;return res;}
VI BM(VI s) {//c[0]s[0]+c[1]s[1]+...=0
VI C(1,1),B(1,1);
int L=0,m=1,rev=1;
rep(n,0,SZ(s)) {
ll d=0;
rep(i,0,L+1) d=(d+(ll)C[i]*s[n-i])%mod;
if (d==0) ++m;
else if (2*L<=n) {
VI T=C;
ll c=mod-d*rev%mod;
C.resize(SZ(B)+m);
rep(i,0,SZ(B)) C[i+m]=(C[i+m]+c*B[i])%mod;
L=n+1-L; B=T; rev=qpow(d,mod-2); m=1;
} else {
ll c=mod-d*rev%mod;
C.resize(SZ(B)+m);
rep(i,0,SZ(B)) C[i+m]=(C[i+m]+c*B[i])%mod;
++m;
}
}
rep(i,0,SZ(C))printf("%dx[%d]%s",C[i],i,i+1==SZ(C)?"=0\n":"+");
return C;
}
调用:BM(VI{31,197,1255,7997})