11.14 Arts Comprehensive Examination
Science comprehensive test in others, we in Covent comprehensive
T1 history
will not.
#include<bits/stdc++.h>
#define int long long
char buf[1<<20],*p1,*p2;
#define uc (p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<20,stdin),p1==p2)?0:*p1++)
#define ph putchar
using namespace std;
const int inf=1e9+7;
inline int in();
inline void out(int x);
int a=in(),k=in(),m=in(),mky,ax,lx=-1,lyr,ayr=-1;
int hea[10000010],lst[10000010],ext[10000010];
bool bkl[10000010],oier;
main(void) {
register int i,mu;
for(i=0; i<k; i++) {
mky=i*i%k;
if(!bkl[mky]) {
bkl[mky]=1;
hea[mky]=i;
} else ext[lst[mky]]=i;
lst[mky]=i,ext[i]=-1;
} register int qt,rt;
while(a) {
if(!bkl[a%k]) {
ax=-1;
if(!oier&&m>=a&&(m-a)%k==0)
ayr=lyr+(m-a)/k;
break;
} qt=ceil(sqrt(a)),mu=qt%k;
for(i=hea[a%k]; i>=0; i=ext[i])
if(i>=mu)break;
if(i<0)i=hea[a%k]+k;
qt+=i-mu,rt=qt*qt;
if(!oier&&m>=a&&m<=rt&&(m-a)%k==0)
ayr=lyr+(m-a)/k,oier=1;
if(lx==rt)break;
lyr+=(rt-a)/k+1,a=qt;
ax=max(ax,rt),lx=rt;
} out(ax),ph('\n'),out(ayr);
}
inline int in() {
register int x=0;
register char s=uc;
register bool f=0;
while((s<'0'||s>'9')&&s!='-')s=uc;
if(s=='-')f=1,s=uc;
while(s>='0'&&s<='9')x=(x<<1)+(x<<3)+(s^48),s=uc;
return f?-x:x;
}
inline void out(int x) {
if(x<0)ph('-'),x=~x+1;
if(x>9)out(x/10);
ph((x%10)^48);
}
T2 Geography
The only made out of the question.Could it be that this indicates that the selection of subjects to choose my geography?
Title effect:
given a (N * N * N \) \ cube, the cube represent obstacles 0, 1 represents the ground, from the ground to \ (1s / \) went to a four-frame adjacent grid speed. May be transferred to another surface along the edge. Exist \ (m \) a wormhole, from the point \ (lS \) to the other point.
There \ (Q \) once asked, ask it went the other point of the shortest time. If you can not reach the output \ (--1 \) .
Data range:
\ (. 1 \ Leq N \ leq10,0 \ Leq m \ Leq 1000,1 \ Leq Q \ ^. 5 Leq 10 \)
do not work the problem difficult to see.
This question is really very simple.
Really.
\ (Qaq \)
In fact, \ (BFS \) After that, it seems to be able to reach the edges to link \ (Floyd \) also run.
Note To ask first before all counted out several programs, for insurance with a 6-dimensional array of memory answer.
Really it is \ (BFS \) only.
Only.
#include<cstdio>
#include<iostream>
#include<cstring>
#include<cmath>
#include<queue>
using namespace std;
int n;
bool Map[7][12][12];//1:U 2:L 3:F 4:R 5:B 6:D
inline int sel(char x){switch(x){case 'U':{return 1;}case 'L':{return 2;}case 'F':{return 3;}case 'R':{return 4;}case 'B':{return 5;}case 'D':{return 6;}}}
struct cd{
int sum;
int to1[601],tox[601],toy[601];
bool ap;
}map1[7][12][12];
char T[100];
bool mark[7][12][12];
int dis[7][12][12][7][12][12];
struct node{
int pm,x,y;
int dis;
};
int dx[4]={1,0,-1,0};
int dy[4]={0,-1,0,1};
node change(int p,int x,int y)
{
node tmp;
switch(p){
case 1:{
if(x==0){
tmp.pm=4;
tmp.x=1;
tmp.y=n-y+1;
}
if(x==n+1){
tmp.pm=2;
tmp.x=1;
tmp.y=y;
}
if(y==0){
tmp.pm=5;
tmp.x=1;
tmp.y=x;
}
if(y==n+1){
tmp.pm=3;
tmp.x=1;
tmp.y=n-x+1;
}
return tmp;
break;
}
case 2:{
if(x==0){
tmp.pm=1;
tmp.x=n;
tmp.y=y;
}
if(x==n+1){
tmp.pm=6;
tmp.x=1;
tmp.y=y;
}
if(y==0){
tmp.pm=5;
tmp.x=x;
tmp.y=n;
}
if(y==n+1){
tmp.pm=3;
tmp.x=x;
tmp.y=1;
}
return tmp;
break;
}
case 3:{
if(x==0){
tmp.pm=1;
tmp.x=n-y+1;
tmp.y=n;
}
if(x==n+1){
tmp.pm=6;
tmp.x=y;
tmp.y=n;
}
if(y==0){
tmp.pm=2;
tmp.x=x;
tmp.y=n;
}
if(y==n+1){
tmp.pm=4;
tmp.x=x;
tmp.y=1;
}
return tmp;
break;
}
case 4:{
if(x==0){
tmp.pm=1;
tmp.x=1;
tmp.y=n-y+1;
}
if(x==n+1){
tmp.pm=6;
tmp.x=n;
tmp.y=n-y+1;
}
if(y==0){
tmp.pm=3;
tmp.x=x;
tmp.y=n;
}
if(y==n+1){
tmp.pm=5;
tmp.x=x;
tmp.y=1;
}
return tmp;
break;
}
case 5:{
if(x==0){
tmp.pm=1;
tmp.x=y;
tmp.y=1;
}
if(x==n+1){
tmp.pm=6;
tmp.x=n-y+1;
tmp.y=1;
}
if(y==0){
tmp.pm=4;
tmp.x=x;
tmp.y=n;
}
if(y==n+1){
tmp.pm=2;
tmp.x=x;
tmp.y=1;
}
return tmp;
break;
}
case 6:{
if(x==0){
tmp.pm=2;
tmp.x=n;
tmp.y=y;
}
if(x==n+1){
tmp.pm=4;
tmp.x=n;
tmp.y=n-y+1;
}
if(y==0){
tmp.pm=5;
tmp.x=n;
tmp.y=n-x+1;
}
if(y==n+1){
tmp.pm=3;
tmp.x=n;
tmp.y=x;
}
return tmp;
break;
}
}
}
int bfs(int sp,int sx,int sy)
{
memset(mark,0,sizeof(mark));
queue<node>q;
q.push((node){sp,sx,sy,0});
while(q.size()){
node tmp=q.front();
int p=tmp.pm,x=tmp.x,y=tmp.y;
q.pop();
if(mark[p][x][y])continue;
mark[p][x][y]=1;
dis[sp][sx][sy][p][x][y]=tmp.dis;
for(int i=0;i<4;i++){
int fx=x+dx[i],fy=y+dy[i];
if(fx==0||fx==n+1||fy==0||fy==n+1){
node p1=change(p,fx,fy);
if(!Map[p1.pm][p1.x][p1.y])continue;
p1.dis=tmp.dis+1;
q.push(p1);
}
else{
node p2;
if(!Map[p][fx][fy])continue;
p2.pm=p;p2.x=fx;p2.y=fy;p2.dis=tmp.dis+1;
q.push(p2);
}
}
if(map1[p][x][y].ap){
cd t=map1[p][x][y];
for(int i=1;i<=t.sum;i++)
{
node p3;
p3.pm=t.to1[i];p3.x=t.tox[i];p3.y=t.toy[i];p3.dis=tmp.dis+1;
if(!Map[p3.pm][p3.x][p3.y])continue;
q.push(p3);
}
}
}
}
int main()
{
memset(dis,0x3f,sizeof(dis));
ios::sync_with_stdio(0);
cin.tie(0);
// freopen("geog.in","r",stdin);
// freopen("geog.out","w",stdout);
int i,j;
cin>>n;
for(i=1;i!=n+1;++i){
cin>>T;
for(j=1;j!=n+1;++j){
Map[1][i][j]=T[j-1]-'0';
}
}
for(i=1;i!=n+1;++i)
{
cin>>T;
for(j=1;j!=4*n+1;++j){
Map[1+(j-1)/n+1][i][(j-1)%n+1]=T[j-1]-'0';
}
}
for(i=1;i!=n+1;++i){
cin>>T;
for(j=1;j!=n+1;++j){
Map[6][i][j]=T[j-1]-'0';
}
}
int m;
cin>>m;
for(i=1;i<=m;++i)
{
char f1,f2;
int x1,y1,x2,y2;
cin>>f1>>x1>>y1>>f2>>x2>>y2;
int ff1=sel(f1),ff2=sel(f2);
map1[ff1][x1][y1].sum+=1;map1[ff2][x2][y2].sum+=1;
map1[ff1][x1][y1].ap=1;map1[ff1][x1][y1].to1[map1[ff1][x1][y1].sum]=ff2;map1[ff1][x1][y1].tox[map1[ff1][x1][y1].sum]=x2;map1[ff1][x1][y1].toy[map1[ff1][x1][y1].sum]=y2;
map1[ff2][x2][y2].ap=1;map1[ff2][x2][y2].to1[map1[ff2][x2][y2].sum]=ff1;map1[ff2][x2][y2].tox[map1[ff2][x2][y2].sum]=x1;map1[ff2][x2][y2].toy[map1[ff2][x2][y2].sum]=y1;
}
for(int p=1;p<=6;p++)
{
for(i=1;i<=n;i++)
{
for(j=1;j<=n;j++){
if(Map[p][i][j]){
bfs(p,i,j);
}
}
}
}
int Q;
cin>>Q;
for(i=1;i<=Q;++i){
char f1,f2;
int x1,y1,x2,y2;
cin>>f1>>x1>>y1>>f2>>x2>>y2;
int d=dis[sel(f1)][x1][y1][sel(f2)][x2][y2];
if(d!=0x3f3f3f3f)
printf("%d\n",d);
else printf("-1\n");
}
return 0;
}Plane Operating good trouble ah ah ah ah ah
Note: There may be multiple wormhole at one point, maybe even their own wormhole
T3 political
Did not see, does not
sum up:
1. What do you think of this question to test the algorithm, in fact, you want to test a code
2.nodgdNMSL