solution
坑点在于前导零对于统计 0 0 0 的个数的影响.
d p [ p o s ] [ n u m ] [ a p p e a r ] dp[pos][num][appear] dp[pos][num][appear]
n u m − > 100000101 num->100000101 num−>100000101 以此为例，表示 0 , 3 , 9 0,3,9 0,3,9 个数均为奇数.
a p p e a r − > 101100101 appear->101100101 appear−>101100101 以此为例，表示 0 , 2 , 5 , 6 , 9 0,2,5,6,9 0,2,5,6,9 均出现过.
code
/*SiberianSquirrel*//*CuteKiloFish*/
#include <bits/stdc++.h>
//#include<bits/extc++.h>
#include<ext/rope>
#include<ext/pb_ds/assoc_container.hpp>
#include<ext/pb_ds/tree_policy.hpp>
using namespace std;
using namespace __gnu_cxx;
using namespace __gnu_pbds;
#define Inv(x) quick_pow(x, mod - 2)
#define Polynomial vector<ll>
#define DEBUG(x, y) cout << x << ": " << y << '\n';
#define mem(a, x) memset(a, x, sizeof a)
#define right_1_pos(x) __builtin_ffs(x)
#define left_0_num(x) __builtin_clz(x)
#define right_0_num(x) __builtin_ctz(x)
#define num_of_1(x) __builtin_popcount(x)
#define mp_(x, y) make_pair(x, y)
#define all(v) (v).begin(), (v).end()
using i16 = short;
using ll = long long;
using ull = unsigned long long;
//using ill = __int128;
using ld = long double;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
using pdd = pair<double, double>;
#define fi first
#define se second
tree<ll, null_type, less<ll>, rb_tree_tag, tree_order_statistics_node_update> tr;
/* rb_tree_tag 和 splay_tree_tag 选择树的类型(红黑树和伸展树) null_type//无映射（g++为null_mapped_type） less<pii>//从小到大排序 tree_order_statistics_node_update//更新方式 tr.insert(mp_(x, y));//插入 tr.erase(mp_(x, y));//删除 tr.order_of_key(pii(x, y));//求排名 tr.find_by_order(x);//找k小值，返回迭代器 tr.join(b);//将b并入tr，前提是两棵树类型一致并且二没有重复元素 tr.split(v, b);//分裂，key小于等于v的元素属于tr，其余属于b tr.lower_bound(x);//返回第一个大于等于x的元素的迭代器 tr.upper_bound(x);//返回第一个大于x的元素的迭代器 以上的所有操作的时间复杂度均为O(logn) 注意，插入的元素会去重，如set */
//-----------------------------------------------------------------IO Template
namespace StandardIO {
template<typename T> inline void read(T &x) {
int mod = 19260817;
x = 0; T f = 1;
char c = getchar();
for (; c < '0' || c > '9'; c = getchar()) if (c == '-') f = -1;
for (; c >= '0' && c <= '9'; c = getchar()) x = (x * 10 % mod + c - '0') % mod;
x = x * f % mod;
}
template<typename T> inline void write(T x) {
if (x < 0) putchar('-'), x *= -1;
if (x >= 10) write(x / 10);
putchar(x % 10 + '0');
}
}
using namespace StandardIO;
// -------------------------------------------------------------------------------BASIC_MATH
namespace BASIC_MATH {
// const ll mod = 998244353, mod_g = 3, img = 86583718;
const ll mod = 998244353;
const ld eps = 1e-8; // when use double gcd, let eps smaller(1e-4)
const ld pi = acos(-1.0);
const int inf = 0x3f3f3f3f;
const int maxp = 1000;
// Very Basic
ll mul(ll a, ll b) {
ll z = (long double) a / mod * b; ll res = (unsigned long long) a * b - (unsigned long long) z * mod; return (res + mod) % mod; }
// O(1) quick_mul, use long double
inline ll quick_pow(ll ans, ll p, ll res = 1) {
for(; p; p >>= 1, ans = mul(ans, ans) % mod)
if(p & 1) res = mul(res, ans) % mod;
return res % mod;
}
double gcd(double a,double b) {
if(fabs(b) < eps) return a;
if(fabs(a) < eps) return b;
return gcd(b, fmod(a,b));
}
int gcd(int a, int b) {
return __gcd(a, b); }
ll gcd(ll a, ll b) {
return __gcd(a, b); }
// Exgcd
ll exgcd(ll A, ll B, ll &x, ll &y) {
// Ax + By = C
if(B == 0) {
x = 1; y = 0; return A; }
ll _gcd_ = exgcd(B, A % B, x, y);//逆推回去求解
ll t = x;
x = y; y = t - A / B * y;
return _gcd_;
}
void call_exgcd(ll A, ll B, ll C, ll &x, ll &y) {
ll d = exgcd(A, B, x, y);
if(C % d != 0) cout << -1 << '\n';
else {
// k = 0 的通解
x = C / d * x;
y = C / d * y;
x %= mod; while(x < 0) x += mod;
y %= mod; while(y < 0) y += mod;
cout << x << ' ' << y << '\n';
}
}
// Geometry
int sgn(double x) {
if(fabs(x) < eps) return 0; return x > 0? 1: -1; }
inline double sqr(double x) {
return x * x; }
struct Point {
double x, y;
Point() {
}
Point(double _x,double _y): x(_x), y(_y) {
}
void input() {
cin >> x >> y; }
void output() {
if(sgn(x) == 0) x = 0; if(sgn(y) == 0) y = 0;
cout << fixed << setprecision(12) << x << ' ' << y << '\n';
}
bool operator == (Point b)const {
return sgn(x - b.x) == 0 && sgn(y - b.y) == 0; }
bool operator < (Point b)const {
return sgn(x - b.x)== 0? sgn(y - b.y) < 0: x < b.x; }
Point operator -(const Point &b)const {
return Point(x - b.x, y - b.y); }
double operator ^(const Point &b)const {
return x * b.y - y * b.x; } // 叉乘
double operator *(const Point &b)const {
return x * b.x + y * b.y; } // 点乘
double len() {
return hypot(x, y); }
double len2() {
return x * x + y * y; }
double distance(Point p) {
return hypot(x - p.x, y - p.y); }
Point operator +(const Point &b)const {
return Point(x + b.x, y + b.y); }
Point operator *(const double &k)const {
return Point(x * k, y * k); }
Point operator /(const double &k)const {
return Point(x / k, y / k); }
//`计算pa 和 pb 的夹角`
double rad(Point a,Point b) {
Point p = *this; return fabs(atan2( fabs((a - p) ^ (b - p)), (a - p) * (b - p) )); }
//`化为长度为r的向量`
Point trunc(double r) {
double l = len(); if(!sgn(l))return *this; r /= l; return Point(x * r, y * r); }
//`逆时针旋转90度`
Point rotleft() {
return Point(-y, x); }
//`顺时针旋转90度`
Point rotright() {
return Point(y, -x); }
//`绕着p点逆时针旋转angle`
Point rotate(Point p, double angle) {
Point v = (*this) - p; double c = cos(angle), s = sin(angle);
return Point(p.x + v.x * c - v.y * s, p.y + v.x * s + v.y * c);
}
};
struct Line {
Point s, e;
Line() {
}
Line(Point _s, Point _e): s(_s), e(_e) {
}
bool operator ==(Line v) {
return (s == v.s) && (e == v.e); }
//`根据一个点和倾斜角angle确定直线, 0 <= angle < pi`
Line(Point p,double angle) {
s = p;
if(sgn(angle-pi/2) == 0) e = (s + Point(0, 1));
else e = (s + Point(1, tan(angle)));
}
//ax + by + c = 0
Line(double a,double b,double c) {
if(sgn(a) == 0) {
s = Point(0, -c / b);
e = Point(1, -c / b);
} else if(sgn(b) == 0) {
s = Point(-c / a, 0);
e = Point(-c / a, 1);
} else {
s = Point(0, -c / b);
e = Point(1, (-c - a) / b);
}
}
void input() {
s.input(); e.input(); }
void adjust() {
if(e < s) swap(s,e); }
double length() {
return s.distance(e); }
//`倾斜角 0 <= angle < pi`
double angle() {
double k = atan2(e.y - s.y, e.x - s.x);
if(sgn(k) < 0)k += pi;
if(sgn(k - pi) == 0) k -= pi;
return k;
}
//`点和直线关系:`1 在左侧; 2 在右侧; 3 在直线上`
int relation(Point p) {
int c = sgn((p - s) ^ (e - s));
if(c < 0) return 1;
else if(c > 0) return 2;
else return 3;
}
// 点在线段上的判断
bool pointonseg(Point p) {
return sgn((p - s) ^ (e - s)) == 0 && sgn((p - s) * (p - e)) <= 0; }
//`直线平行或重合`
bool parallel(Line v) {
return sgn((e - s) ^ (v.e - v.s)) == 0; }
//`两线段相交判断:`2 规范相交; 1 非规范相交; 0 不相交`
int segcrossseg(Line v) {
int d1 = sgn((e-s)^(v.s-s));
int d2 = sgn((e-s)^(v.e-s));
int d3 = sgn((v.e-v.s)^(s-v.s));
int d4 = sgn((v.e-v.s)^(e-v.s));
if( (d1^d2)==-2 && (d3^d4)==-2 )return 2;
return (d1==0 && sgn((v.s-s)*(v.s-e))<=0) ||
(d2==0 && sgn((v.e-s)*(v.e-e))<=0) ||
(d3==0 && sgn((s-v.s)*(s-v.e))<=0) ||
(d4==0 && sgn((e-v.s)*(e-v.e))<=0);
}
//`直线和线段相交判断:`2 规范相交; 1 非规范相交; 0 不相交`
int linecrossseg(Line v) {
int d1 = sgn((e - s) ^ (v.s - s));
int d2 = sgn((e - s) ^ (v.e - s));
if((d1 ^ d2) == -2) return 2;
return (d1 == 0 || d2 == 0);
}
//`两直线关系:`0 平行; 1 重合; 2 相交`
int linecrossline(Line v) {
if((*this).parallel(v))
return v.relation(s)==3;
return 2;
}
//`求两直线的交点``要保证两直线不平行或重合`
Point crosspoint(Line v) {
double a1 = (v.e-v.s)^(s-v.s);
double a2 = (v.e-v.s)^(e-v.s);
return Point((s.x*a2-e.x*a1)/(a2-a1),(s.y*a2-e.y*a1)/(a2-a1));
}
//点到直线的距离
double dispointtoline(Point p) {
return fabs((p-s)^(e-s))/length(); }
//点到线段的距离
double dispointtoseg(Point p) {
if(sgn((p-s)*(e-s))<0 || sgn((p-e)*(s-e))<0)
return min(p.distance(s),p.distance(e));
return dispointtoline(p);
}
//`返回线段到线段的距离``相交距离为0`
double dissegtoseg(Line v) {
return min(min(dispointtoseg(v.s),dispointtoseg(v.e)),min(v.dispointtoseg(s),v.dispointtoseg(e)));
}
//`返回点p在直线上的投影`
Point lineprog(Point p) {
return s + ( ((e-s)*((e-s)*(p-s)))/((e-s).len2()) ); }
//`返回点p关于直线的对称点`
Point symmetrypoint(Point p) {
Point q = lineprog(p); return Point(2*q.x-p.x,2*q.y-p.y); }
};
struct circle {
Point p; double r;
circle() {
}
circle(Point _p, double _r): p(_p), r(_r) {
}
circle(double x, double y, double _r) {
p = Point(x,y); r = _r; }
//`三角形的外接圆``利用两条边的中垂线得到圆心`
circle(Point a, Point b, Point c) {
Line u = Line((a+b)/2,((a+b)/2)+((b-a).rotleft()));
Line v = Line((b+c)/2,((b+c)/2)+((c-b).rotleft()));
p = u.crosspoint(v);
r = p.distance(a);
}
//`三角形的内切圆``参数bool t没有作用`
circle(Point a, Point b, Point c, bool t) {
Line u,v;
double m = atan2(b.y-a.y,b.x-a.x), n = atan2(c.y-a.y,c.x-a.x);
u.s = a;
u.e = u.s + Point(cos((n+m)/2),sin((n+m)/2));
v.s = b;
m = atan2(a.y-b.y,a.x-b.x) , n = atan2(c.y-b.y,c.x-b.x);
v.e = v.s + Point(cos((n+m)/2),sin((n+m)/2));
p = u.crosspoint(v);
r = Line(a,b).dispointtoseg(p);
}
void input() {
p.input(); cin >> r; }
void output() {
cout << fixed << setprecision(9) << "x:" << p.x << ", y:" << p.y << ", r:" << r << '\n'; }
bool operator == (circle v) {
return (p == v.p) && sgn(r - v.r) == 0; }
bool operator < (circle v)const {
return ((p<v.p)||((p==v.p)&&sgn(r-v.r)<0)); }
double area() {
return pi * r * r; }
double circumference() {
return 2 * pi * r; }
//`点和圆的关系: 0 圆外; 1 圆上; 2 圆内`
int relation(Point b) {
double dst = b.distance(p);
if(sgn(dst-r) < 0)return 2;
else if(sgn(dst-r)==0)return 1;
return 0;
}
//`线段和圆的关系``比较的是圆心到线段的距离和半径的关系`
int relationseg(Line v) {
double dst = v.dispointtoseg(p);
if(sgn(dst-r) < 0)return 2;
else if(sgn(dst-r) == 0)return 1;
return 0;
}
//`直线和圆的关系``比较的是圆心到直线的距离和半径的关系`
int relationline(Line v) {
double dst = v.dispointtoline(p);
if(sgn(dst-r) < 0)return 2;
else if(sgn(dst-r) == 0)return 1;
return 0;
}
//`两圆的关系: 5 相离; 4 外切; 3 相交; 2 内切; 1 内含`
int relationcircle(circle v) {
double d = p.distance(v.p);
if(sgn(d-r-v.r) > 0)return 5;
if(sgn(d-r-v.r) == 0)return 4;
double l = fabs(r-v.r);
if(sgn(d-r-v.r)<0 && sgn(d-l)>0)return 3;
if(sgn(d-l)==0)return 2;
if(sgn(d-l)<0)return 1;
}
//`求两个圆的交点，返回0表示没有交点，返回1是一个交点，2是两个交点`
int pointcrosscircle(circle v,Point &p1,Point &p2) {
int rel = relationcircle(v);
if(rel == 1 || rel == 5)return 0;
double d = p.distance(v.p);
double l = (d*d+r*r-v.r*v.r)/(2*d);
double h = sqrt(r*r-l*l);
Point tmp = p + (v.p-p).trunc(l);
p1 = tmp + ((v.p-p).rotleft().trunc(h));
p2 = tmp + ((v.p-p).rotright().trunc(h));
if(rel == 2 || rel == 4)
return 1;
return 2;
}
//`求直线和圆的交点，返回交点个数`
int pointcrossline(Line v,Point &p1,Point &p2) {
if(!(*this).relationline(v))return 0;
Point a = v.lineprog(p);
double d = v.dispointtoline(p);
d = sqrt(r*r-d*d);
if(sgn(d) == 0) {
p1 = a;
p2 = a;
return 1;
}
p1 = a + (v.e-v.s).trunc(d);
p2 = a - (v.e-v.s).trunc(d);
return 2;
}
//`得到过a,b两点，半径为r1的两个圆`
int gercircle(Point a,Point b,double r1,circle &c1,circle &c2) {
circle x(a,r1),y(b,r1);
int t = x.pointcrosscircle(y,c1.p,c2.p);
if(!t)return 0;
c1.r = c2.r = r1;
return t;
}
//`得到与直线u相切，过点q,半径为r1的圆`
int getcircle(Line u,Point q,double r1,circle &c1,circle &c2) {
double dis = u.dispointtoline(q);
if(sgn(dis-r1*2)>0)return 0;
if(sgn(dis) == 0) {
c1.p = q + ((u.e-u.s).rotleft().trunc(r1));
c2.p = q + ((u.e-u.s).rotright().trunc(r1));
c1.r = c2.r = r1;
return 2;
}
Line u1 = Line((u.s + (u.e-u.s).rotleft().trunc(r1)),(u.e + (u.e-u.s).rotleft().trunc(r1)));
Line u2 = Line((u.s + (u.e-u.s).rotright().trunc(r1)),(u.e + (u.e-u.s).rotright().trunc(r1)));
circle cc = circle(q,r1);
Point p1,p2;
if(!cc.pointcrossline(u1,p1,p2))cc.pointcrossline(u2,p1,p2);
c1 = circle(p1,r1);
if(p1 == p2) {
c2 = c1;
return 1;
}
c2 = circle(p2,r1);
return 2;
}
//`同时与直线u,v相切，半径为r1的圆`
int getcircle(Line u,Line v,double r1,circle &c1,circle &c2,circle &c3,circle &c4) {
if(u.parallel(v))return 0;//两直线平行
Line u1 = Line(u.s + (u.e-u.s).rotleft().trunc(r1),u.e + (u.e-u.s).rotleft().trunc(r1));
Line u2 = Line(u.s + (u.e-u.s).rotright().trunc(r1),u.e + (u.e-u.s).rotright().trunc(r1));
Line v1 = Line(v.s + (v.e-v.s).rotleft().trunc(r1),v.e + (v.e-v.s).rotleft().trunc(r1));
Line v2 = Line(v.s + (v.e-v.s).rotright().trunc(r1),v.e + (v.e-v.s).rotright().trunc(r1));
c1.r = c2.r = c3.r = c4.r = r1;
c1.p = u1.crosspoint(v1);
c2.p = u1.crosspoint(v2);
c3.p = u2.crosspoint(v1);
c4.p = u2.crosspoint(v2);
return 4;
}
//`同时与不相交圆cx,cy相切，半径为r1的圆`
int getcircle(circle cx,circle cy,double r1,circle &c1,circle &c2) {
circle x(cx.p,r1+cx.r),y(cy.p,r1+cy.r);
int t = x.pointcrosscircle(y,c1.p,c2.p);
if(!t)return 0;
c1.r = c2.r = r1;
return t;
}
//`过一点作圆的切线(先判断点和圆的关系)`
int tangentline(Point q,Line &u,Line &v) {
int x = relation(q);
if(x == 2)return 0;
if(x == 1) {
u = Line(q,q + (q-p).rotleft());
v = u;
return 1;
}
double d = p.distance(q);
double l = r*r/d;
double h = sqrt(r*r-l*l);
u = Line(q,p + ((q-p).trunc(l) + (q-p).rotleft().trunc(h)));
v = Line(q,p + ((q-p).trunc(l) + (q-p).rotright().trunc(h)));
return 2;
}
//`求两圆相交的面积`
double areacircle(circle v) {
int rel = relationcircle(v);
if(rel >= 4)return 0.0;
if(rel <= 2)return min(area(),v.area());
double d = p.distance(v.p);
double hf = (r+v.r+d)/2.0;
double ss = 2*sqrt(hf*(hf-r)*(hf-v.r)*(hf-d));
double a1 = acos((r*r+d*d-v.r*v.r)/(2.0*r*d));
a1 = a1*r*r;
double a2 = acos((v.r*v.r+d*d-r*r)/(2.0*v.r*d));
a2 = a2*v.r*v.r;
return a1+a2-ss;
}
//`求两圆相交的面积(精度更高)(需要long double)`
double areacircle2(circle v) {
double a=hypot(p.x-v.p.x,p.y-v.p.y),b=r,c=v.r;
double s1=pi*r*r,s2=pi*v.r*v.r;
if(sgn(a-b-c)>=0)
return 0;
if(sgn(a+min(b,c)-max(b,c))<=0)
return min(s1,s2);
else {
double cta1=2*acos((a*a+b*b-c*c)/(2*a*b));
double cta2=2*acos((a*a+c*c-b*b)/(2*a*c));
return cta1/(2*pi)*s1-0.5*sin(cta1)*b*b+cta2/(2*pi)*s2-0.5*sin(cta2)*c*c;
}
}
//`求圆和三角形pab的相交面积`
double areatriangle(Point a,Point b) {
if(sgn((p-a)^(p-b)) == 0)return 0.0;
Point q[5]; int len = 0;
q[len++] = a;
Line l(a,b); Point p1,p2;
if(pointcrossline(l,q[1],q[2])==2) {
if(sgn((a-q[1])*(b-q[1]))<0)q[len++] = q[1];
if(sgn((a-q[2])*(b-q[2]))<0)q[len++] = q[2];
}
q[len++] = b;
if(len == 4 && sgn((q[0]-q[1])*(q[2]-q[1]))>0)swap(q[1],q[2]);
double res = 0; for(int i = 0; i < len-1; i++) {
if(relation(q[i])==0||relation(q[i+1])==0) {
double arg = p.rad(q[i],q[i+1]);
res += r*r*arg/2.0;
} else {
res += fabs((q[i]-p)^(q[i+1]-p))/2.0;
}
}
return res;
}
};
struct polygon {
int n; Point p[maxp]; Line l[maxp];
void input(int _n) {
n = _n; for(int i = 0; i < n; i++) p[i].input(); }
void add(Point q) {
p[n++] = q; }
void getline() {
for(int i = 0; i < n; i++) l[i] = Line(p[i],p[(i + 1) % n]); }
struct cmp {
Point p;
cmp(const Point &p0) {
p = p0; }
bool operator()(const Point &aa,const Point &bb) {
Point a = aa, b = bb;
int d = sgn((a-p)^(b-p));
if(d == 0) {
return sgn(a.distance(p)-b.distance(p)) < 0;
}
return d > 0;
}
};
//`进行极角排序``首先需要找到最左下角的点`
void norm() {
Point mi = p[0]; for(int i = 1; i < n; i++)mi = min(mi,p[i]); sort(p,p+n,cmp(mi)); }
//`得到的凸包里面的点编号是0~n-1``注意如果有影响，要特判下所有点共点，或者共线的特殊情况`
void getconvex(polygon &convex) {
sort(p,p + n); convex.n = n;
for(int i = 0; i < min(n,2); i++) convex.p[i] = p[i];
if(convex.n == 2 && (convex.p[0] == convex.p[1]))convex.n--;//特判
if(n <= 2)return; int &top = convex.n; top = 1;
for(int i = 2; i < n; i++) {
while(top && sgn((convex.p[top]-p[i])^(convex.p[top-1]-p[i])) <= 0) top--;
convex.p[++top] = p[i];
}
int temp = top; convex.p[++top] = p[n-2];
for(int i = n-3; i >= 0; i--) {
while(top != temp && sgn((convex.p[top]-p[i])^(convex.p[top-1]-p[i])) <= 0) top--;
convex.p[++top] = p[i];
}
if(convex.n == 2 && (convex.p[0] == convex.p[1])) convex.n--; convex.norm();//`原来得到的是顺时针的点，排序后逆时针`
}
//`得到凸包的另外一种方法`
void Graham(polygon &convex) {
norm();
int &top = convex.n; top = 0;
if(n == 1) {
top = 1; convex.p[0] = p[0]; return; }
if(n == 2) {
top = 2;
convex.p[0] = p[0];
convex.p[1] = p[1];
if(convex.p[0] == convex.p[1])top--;
return;
}
convex.p[0] = p[0]; convex.p[1] = p[1];
top = 2;
for(int i = 2; i < n; i++) {
while( top > 1 && sgn((convex.p[top-1]-convex.p[top-2])^(p[i]-convex.p[top-2])) <= 0 )
top--;
convex.p[top++] = p[i];
}
if(convex.n == 2 && (convex.p[0] == convex.p[1]))convex.n--;//特判
}
//`判断是不是凸的`
bool isconvex() {
bool s[3];
memset(s,false,sizeof(s));
for(int i = 0; i < n; i++) {
int j = (i+1)%n;
int k = (j+1)%n;
s[sgn((p[j]-p[i])^(p[k]-p[i]))+1] = true;
if(s[0] && s[2])return false;
}
return true;
}
//`判断点和任意多边形的关系`
//` 3 点上; 2 边上; 1 内部; 0 外部`
int relationpoint(Point q) {
for(int i = 0; i < n; i++) {
if(p[i] == q)return 3;
}
getline();
for(int i = 0; i < n; i++) {
if(l[i].pointonseg(q))return 2;
}
int cnt = 0;
for(int i = 0; i < n; i++) {
int j = (i+1)%n;
int k = sgn((q-p[j])^(p[i]-p[j]));
int u = sgn(p[i].y-q.y);
int v = sgn(p[j].y-q.y);
if(k > 0 && u < 0 && v >= 0)cnt++;
if(k < 0 && v < 0 && u >= 0)cnt--;
}
return cnt != 0;
}
//`直线u切割凸多边形左侧`
//`注意直线方向`
void convexcut(Line u,polygon &po) {
int &top = po.n;//注意引用
top = 0;
for(int i = 0; i < n; i++) {
int d1 = sgn((u.e-u.s)^(p[i]-u.s));
int d2 = sgn((u.e-u.s)^(p[(i+1)%n]-u.s));
if(d1 >= 0)po.p[top++] = p[i];
if(d1*d2 < 0)po.p[top++] = u.crosspoint(Line(p[i],p[(i+1)%n]));
}
}
//`得到周长`
double getcircumference() {
double sum = 0;
for(int i = 0; i < n; i++) {
sum += p[i].distance(p[(i+1)%n]);
}
return sum;
}
//`得到面积`
double getarea() {
double sum = 0;
for(int i = 0; i < n; i++) {
sum += (p[i]^p[(i+1)%n]);
}
return fabs(sum)/2;
}
//`得到方向: 1 表示逆时针; 0表示顺时针`
bool getdir() {
double sum = 0;
for(int i = 0; i < n; i++)
sum += (p[i]^p[(i+1)%n]);
if(sgn(sum) > 0)return 1;
return 0;
}
//`得到重心`
Point getbarycentre() {
Point ret(0,0);
double area = 0;
for(int i = 1; i < n-1; i++) {
double tmp = (p[i]-p[0])^(p[i+1]-p[0]);
if(sgn(tmp) == 0)continue;
area += tmp;
ret.x += (p[0].x+p[i].x+p[i+1].x)/3*tmp;
ret.y += (p[0].y+p[i].y+p[i+1].y)/3*tmp;
}
if(sgn(area)) ret = ret/area;
return ret;
}
//`多边形和圆交的面积`
double areacircle(circle c) {
double ans = 0;
for(int i = 0; i < n; i++) {
int j = (i+1)%n;
if(sgn( (p[j]-c.p)^(p[i]-c.p) ) >= 0)
ans += c.areatriangle(p[i],p[j]);
else ans -= c.areatriangle(p[i],p[j]);
}
return fabs(ans);
}
//`多边形和圆关系: 2 圆完全在多边形内; 1 圆在多边形里面，碰到了多边形边界; 0 其它`
int relationcircle(circle c) {
getline();
int x = 2;
if(relationpoint(c.p) != 1)return 0;//圆心不在内部
for(int i = 0; i < n; i++) {
if(c.relationseg(l[i])==2)return 0;
if(c.relationseg(l[i])==1)x = 1;
}
return x;
}
};
//`AB * AC`
double dot(Point A,Point B,Point C) {
return (B-A)*(C-A);
}
double cross(Point A,Point B,Point C) {
return (B-A)^(C-A);
}
//`最小矩形面积覆盖`A 必须是凸包(而且是逆时针顺序)`
double minRectangleCover(polygon A) {
//`要特判A.n < 3的情况`
if(A.n < 3)return 0.0;
A.p[A.n] = A.p[0];
double ans = -1;
int r = 1, p = 1, q;
for(int i = 0; i < A.n; i++) {
//`卡出离边A.p[i] - A.p[i+1]最远的点`
while( sgn( cross(A.p[i],A.p[i+1],A.p[r+1]) - cross(A.p[i],A.p[i+1],A.p[r]) ) >= 0 )
r = (r+1)%A.n;
//`卡出A.p[i] - A.p[i+1]方向上正向n最远的点`
while(sgn( dot(A.p[i],A.p[i+1],A.p[p+1]) - dot(A.p[i],A.p[i+1],A.p[p]) ) >= 0 )
p = (p+1)%A.n;
if(i == 0)q = p;
//`卡出A.p[i] - A.p[i+1]方向上负向最远的点`
while(sgn(dot(A.p[i],A.p[i+1],A.p[q+1]) - dot(A.p[i],A.p[i+1],A.p[q])) <= 0)
q = (q+1)%A.n;
double d = (A.p[i] - A.p[i+1]).len2();
double tmp = cross(A.p[i],A.p[i+1],A.p[r]) *
(dot(A.p[i],A.p[i+1],A.p[p]) - dot(A.p[i],A.p[i+1],A.p[q]))/d;
if(ans < 0 || ans > tmp)ans = tmp;
}
return ans;
}
circle minCircleCover(int n, Point p[], Point P = Point(0, 0)) {
random_shuffle(p, p + n); double r2 = 0;
for(int i = 0; i < n; ++ i) {
if((p[i] - P).len2() > r2) {
P = p[i], r2 = 0;
for(int j = 0; j < i; ++ j) {
if((p[j]-P).len2() > r2) {
P = (p[i]+p[j])/2, r2 = (p[j]-P).len2();
for(int k = 0; k < j; ++ k) {
if((p[k]-P).len2() > r2) {
P = circle(p[i], p[j], p[k]).p, r2 = (p[k] - P).len2();
}
}
}
}
}
}
return circle(P, sqrt(r2));
}
}
using namespace BASIC_MATH;
namespace MODINT {
template<unsigned M_> struct ModInt {
static constexpr unsigned M = M_;
unsigned x;
constexpr ModInt() : x(0U) {
}
constexpr ModInt(unsigned x_) : x(x_ % M) {
}
constexpr ModInt(unsigned long long x_) : x(x_ % M) {
}
constexpr ModInt(int x_) : x(((x_ %= static_cast<int>(M)) < 0) ? (x_ + static_cast<int>(M)) : x_) {
}
constexpr ModInt(long long x_) : x(
((x_ %= static_cast<long long>(M)) < 0) ? (x_ + static_cast<long long>(M)) : x_) {
}
ModInt &operator+=(const ModInt &a) {
x = ((x += a.x) >= M) ? (x - M) : x;
return *this;
}
ModInt &operator-=(const ModInt &a) {
x = ((x -= a.x) >= M) ? (x + M) : x;
return *this;
}
ModInt &operator*=(const ModInt &a) {
x = (static_cast<unsigned long long>(x) * a.x) % M;
return *this;
}
ModInt &operator/=(const ModInt &a) {
return (*this *= a.inv()); }
ModInt quick_pow(long long e) const {
if (e < 0) return inv().quick_pow(-e);
ModInt a = *this, b = 1U;
for (; e; e >>= 1) {
if (e & 1) b *= a;
a *= a;
}
return b;
}
ModInt inv() const {
unsigned a = M, b = x;
int y = 0, z = 1;
for (; b;) {
const unsigned q = a / b;
const unsigned c = a - q * b;
a = b; b = c;
const int w = y - static_cast<int>(q) * z;
y = z; z = w;
}
assert(a == 1U);
return ModInt(y);
}
ModInt operator+() const {
return *this; }
ModInt operator-() const {
ModInt a; a.x = x ? (M - x) : 0U; return a; }
ModInt operator+(const ModInt &a) const {
return (ModInt(*this) += a); }
ModInt operator-(const ModInt &a) const {
return (ModInt(*this) -= a); }
ModInt operator*(const ModInt &a) const {
return (ModInt(*this) *= a); }
ModInt operator/(const ModInt &a) const {
return (ModInt(*this) /= a); }
template<class T> friend ModInt operator+(T a, const ModInt &b) {
return (ModInt(a) += b); }
template<class T> friend ModInt operator-(T a, const ModInt &b) {
return (ModInt(a) -= b); }
template<class T> friend ModInt operator*(T a, const ModInt &b) {
return (ModInt(a) *= b); }
template<class T> friend ModInt operator/(T a, const ModInt &b) {
return (ModInt(a) /= b); }
explicit operator bool() const {
return x; }
bool operator==(const ModInt &a) const {
return (x == a.x); }
bool operator!=(const ModInt &a) const {
return (x != a.x); }
friend std::ostream &operator<<(std::ostream &os, const ModInt &a) {
return os << a.x; }
};
// constexpr unsigned MO = 1000000007;
constexpr unsigned MO = 998244353;
using Mint = ModInt<MO>;
}
using namespace MODINT;
int a[23];
ll dp[23][1 << 10 | 1][1 << 10 | 1];
ll dfs(int pos, bool limit, bool lead, int now, int appear) {
if(pos == 0) {
int cnt = 1; while(appear) {
if(appear & 1) if ((now & 1) != cnt) return 0;
now >>= 1; appear >>= 1;
cnt = 1 - cnt;
}
return 1;
}
if(!lead && !limit && dp[pos][now][appear] != -1) return dp[pos][now][appear];
int up = limit? a[pos]: 9;
ll ans = 0;
for(int i = 0; i <= up; ++ i) {
bool st = lead && i == 0;
ans += dfs(pos - 1, limit && i == up, lead && i == 0, st? now: (now ^ (1 << i)), st? appear: (appear | (1 << i)));
}
if(!lead && !limit) dp[pos][now][appear] = ans;
return ans;
}
inline ll cala(ll n) {
int pos = 0; while(n) a[++ pos] = n % 10, n /= 10;
return dfs(pos, true, true, 0, 0);
}
inline void solve() {
mem(dp, -1);
int o; cin >> o; while(o --) {
ll A, B; cin >> A >> B;
cout << cala(B) - cala(A - 1) << '\n';
}
}
signed main() {
ios_base::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
#ifdef ACM_LOCAL
freopen("input", "r", stdin);
freopen("output", "w", stdout);
signed test_index_for_debug = 1;
char acm_local_for_debug = 0;
do {
if (acm_local_for_debug == '$') exit(0);
if (test_index_for_debug > 20)
throw runtime_error("Check the stdin!!!");
auto start_clock_for_debug = clock();
solve();
auto end_clock_for_debug = clock();
cout << "\nTest " << test_index_for_debug << " successful!" << endl;
cerr << "Test " << test_index_for_debug++ << " Run Time: "
<< double(end_clock_for_debug - start_clock_for_debug) / CLOCKS_PER_SEC << "s" << endl;
cout << "--------------------------------------------------" << endl;
} while (cin >> acm_local_for_debug && cin.putback(acm_local_for_debug));
#else
solve();
#endif
return 0;
}
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