A. Base $i - 1$ Notation
两个性质：
利用这两条性质实现高精度加法即可。
时间复杂度$O(n)$。
#include&stdio.h&
#include&iostream&
#include&string.h&
#include&string&
#include&ctype.h&
#include&math.h&
#include&set&
#include&map&
#include&vector&
#include&queue&
#include&bitset&
#include&algorithm&
#include&time.h&
const int N = 6e6 + 10;
const int inf = 1e9;
int casenum,
char a[N], b[N];
int main()
freopen("base-i-1.in","r",stdin); freopen("base-i-1.out","w",stdout);
while(~scanf("%s%s", a, b))
int n = strlen(a); reverse(a, a + n);
int m = strlen(b); reverse(b, b + m);
//int n = 5e5; for(int i = 0; i & ++i)a[i] = '1'; a[n] = 0;
//int m = 5e5; for(int i = 0; i & ++i)b[i] = '1'; b[m] = 0;
int g = max(n, m);
for(int i = 0; i & ++i)
int p = i / 2;
if(i & n)c[i] += a[i] - 48;
if(i & m)c[i] += b[i] - 48;
for(int i = 0; i & ++i)
int t=min(c[i]/2,min(c[i+1]/2,c[i+2]));
c[i]-=t*2;
c[i+1]-=t*2;
c[i+2]-=t;
int w = c[i] / 2;
c[i + 2] +=
c[i + 3] +=
g = max(g, i + 4);
c[i] %= 2;
while(g&1&&!c[g-1])g--;
for(int i = g - 1; i &= 0; --i)printf("%d", c[i]);
B. Squaring a Bit
直接暴力枚举平方数即可通过，需要手写popcount。
#include&stdio.h&
#include&iostream&
#include&string.h&
#include&string&
#include&ctype.h&
#include&math.h&
#include&set&
#include&map&
#include&vector&
#include&queue&
#include&bitset&
#include&algorithm&
#include&time.h&
const int N = 3e5 + 10;
typedef long long LL;
int v[65555];
int popcount(LL x){
return v[x&65535]+v[x&&16&65535]+v[x&&32&65535]+v[x&&48];
int main()
for(int i=1;i&65536;i++)v[i]=v[i&&1]+(i&1);
freopen("bit-squares.in","r",stdin); freopen("bit-squares.out","w",stdout);
while(~scanf("%lld", &n))
LL len = 0;
int one = 0;
one += n & 1;
LL bott = 1ll && (len - 1);
int bot = sqrt(bott) + 0.;
LL topp = (1ll && len) - 1;
int top = sqrt(min(topp, (LL)1e18));
//printf("%lld %lld\n", bott, topp);
//printf("%d %d\n", bot, top);
int ans = 0;
LL val = (LL)bot * bot, add = bot * 2 + 1;
for(;i+4&=i+=4){
if(popcount(val) == one)++
val += add, add += 2;
if(popcount(val) == one)++
val += add, add += 2;
if(popcount(val) == one)++
val += add, add += 2;
if(popcount(val) == one)++
val += add, add += 2;
for(; i &= ++i, val += add, add += 2)
//printf("%d %lld\n", i, val);
//val = (LL)i *
if(popcount(val) == one)++
printf("%d\n", ans);
C. Chickens
状压DP求方案数。
#include&stdio.h&
#include&iostream&
#include&string.h&
#include&string&
#include&ctype.h&
#include&math.h&
#include&set&
#include&map&
#include&vector&
#include&queue&
#include&bitset&
#include&algorithm&
#include&time.h&
const int N = 3e5 + 10;
const int inf = 1e9;
int casenum,
int c[15], e[15];
int f[15][1 && 13];
int main()
freopen("chickens.in","r",stdin);
freopen("chickens.out","w",stdout);
while(~scanf("%d", &n))
for(int i = 0; i & ++i)scanf("%d", &c[i]);
for(int i = 0; i & ++i)scanf("%d", &e[i]);
memset(f, 0, sizeof(f)); f[0][0] = 1;
int top = (1 && n) - 1;
for(int i = 0; i & ++i)
for(int j = 0; j &= ++j)if(f[i][j])
for(int k = 0; k & ++k)if((~j && k & 1) && c[i] &= e[k])
f[i + 1][j | 1 && k] += f[i][j];
printf("%d\n", f[n][top]);
D. Lights at a Crossing
对于每个观察，枚举两盏红灯，由剩余时间大的向小的连权值为差值的有向边，则最小周期为这个有向图的最小环。
时间复杂度$O(n^3+mn^2)$。
#include&cstdio&
const int N=30,inf=;
int n,m,i,j,k,f[N][N],a[N],ans=char s[99];
inline void up(int&x,int y){x&y?(x=y):0;}
int main(){
scanf("%d%d",&n,&m);
for(i=1;i&=n;i++)for(j=1;j&=n;j++)f[i][j]=
while(m--){
for(i=1;i&=n;i++){
scanf("%s",s);
if(s[0]=='X')a[i]=-1;else sscanf(s,"%d",&a[i]);
for(i=1;i&=n;i++)if(~a[i])for(j=i+1;j&=n;j++)if(~a[j]){
if(a[i]&a[j])up(f[i][j],a[j]-a[i]);
if(a[i]&a[j])up(f[j][i],a[i]-a[j]);
for(k=1;k&=n;k++)for(i=1;i&=n;i++)for(j=1;j&=n;j++)up(f[i][j],f[i][k]+f[k][j]);
for(i=1;i&=n;i++)up(ans,f[i][i]);
if(ans==inf)ans=-1;
printf("%d",ans);
E. Decimal Form
法雷序列求两分数之间分母最小的分数。
时间复杂度$O(T\log w)$。
#include&cstdio&
#include&iostream&
#include&algorithm&
#include&cmath&
#include&cstring&
typedef __int128
typedef pair&lll,lll&P;
lll n,r,a,b,c,d,t;
const int lim=;
const ld eps=1e-15;
lll gcd(lll a,lll b){return b?gcd(b,a%b):a;}
P cal(lll a,lll b,lll c,lll d){
lll x=a/b+1;
if(x*d&c)return P(x,1);
if(!a)return P(1,d/c+1);
if(a&=b&&c&=d){
P t=cal(d,c,b,a);
swap(t.first,t.second);
P t=cal(a-b*x,b,c-d*x,d);
t.first+=t.second*x;
void write(lll x){
if(x&=10)write(x/10);
printf("%d",(int)x);
//0.666667
int main(){
freopen("decimal-form.in","r",stdin);
freopen("decimal-form.out","w",stdout);
scanf("%d",&Case);
while(Case--){
static char s[100];
scanf("%s",s);
int len=strlen(s);
for(int i=2;i&i++)r=r*10+s[i]-'0';
puts("0 1");
for(t=10;n--;t*=10);
a=r*10-5,b=t;
c=r*10+5,d=t;
lll o=gcd(a,b);
a/=o,b/=o;
o=gcd(c,d);
c/=o,d/=o;
P p=cal(a,b,c,d);
if(b&p.second)p=P(a,b);
write(p.first);
putchar(' ');
write(p.second);
F. Martian Maze
G. Wet Mole
Floodfill求出所有水能灌到的点即可。
时间复杂度$O(nm)$。
#include&cstdio&
const int N=1010;
int n,m,i,j,char a[N][N];bool v[N][N];
inline bool check(int x,int y){
if(x&1||x&n||y&1||y&m)return 0;
if(a[x][y]=='#')return 0;
void dfs(int x,int y){
if(v[x][y])
v[x][y]=1;
if(check(x+1,y))dfs(x+1,y);
if(check(x,y-1))dfs(x,y-1);
if(check(x,y+1))dfs(x,y+1);
int main(){
freopen("mole.in","r",stdin);
freopen("mole.out","w",stdout);
scanf("%d%d",&n,&m);
for(i=1;i&=n;i++)scanf("%s",a[i]+1);
for(i=1;i&=m;i++)if(a[1][i]=='.')dfs(1,i);
for(i=1;i&=n;i++)for(j=1;j&=m;j++)if(a[i][j]=='.'&&!v[i][j]){
if(!flag){
a[i][j]='X';
puts("Yes");
for(i=1;i&=n;i++)puts(a[i]+1);
}else puts("No");
H. Oddities
I. Sorting on the Plane
将所有向量分成两类：在$1$号向量左侧和右侧的。然后分别排序即可。
时间复杂度$O(n\log n)$。
#include&cstdio&
#include&algorithm&
#include&vector&
const int N=100010;
int n,i,f[N],m,t;
vector&int&q[2];
int ask(int x,int y){
printf("? %d %d\n",x,y);
fflush(stdout);
scanf("%d",&t);
bool cmp(int x,int y){return ask(x,y);}
int main(){
scanf("%d",&n);
puts("! YES");
puts("1");
fflush(stdout);
for(i=2;i&=n;i++){
q[ask(1,i)].push_back(i);
for(i=0;i&2;i++)sort(q[i].begin(),q[i].end(),cmp);
for(i=0;i&q[0].size();i++)f[++m]=q[0][i];
for(i=0;i&q[1].size();i++)f[++m]=q[1][i];
if(t=ask(f[1],f[m]))puts("! YES");else puts("! NO");
if(!t)printf("%d ",m);
for(i=1;i&=m;i++)printf("%d ",f[i]);
fflush(stdout);
J. Center of List of Sums
二分求出上下界以及上下界的排名，然后将中间的数暴力取出即可。
时间复杂度$O(n\log n)$。
#include&cstdio&
#include&algorithm&
const int N=2000010;
int n,m,i;ll a[N],b[N],q[N],L,R,down,up,cntdown,
inline ll cal(ll lim){//&=lim
int i=1,j=n;
for(i=1;i&=n;i++){
while(j&&a[i]+b[j]&lim)j--;
inline ll getkth(ll k){
ll l=0,r=,mid,t;
while(l&=r){
mid=(l+r)&&1;
if(cal(mid)&=k)r=(t=mid)-1;else l=mid+1;
void push(ll A,ll B){
int i,j=n,k=n;
for(i=1;i&=n;i++){
while(j&&a[i]+b[j]&=A)j--;
while(k&&a[i]+b[k]&B)k--;
for(int o=j+1;o&=k;o++)q[++m]=a[i]+b[o];
int main(){
freopen("sums-center.in","r",stdin);
freopen("sums-center.out","w",stdout);
scanf("%d",&n);
for(i=1;i&=n;i++)scanf("%lld",&a[i]);
for(i=1;i&=n;i++)scanf("%lld",&b[i]);
sort(a+1,a+n+1);
sort(b+1,b+n+1);
L=1LL*n*(n-1)/2+1;
R=1LL*n*(n+1)/2;
down=getkth(L);
up=getkth(R);
//printf("L=%lld R=%lld down=%lld up=%lld\n",L,R,down,up);
if(down==up){
for(ll o=L;o&=R;o++)printf("%lld ",down);
cntdown=cal(down);//&=down
cntup=cal(up-1);//&up
//printf("cnt=%lld %lld\n",cntdown,cntup);
push(down+1,up-1);
for(ll o=L;o&=cntdown&&o&=R;o++)q[++m]=
for(ll o=R;o&cntup&&o&=L;o--)q[++m]=
sort(q+1,q+m+1);
for(i=1;i&=m;i++)printf("%lld ",q[i]);
K. Cookies
将每个串排序，然后直接DP求出最长链即可。
#include&cstdio&
#include&string&
#include&algorithm&
#include&map&
#include&iostream&
const int N=180000;
int n,m,i;
int len[N];
int f[N],g[N];
string name[N],show[N],need[N];
map&string,int&
string q[N];
int dp(int x){
if(!x)return 0;
if(f[x])return f[x];
int A=f[x],B=0;
for(int i=0;i&name[x].size();i++){
string now=name[x];
now.erase(i,1);
int o=pos[now];
int t=dp(o);
if(t&A)A=t,B=o;
f[x]=A+1,g[x]=B;
return f[x];
int main(){
freopen("word-chains.in","r",stdin); freopen("word-chains.out","w",stdout);
cin&&n;//n=10000
for(i=0;i&n;i++)cin&&need[i];
cin&&m;//173554
name[1]="";
show[1]=".";
for(i=2;i&=m;i++){
cin&&name[i];
show[i]=name[i];
for(i=1;i&=m;i++){
sort(name[i].begin(),name[i].end());
pos[name[i]]=i;//sorted
len[i]=name[i].size();
for(i=0;i&n;i++){
string now=need[i];
sort(now.begin(),now.end());
int x=pos[now];
int cnt=0;
q[++cnt]=show[x];
q[1]=need[i];
cout&&cnt&&
for(int j=1;j&=j++){
cout&&q[j];
if(j&cnt)cout&&" -& ";else cout&&
university
university
L. Xor-fair Division
若所有数异或和非$0$，则答案为$0$，否则答案为$2^n-2$。
#include&cstdio&
long long ans,sum,x;
int main(){
freopen("xorseq.in","r",stdin);
freopen("xorseq.out","w",stdout);
scanf("%d",&n);
ans=1LL&&n;
while(n--){
scanf("%lld",&x);
if(sum)ans=0;
else if(n==1)ans=0;
else ans-=2;
printf("%lld",ans);
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