T1
记
表示走i步，第i步结尾是向北的方案数
记
表示走i步的方案数
那么
然后高精搞一搞
#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std;
int m[2333][2333];
int sx,sy,n,ans;
const int dx[]={1,-1,0};
const int dy[]={0,0,-1};
int px[2333],py[2333],f[2333],g[2333];
void dfs(int x,int y,int tm,int px[],int py[]){
if(tm==n){
printf("Found path: ");
for(int i=1;i<=tm;i++)printf("(%d,%d)",px[i],py[i]);printf("(%d,%d)",x,y);puts("");
ans++;
}else
for(int i=0;i<3;i++){
int nx=x+dx[i],ny=y+dy[i];
if(!m[nx][ny]&&(nx!=px[tm]||ny!=py[tm])){
//printf("(%d,%d)->(%d,%d)\n",x,y,nx,ny);
m[nx][ny]=1;
px[tm+1]=x;py[tm+1]=y;
dfs(nx,ny,tm+1,px,py);
m[nx][ny]=0;
}
}
}
struct Bn{
int f[2333],len;
};
Bn cheng2(Bn a){
Bn c;c.len=a.len;
memset(c.f,0,sizeof(c.f));
for(int i=1;i<=c.len;i++)c.f[i]=a.f[i]*2;
for(int i=1;i<=c.len;i++)c.f[i+1]=c.f[i]/10+c.f[i+1],c.f[i]%=10;
while(c.f[c.len+1]>0){
c.f[c.len+1]+=c.f[c.len]/10;
c.f[c.len]%=10;
++c.len;
}
return c;
}
Bn jia(Bn a,Bn b){
Bn c;c.len=max(a.len,b.len);
memset(c.f,0,sizeof(c.f));
for(int i=1;i<=c.len;i++){
c.f[i]+=a.f[i]+b.f[i];
c.f[i+1]=c.f[i]/10;
c.f[i]%=10;
}
while(c.f[c.len+1]>0){
c.f[c.len+1]+=c.f[c.len]/10;
c.f[c.len]%=10;
++c.len;
}
return c;
}
Bn F[101];
void test(){
F[1].f[1]=3;F[1].len=1;
F[0].f[1]=1;F[0].len=1;
for(int i=2;i<=n;i++){
F[i]=jia(cheng2(F[(i-1)]),F[(i-2)]);
}
for(int i=F[n].len;i>=1;i--)cout<<F[n].f[i];puts("");
}
int main(){
freopen("game.in","r",stdin);
freopen("game.out","w",stdout);
scanf("%d",&n);
//f[1]=3;f[0]=1;
// for(int i=2;i<=n;i++){
// f[i]=f[i-1]*2+f[i-2];
// }
//printf("%d\n",f[n]);
//shorter digits
//Bigenumber
test();
return 0;
}
T2
随便建下树就好了
#include<stack>
#include<queue>
#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std;
char s[350000];
struct edge{
int to,next;
}e[66666];int cnt,last[33333]; int boolean[33333];
void link(int a,int b){
//printf("~~~%d %d\n",a,b);
e[++cnt]=(edge){b,last[a]};last[a]=cnt;
e[++cnt]=(edge){a,last[b]};last[b]=cnt;
}
void initial(){
memset(last,0,sizeof(last));
cnt=0;
}
int L[32003],R[32003],qzh[32003],isl[32003];
bool q(int r,int l){
}
void Cpare(){
puts("--------------------");
int n=strlen(s+1),_L=0,_R=0;
stack<int>st;
for(int i=1;i<=n;i++){
if(s[i]==‘(‘){
st.push(i);
}
if(s[i]==‘)‘){
L[st.top()]=i;
st.pop();
}
}
for(int i=1;i<=n;i++)cout<<L[i]<<endl;
for(int i=1;i<=n;i++)qzh[i]=qzh[i-1]+(s[i]==‘(‘||s[i]==‘)‘);
for(int i=1;i<=n;i++)if(L[i])isl[i]=qzh[L[i]]-qzh[i-1]-2;
for(int i=1;i<=n;i++)if(L[i])cout<<"~"<<isl[i]<<endl;cout<<endl;
puts("--------------------");
}//debug
int CaseCnt;
struct data{
int u,d,fa;
};
int dep[333333];
void bfs(){
queue<data>q;
q.push((data){1,dep[1]=1,-1});
while(!q.empty()){
data c=q.front();q.pop();
for(int i=last[c.u];i;i=e[i].next){
if(e[i].to!=c.fa){
dep[e[i].to]=dep[c.u]+1;
q.push((data){
e[i].to,c.d+1,c.u});
}
}
}
}
int main(){
freopen("form.in","r",stdin);
freopen("form.out","w",stdout);
while(scanf("%s",s+1)!=EOF){
initial();
if(strlen(s+1)==2&&s[1]==‘(‘&&s[2]==‘)‘)return 0;
int n=strlen(s+1);
s[0]=‘R‘;
stack<char>st,st2;
//Cpare();
int _n=0;
memset(boolean,-1,sizeof(boolean));
for(int i=1;i<=n;i++){
if(s[i]==‘(‘ ){
if(!st.empty()){
link(st.top(),_n+1);
}
st.push(++_n);
}
if(s[i]==‘F‘||s[i]==‘T‘){
if((s[i-1]!=‘F‘&&s[i-1]!=‘T‘)&&
(s[i+1]!=‘F‘&&s[i+1]!=‘T‘)){
//puts("GS");//get stuck
boolean[_n+1]=(s[i]==‘T‘);
link(st.top(),_n+1);_n++;
}else{
boolean[_n+1]=(s[i]==‘T‘);
link(st.top(),++_n);
}
}
if(s[i]==‘)‘)st.pop();
}
n=_n;
bfs();
// cerr<<"GS";//got safely
for(int i=n;i>=1;i--){
if(boolean[i]==-1){
if(last[i])boolean[i]=boolean[e[last[i]].to];
if(dep[i]&1){
for(int v=e[last[i]].next;v;v=e[v].next)
if(dep[e[v].to]>dep[i])
boolean[i]=boolean[i]&&boolean[e[v].to];
}
else
for(int v=e[last[i]].next;v;v=e[v].next)
if(dep[e[v].to]>dep[i]){
boolean[i]=boolean[i]||boolean[e[v].to];
}
}
//cout<<boolean[i]<<‘ ‘;
}//puts("");
printf("%d. ",++CaseCnt);
if(boolean[1])printf("true");
else printf("false");
puts("");
}
return 0;
}
T3
同：BZOJ1041圆上的整点
#include<map>
#include<stack>
#include<queue>
#include<cstdio>
#include<string>
#include<vector>
#include<cstring>
#include<complex>
#include<iostream>
#include<assert.h>
#include<algorithm>
using namespace std;
#define inf 1001001001
#define infll 1001001001001001001LL
#define FOR0(i,n) for(int (i)=0;(i)<(n);++(i))
#define FOR1(i,n) for(int (i)=1;(i)<=(n);++(i))
#define ll long long
#define dbg(vari) cerr<<#vari<<" = "<<(vari)<<endl
#define gmax(a,b) (a)=max((a),(b))
#define gmin(a,b) (a)=min((a),(b))
#define ios0 ios_base::sync_with_stdio(0)
#define Ri register int
#define gc getchar()
#define il inline
il int read(){
bool f=true;
ll x=0;char ch;
while(!isdigit(ch=gc))if(ch==‘-‘)f=false;
while(isdigit(ch)){x=(x<<1)+(x<<3)+ch-‘0‘;ch=gc;}
return f?x:-x;
}
#define gi read()
#define FO(x) freopen(#x".in","r",stdin),freopen(#x".out","w",stdout);
double eps=1e-10;
ll n,m,r,ans;
ll pf(ll x){
return x*x;
}
void get(ll d,ll a,double b){
ll t=b;
if(b-t!=0||__gcd(a,t)!=1||a==t){
return;
}
ll y=d*a*t,x;x=(ll)sqrt(pf(r)-pf(y));
if(x>=n||y>=m)return;
ans+=(n-x)*(m-y)*2;
}
int main(){
FO(dist);
n=gi;m=gi;
int t=gi;
while(t--){
r=gi;
ans=0;
if(pf(r)>pf(n-1)+pf(m-1)){
printf("0 ");continue;
}
if(r<n)ans=ans+m*n-m*r;
if(r<m)ans=ans+m*n-n*r;
for(ll d=1;pf(d)<=2*r;d++){
if(2*r%d==0){
for(ll a=1;pf(a)<=r/d;a++)get(d,a,sqrt(2*r/d-pf(a)));
if(d!=2*r/d)
for(ll a=1;pf(a)<=d/2;a++){
get(2*r/d,a,sqrt(d-pf(a)));
}
}
}
cout<<ans<<‘ ‘;
}
}
时间： 2024-12-14 18:49:46