A. Add and Reverse
要么全部都选择$+1$，要么加出高$16$位后翻转位序然后再补充低$16$位。
#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>
using namespace std;
void fre() { }
#define MS(x, y) memset(x, y, sizeof(x))
#define ls o<<1
#define rs o<<1|1
typedef long long LL;
typedef unsigned long long UL;
typedef unsigned int UI;
template <class T1, class T2>inline void gmax(T1 &a, T2 b) { if (b > a)a = b; }
template <class T1, class T2>inline void gmin(T1 &a, T2 b) { if (b < a)a = b; }
const int N = 1<<16, M = 0, Z = 1e9 + 7, inf = 0x3f3f3f3f;
template <class T1, class T2>inline void gadd(T1 &a, T2 b) { a = (a + b) % Z; }
int casenum, casei;
LL n;
int f[N];
int main()
{
scanf("%lld",&n);
LL A=n>>16;
LL B=n&65535;
LL ans=B;
if(A)ans++;
for(LL i=0;i<16;i++)if(A>>(15-i)&1)ans+=1LL<<i;
printf("%lld",min(ans,n));
return 0;
if(0)while(~scanf("%lld",&n))
{
LL top = 1ll << 32;
LL i = n;
LL w =
printf("%lld\n", w);
}
//return 0;
{
MS(f, 63);
f[0] = 0;
queue<int>q;
q.push(0);
int top = 1ll << 16;
while(!q.empty())
{
int x = q.front(); q.pop();
int y = x + 1;
if(y <= top && f[x] + 1 < f[y])
{
f[y] = f[x] + 1;
q.push(y);
}
int z = top - x;
if(f[x] + 1 < f[z])
{
f[z] = f[x] + 1;
q.push(y);
}
}
for(int i = 0; i < top; ++i)
{
int w = i <= top / 2 ? i : top + 1 - i;
printf("%d: %d %d\n", i, f[i], w);
if(w != f[i])
{
puts("no!!!");
while(1);
}
}
}
return 0;
} /*
【trick&&吐槽】 【题意】 【分析】 【时间复杂度&&优化】 */
B. Analyze This
bitset加速暴力。
#include<cstdio>
const int N=400010;
int n,m,mx,lim,i,j,x,a[N],f[N][2];unsigned long long v[64][N/64+5];
inline void solve(int d){
int i,j,A=d>>6,B=d&63;
for(i=0;i+A<=lim;i++)if(v[0][i]&v[B][i+A])for(j=i<<6;;j++)if(a[j]&&a[j+d]){
f[d][0]=a[j];
f[d][1]=a[j+d];
return;
}
}
int main(){
scanf("%d%d",&n,&m);
for(i=1;i<=n;i++){
scanf("%d",&x);
a[x]=i;
if(x>mx)mx=x;
}
lim=mx/64;
for(i=0;i<=mx;i++)if(a[i])for(j=0;j<64&&j<=i;j++)v[j][(i-j)>>6]|=1ULL<<((i-j)&63);
for(f[0][1]=i=1;i<=mx;i++){
f[i][0]=f[i-1][0];
f[i][1]=f[i-1][1];
solve(i);
}
while(m--){
scanf("%d",&x);
if(x>mx)x=mx;
printf("%d %d\n",f[x][0],f[x][1]);
}
}
C. Bipartite Graph
每个点向附近$3$个点连边即可。
#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>
using namespace std;
void fre() { }
#define MS(x, y) memset(x, y, sizeof(x))
#define ls o<<1
#define rs o<<1|1
typedef long long LL;
typedef unsigned long long UL;
typedef unsigned int UI;
template <class T1, class T2>inline void gmax(T1 &a, T2 b) { if (b > a)a = b; }
template <class T1, class T2>inline void gmin(T1 &a, T2 b) { if (b < a)a = b; }
const int N = 0, M = 0, Z = 1e9 + 7, inf = 0x3f3f3f3f;
template <class T1, class T2>inline void gadd(T1 &a, T2 b) { a = (a + b) % Z; }
int casenum, casei;
int T;
int main()
{
scanf("%d",&T);
while(T--)
{
int n; scanf("%d", &n);
printf("%d\n", (n - 2) * 3);
for(int d = 0; d <= 2; ++d)
{
for(int i = 0; i < n - 2; ++i)
{
printf("%d %d\n", d + i, i);
}
}
}
return 0;
} /*
【trick&&吐槽】 【题意】 【分析】 【时间复杂度&&优化】 */
D. Bridge Building
二分答案，那么对于$x$个物品$a$，$y$的数量是定的，设$f[i][j]$表示$i$个$a$和$j$个$b$最多拼出几列，按性价比从小到大背包，一旦有解则返回。
时间复杂度$O(n^3\log n)$。
#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>
using namespace std;
const int N=505;
int X,A,Y,B,L,mid,f[N][N];
inline void up(int&a,int b){a<b?(a=b):0;}
pair<int,int>a[N];
bool cmp(pair<int,int> a, pair<int,int> b)
{
int val1 = a.first * A + a.second * B;
int val2 = b.first * A + b.second * B;
return val1 < val2;
}
inline bool check(){
int i,j,k;
for(i=0;i<=X;i++)for(j=0;j<=Y;j++)f[i][j]=0; int g = 0;
int pre=10000000;
for(i=0;i<=X;i++){
//A*i+B*j>=mid
int ret=mid-A*i;
int tmp=0;
if(ret>0)tmp=ret/B+((ret%B)>0);
if(tmp>=pre)continue;
pre=tmp;
if(tmp>Y)continue;
a[++g].first = i; a[g].second = tmp;
if(ret <= 0)break;
}
//x++, y-- /*
for(int j = X; j >= 0; --j)
{
for(int k = Y; k >= 0; --k)
{
int st = g + 1;
int l = 1;
int r = g;
while(l <= r)
{
int mid = (l+r)/2;
if(a[mid].second <= k)
{
st = mid;
r = mid - 1;
}
else l = mid + 1;
}
if(st > g || a[st].first > j)break;
if(f[j][k] >= L - 1)return 1;
for(int i = st; i <= g && a[i].first <= j; ++i)
{
up(f[j - a[i].first][k - a[i].second], f[j][k] + 1);
}
}
}
*/ sort(a + 1, a + g + 1, cmp); for(int o=1;o<=g;++o){ int i = a[o].first;
int tmp = a[o].second; for(j=X;j>=i;j--)for(k=Y;k>=tmp;k--){
if(f[j][k]>=L-1)return 1;
up(f[j-i][k-tmp],f[j][k]+1);
}
}
//for(i=0;i<=X;i++)for(j=0;j<=Y;j++)if(f[i][j]>=L)return 1; return 0;
}
int main(){
while(~scanf("%d%d%d%d%d",&X,&A,&Y,&B,&L)){
int l=1,r=(A*X+Y*B)/L,ans=0;
while(l<=r){
mid=(l+r)>>1;
if(check())l=(ans=mid)+1;else r=mid-1;
}
printf("%d\n",ans);
}
}
/*
1 1 1 1 1
*/
E. Child’s Game with Robot
顺时针走一圈，若目的地在中心则回来，否则在目的地两侧来回挪动，若奇偶性不同则原地不动一次进行调整。
#include<cstdio>
int movenorth(){//founded?
puts("move north");
fflush(stdout);
char s[100];
scanf("%s",s);
return s[0]=='f';
}
int movewest(){//founded?
puts("move west");
fflush(stdout);
char s[100];
scanf("%s",s);
return s[0]=='f';
}
int moveeast(){//founded?
puts("move east");
fflush(stdout);
char s[100];
scanf("%s",s);
return s[0]=='f';
}
int movesouth(){//founded?
puts("move south");
fflush(stdout);
char s[100];
scanf("%s",s);
return s[0]=='f';
}
int stay(){//founded?
puts("echo Ready!");
fflush(stdout);
char s[100];
scanf("%s",s);
return s[0]=='f';
}
int main(){
if(movenorth()){
stay();
for(int i=1;i<=4;i++){
movesouth();
movenorth();
}
return 0;
}
if(movewest()){
for(int i=1;i<=4;i++){
moveeast();
movewest();
}
return 0;
}
if(movesouth()){
stay();
for(int i=1;i<=3;i++){
movesouth();
movenorth();
}
return 0;
}
if(movesouth()){
for(int i=1;i<=3;i++){
movenorth();
movesouth();
}
return 0;
}
if(moveeast()){
stay();
for(int i=1;i<=2;i++){
movenorth();
movesouth();
}
return 0;
}
if(moveeast()){
for(int i=1;i<=2;i++){
movenorth();
movesouth();
}
return 0;
}
if(movenorth()){
stay();
movenorth();
movesouth();
return 0;
}
if(movenorth()){
movesouth();
movenorth();
return 0;
}
movesouth();
movewest();
return 0;
}
F. Quadruples of Points
给每个集合一个unsigned long long的随机权值，扫描线+树状数组查询矩形内所有点的权值和即可。
#include<cstdio>
#include<algorithm>
using namespace std;
typedef unsigned long long ll;
const int N=2000010;
ll ran,goal;
int n,m,i,j,ce;
int a[N],cnt;
ll bit[N],ans[N];
struct P{
int x,l,r,t;ll v;
P(){}
P(int _x,int _l,int _r,int _t,ll _v){x=_x,l=_l,r=_r,t=_t,v=_v;}
}e[N];
inline bool cmp(const P&a,const P&b){
return a.x!=b.x?a.x<b.x:a.t<b.t;
}
inline void ins(int x,ll p){for(;x<=cnt;x+=x&-x)bit[x]+=p;}
inline ll ask(int x){ll t=0;for(;x;x-=x&-x)t+=bit[x];return t;}
inline ll sum(int l,int r){return ask(r)-ask(l-1);}
int main(){
scanf("%d%d",&n,&m);
for(i=1;i<=n;i++){
ran=ran*233+17;
goal+=ran;
goal+=ran;
for(j=0;j<4;j++){
int x,y;
scanf("%d%d",&x,&y);
e[++ce]=P(x,y,0,0,ran);
a[++cnt]=y;
}
}
for(i=1;i<=m;i++){
int x1,y1,x2,y2;
scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
e[++ce]=P(x2,y1,y2,1,i);
e[++ce]=P(x1-1,y1,y2,2,i);
a[++cnt]=y1;
a[++cnt]=y2;
}
sort(a+1,a+cnt+1);
sort(e+1,e+ce+1,cmp);
for(i=1;i<=ce;i++){
if(e[i].t==0){
ins(lower_bound(a+1,a+cnt+1,e[i].l)-a,e[i].v);
}
if(e[i].t==1){
ans[e[i].v]+=sum(lower_bound(a+1,a+cnt+1,e[i].l)-a,lower_bound(a+1,a+cnt+1,e[i].r)-a);
}
if(e[i].t==2){
ans[e[i].v]-=sum(lower_bound(a+1,a+cnt+1,e[i].l)-a,lower_bound(a+1,a+cnt+1,e[i].r)-a);
}
}
for(i=1;i<=m;i++)puts(ans[i]==goal?"YES":"NO");
}
/*
2 3
0 0
0 1
1 0
1 2
2 0
2 -1
2 -2
2 -3
0 -1 2 0
0 -1 2 1
0 0 0 1 1 1
0 0
0 1
1 0
1 2
0 -1 2 0
*/
G. Mosaic Tracery
找到度数为$2$的点作为角落然后开始构造。
H. List of Powers
若$r-l$比较小，那么可以枚举其中每个数，用BSGS检查。
否则$r-l$比较大，因为$a^k\bmod p$分布非常随机，而答案不超过$100$个，故周期很短，暴力枚举循环节内所有$k$即可。
注意BSGS检查次数比较多，可以通过增加步长，牺牲预处理复杂度来减少每次查询的复杂度。
#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>
#include<tr1/unordered_set>
using namespace std;
using namespace std::tr1;
void fre() { }
#define MS(x, y) memset(x, y, sizeof(x))
#define ls o<<1
#define rs o<<1|1
typedef long long LL;
typedef unsigned long long UL;
typedef unsigned int UI;
template <class T1, class T2>inline void gmax(T1 &a, T2 b) { if (b > a)a = b; }
template <class T1, class T2>inline void gmin(T1 &a, T2 b) { if (b < a)a = b; }
const int N = 1e5 + 10, M = 0, Z = 1e9 + 7, inf = 0x3f3f3f3f;
template <class T1, class T2>inline void gadd(T1 &a, T2 b) { a = (a + b) % Z; }
int casenum, casei; LL pow_mod(LL x, int y, int Z)
{
LL ans = 1;
while(y){
if(y & 1) ans = ans * x % Z;
y >>= 1;
x = x * x % Z;
}
return ans;
} int s;
unordered_set<int> rec;
int cur[N], powmod[N];
void init(int m, int x)
{
s = (int) (sqrt((double)m));
for(; (LL) s * s <= m;) s ++;
int Cur = 1;
for(int i = 0; i < s; i ++){
rec.insert(Cur);
Cur = 1LL * Cur * x % m;
}
int mul = Cur;
cur[0] = 1; powmod[0] = pow_mod(cur[0], m - 2, m);
for(int i = 1; i < s; i ++){
cur[i] = 1LL * cur[i - 1] * mul % m;
powmod[i] = pow_mod(cur[i], m - 2, m);
} } LL discrete_log(int x, int n, int m)
{
for(int i = 0; i < s; i ++){
int more = 1LL * n * powmod[i] % m;
if(rec.find(more)!=rec.end()){
return 1;
}
}
return -1;
} int P, A, L, R;
set<int>sot;
set<int> :: iterator it; inline LL po(LL a,LL b,LL p){
b=(b%(p-1)+p-1)%(p-1);
a%=p;
LL t=1;
for(;b;b>>=1LL,a=a*a%p)if(b&1)t=t*a%p;
return t;
} int main()
{
scanf("%d%d%d%d", &P, &A, &L, &R);
int phi=P-1;
int per=phi;
for(int i=1;i*i<=phi;i++)if(phi%i==0){
if(po(A,i,P)==1)per=min(per,(int)(i));
if(po(A,phi/i,P)==1)per=min(per,(int)(phi/i));
}
sot.clear();
int lim = 1e8;
if(R-L>2000){
LL t = 1;
for(int i = 1; i <= per; i ++){
t = t * A;
if(t==1)break;
if(t >= P) t %= P;
if(t <= R && t >= L){
sot.insert(t);
}
//if(sot.size() >= 100) break;
} for(it = sot.begin(); it != sot.end(); it ++){
printf("%d ", *it);
}puts("");
}
else{
vector<int> sot;
init(P, A);
for(int i = L; i <= R; i ++){
if(discrete_log(A, i, P) != -1) sot.push_back(i);
//if(sot.size() >= 100) break;
} for(auto it = sot.begin(); it != sot.end(); it ++){
printf("%d ", *it);
}puts("");
}
return 0;
} /*
【trick&&吐槽】 【题意】 【分析】 【时间复杂度&&优化】 */
I. Potential Well
答案就是边权平均数最小的环，求出答案后就是差分约束系统模型，Bellman-Ford即可。
对于边权平均数最小的环，设$f[i][j]$表示经过$i$条边到达$j$的最短路，则：
\[ans=\min_{i=1}^n\{\max_{j=0}^{n-1}\frac{f[n][i]-f[j][i]}{n-j}\}\]
时间复杂度$O(nm)$。
#include<cstdio>
typedef long long ll;
const int N=1005,M=100005;
const double inf=1e18;
int n,m,i,j,u[M],v[M];double w[M],f[N][N],d[N],ans=inf,now,tmp;
inline void up(double&a,double b){a>b?(a=b):0;}
int main(){
scanf("%d%d",&n,&m);
for(i=1;i<=m;i++)scanf("%d%d%lf",&u[i],&v[i],&w[i]);
for(i=1;i<=n;i++)for(j=1;j<=n;j++)f[i][j]=inf;
for(i=0;i<n;i++)for(j=1;j<=m;j++)up(f[i+1][v[j]],f[i][u[j]]+w[j]);
for(i=1;i<=n;i++)if(f[n][i]<inf/2){
now=-inf;
for(j=0;j<n;j++)if(f[j][i]<inf/2){
tmp=1.0*(f[n][i]-f[j][i])/(n-j);
if(now<tmp)now=tmp;
}
up(ans,now);
}
if(ans>inf/2)return puts("+inf"),0;
for(i=1;i<=n;i++)for(j=1;j<=m;j++)up(d[v[j]],d[u[j]]+w[j]-ans);
printf("%.10f\n",ans);
for(i=1;i<=n;i++)printf("%.10f ",d[i]);
}
J. Steiner Tree in Random Graph
留坑。
K. Rotation Transformation
列方程推公式即可，取精度误差最小的作为解。
#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>
using namespace std;
void fre() { }
#define MS(x, y) memset(x, y, sizeof(x))
#define ls o<<1
#define rs o<<1|1
typedef long long LL;
typedef unsigned long long UL;
typedef unsigned int UI;
template <class T1, class T2>inline void gmax(T1 &a, T2 b) { if (b > a)a = b; }
template <class T1, class T2>inline void gmin(T1 &a, T2 b) { if (b < a)a = b; }
const int N = 0, M = 0, Z = 1e9 + 7, inf = 0x3f3f3f3f;
template <class T1, class T2>inline void gadd(T1 &a, T2 b) { a = (a + b) % Z; }
int casenum, casei;
double a[4][4], b[4][4];
const double PI = acos(-1.0), eps = 1e-8;
int sgn(double x)
{
if(fabs(x) < eps) return 0;
return x > 0 ? 1 : -1;
}
double th, sinth, costh, ux, uy, uz;
int main()
{ /*
scanf("%lf", &th);
th = th / 180 * PI;
scanf("%lf%lf%lf", &ux, &uy, &uz);
sinth = sin(th);
costh = cos(th);
b[1][1] = costh + ux * ux * (1 - costh);
b[1][2] = ux * uy * (1 - costh) - uz * sinth;
b[1][3] = ux * uz * (1 - costh) + uy * sinth;
b[2][1] = uy * ux * (1 - costh) + uz * sinth;
b[2][2] = costh + uy * uy * (1 - costh);
b[2][3] = uy * uz * (1 - costh) - ux * sinth;
b[3][1] = uz * ux * (1 - costh) - uy * sinth;
b[3][2] = uz * uy * (1 - costh) + ux * sinth;
b[3][3] = costh + uz * uz * (1 - costh);
for(int i = 1; i <= 3; i ++){
for(int j = 1; j <= 3; j ++){
//if(sgn(a[i][j] - b[i][j])){
//printf("%d %d %lf %lf", i, j, a[i][j], b[i][j]);
//} }
}
*/ for(int i = 1; i <= 3; i ++){
for(int j = 1; j <= 3; j ++){
scanf("%lf", &a[i][j]);
//a[i][j] = b[i][j];
}
} uz = (a[2][1] - a[1][2]) / 2;
uy = (a[1][3] - a[3][1]) / 2;
ux = (a[3][2] - a[2][3]) / 2;
sinth = uz * uz + uy * uy + ux * ux;
sinth = sqrt(sinth);
th = asin(sinth);
if(sgn(sinth)){
uz /= sinth;
uy /= sinth;
ux /= sinth;
}
else{
ux = 0;
uy = 0;
uz = 1;
} double t[5];
t[1] = th; t[2] = -th; t[3] = PI - th; t[4] = th - PI; double differ = 1e9; int oo = 1;
for(int o = 1; o <= 4; o ++){
double tmp = 0; sinth = sin(t[o]);
costh = cos(t[o]);
b[1][1] = costh + ux * ux * (1 - costh);
b[1][2] = ux * uy * (1 - costh) - uz * sinth;
b[1][3] = ux * uz * (1 - costh) + uy * sinth;
b[2][1] = uy * ux * (1 - costh) + uz * sinth;
b[2][2] = costh + uy * uy * (1 - costh);
b[2][3] = uy * uz * (1 - costh) - ux * sinth;
b[3][1] = uz * ux * (1 - costh) - uy * sinth;
b[3][2] = uz * uy * (1 - costh) + ux * sinth;
b[3][3] = costh + uz * uz * (1 - costh);
for(int i = 1; i <= 3; i ++){
for(int j = 1; j <= 3; j ++){
tmp += fabs(a[i][j] - b[i][j]);
}
}
//printf("tmp %d %.10f\n", o, tmp);
if(sgn(tmp - differ) < 0){
differ = tmp;
oo = o;
}
}
th = t[oo];
//printf("differ = %.10f\n", differ);
th = th / PI * 180; printf("%.10f\n", th);
printf("%.10f %.10f %.10f\n", ux, uy, uz);
return 0;
} /*
【trick&&吐槽】 【题意】 【分析】 【时间复杂度&&优化】 */