| Solved | Pro.ID | Title | Ratio(Accepted / Submitted) | |
|---|---|---|---|---|
| 1001 | fraction | 4.17%(7/168) | ||
| 已补 | 1002 | three arrays | 12.69%(76/599) | 字典树 |
| 1003 | geometric problem | 1.59%(1/63) | ||
| 已补 | 1004 | equation | 20.65%(310/1501) | 思维 |
| 已补 | 1005 | permutation 1 | 24.77%(407/1643) | dfs+思维 |
 |
1006 | string matching | 23.12%(724/3131) | exkmp |
|
1007 | permutation 2 | 47.19%(688/1458) | DP |
|
1008 | line symmetric | 1.87%(11/588) | 计算几何 |
| 1009 | discrete logarithm problem | 18.42%(7/38) | ||
| 1010 | find hidden array | 6.25%(2/32) |
题目
1002
将a序列和b序列存放到两个字典树里面，然后每次贪心的匹配a中的0和b中0，或者a中的1和b中的1，如果没有，那就01或者10。最后获得的数字不一定是当前最小的，不过一定是在C中的，所以只需要在最后排个序就行了。
#include <bits/stdc++.h>
using namespace std;
const int N = 100010;
int son[31*N][2],cnt[62*N],tot,n,x;
vector<int> res;
inline void insert(int head,int x){
int p = head;
for(int i=30;i>=0;i--){
int u = x >> i & 1;
if(!son[p][u])son[p][u] = ++tot;
p = son[p][u];
cnt[p]++;
}
}
inline int getVal(int p1,int p2){
int res = 0;
for(int i=30;i>=0;i--){
if(son[p1][0] && son[p2][0] && cnt[son[p1][0]] && cnt[son[p2][0]]){
p1 = son[p1][0];
p2 = son[p2][0];
cnt[p1]--;cnt[p2]--;
}
else if(son[p1][1] && son[p2][1] && cnt[son[p1][1]] && cnt[son[p2][1]]){
p1 = son[p1][1];
p2 = son[p2][1];
cnt[p1]--;cnt[p2]--;
}
else if(son[p1][1] && son[p2][0] && cnt[son[p1][1]] && cnt[son[p2][0]]){
p1 = son[p1][1];
p2 = son[p2][0];
cnt[p1]--;cnt[p2]--;
res |= (1<<i);
}
else if(son[p1][0] && son[p2][1] && cnt[son[p1][0]] && cnt[son[p2][1]]){
p1 = son[p1][0];
p2 = son[p2][1];
cnt[p1]--;cnt[p2]--;
res |= (1<<i);
}
}
return res;
}
int main(){
int T;scanf("%d",&T);
while(T--){
tot = 1;
res.clear();
scanf("%d",&n);
for(int i=1;i<=n;i++){
scanf("%d",&x);
insert(0,x);
}
for(int i=1;i<=n;i++){
scanf("%d",&x);
insert(1,x);
}
for(int i=1;i<=n;i++)res.push_back(getVal(0,1));
sort(res.begin(),res.end());
for(int i=0;i<n;i++){
if(i)printf(" ");
printf("%d",res[i]);
}
puts("");
for(int i=0;i<=tot;i++){
son[i][0] = son[i][1] = 0;
cnt[i] = 0;
}
}
return 0;
}
1004
把n个等式按照 (-bover a) 从小到大排序，然后可以发现把n条线(x={-bover a})画到坐标轴上面可以把x分为n+1个区域，针对每个区域，可以将每个等式的绝对值去掉，然后加到一起变成一个等式，直接解出来看x在不在这个范围即可
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll N = 100010;
ll n;
int t;
ll C;
ll gcd(ll a,ll b){
return b == 0 ? a : gcd(b,a%b);
}
struct node{
int a,b;
bool operator <(const node y){
return (-b) * y.a < -y.b * a;
}
}a[N];
ll aa[N],bb[N],tot;
int main(){
scanf("%d",&t);
while(t--){
tot = 0;
scanf("%lld%lld",&n,&C);
ll A = 0,B = 0;
for(ll i=1;i<=n;i++){
scanf("%lld%lld",&a[i].a,&a[i].b);
A += a[i].a;
B += a[i].b;
}
sort(a+1,a+n+1);
ll resa = 0, resb = 0;
bool flag = false;
for(ll i=n;i>=0;i--){
if(A == 0){
if(B == C){
flag = true;
break;
}
else {
continue;
}
}
resa = C-B;
resb = A;
if(resb < 0){
resa = - resa;
resb = - resb;
}
ll g = gcd(resa > 0 ? resa : -resa,resb);
if(g > 1){
resa/=g;
resb/=g;
}
if(i >= 1){
A -= 2 * a[i].a;
B -= 2 * a[i].b;
}
if(i >= 1 && resa * a[i].a < (-a[i].b * resb))continue;
if(i < n && resa * a[i+1].a >= (-a[i+1].b * resb))continue;
aa[++tot] = resa;
bb[tot] = resb;
}
if(flag)puts("-1");
else{
printf("%lld",tot);
for(ll i = tot; i>= 1;i--){
printf(" %lld/%lld",aa[i],bb[i]);
}
puts("");
}
}
return 0;
}
1005
企图用组合数学，一直wa找不着北，场后先找到前两个数字然后再dfs还是wa，看题解明白了，但是并不知道自己错在哪里
#include <bits/stdc++.h>
using namespace std;
const int N = 50;
int n,k;
int d[N],u[N];
//d中存在n个数字，他们表示着相对大小，实际大小由整个数组的low和hi决定。
//low表示现在填过的数字中最大的和最小的数字是多少。
int dfs(int x,int low,int hi){
if(x == n){
k --;
if(k == 0){
for(int i=0;i<n;i++){
if(i)printf(" ");
printf("%d",d[i] - low + 1);
}
puts("");
return 1;
}
return 0;
}
for(int i = hi - n + 1;i <= low + n -1;i++){
if(u[i])continue;
u[i] = 1;
d[x] = i;
if(dfs(x+1,min(low,i),max(hi,i)) == 1){
u[i] = 0;
return 1;
}
u[i] = 0;
}
return 0;
}
int main(){
int T;scanf("%d",&T);
while(T--){
scanf("%d%d",&n,&k);
d[0] = n;
u[n] = 1;
dfs(1,n,n);
u[n] = 0;
}
return 0;
}
1006
开场SA+LCP获得一T（看SA看晕了），回过头来发现就是个exkmp，直接上板子
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 1e6+10;
char s[N];
int nxt[N];
void pre_EKMP(char x[],int m){
nxt[0] = m;
int j = 0;
while(j+1 < m && x[j] == x[j+1])j++;
nxt[1] = j;
int k = 1;
for(int i=2;i<m;i++){
int p = nxt[k] + k - 1;
int l = nxt[i-k];
if(i + l < p + 1)nxt[i] = l;
else{
j = max(0,p-i+1);
while(i + j < m && x[i+j] == x[j])j++;
nxt[i] = j;
k = i;
}
}
}
int main(){
int T;scanf("%d",&T);
while(T--){
scanf("%s",s);
int len = strlen(s);
pre_EKMP(s,len);
ll res = 0;
for(int i=1;i<len;i++){
if(nxt[i] == len - i){
res += nxt[i];
}
else res += nxt[i] + 1;
}
printf("%lld
",res);
}
return 0;
}
1007
学长过的，是我不会的DP
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll mod = 998244353;
ll dp[100005][2];
int main() {
dp[3][0] = 1LL;
dp[3][1] = 0LL;
dp[4][0] = 1LL;
dp[4][1] = 1LL;
for(int i = 5; i <= 100000; i++) {
dp[i][0] = (dp[i - 1][0] + dp[i - 1][1]) % mod;
dp[i][1] = dp[i - 2][0];
}
int T;
scanf("%d", &T);
while(T--) {
int n, x, y;
scanf("%d%d%d", &n, &x, &y);
if(x + 1 == y && n > 3) {
if(y == n || x == 1) puts("1");
else puts("0");
continue;
}
ll ans = 0;
int l, r;
if(x == 1) l = 1;
else l = x + 1;
if(y == n) r = n;
else r = y - 1;
r -= l - 1;
//cout << r << endl;
if(r <= 3) ans = 1LL;
else ans = (dp[r][0] + dp[r][1]) % mod;
printf("%lld
", ans);
}
return 0;
}
1008
学长过的，我什么都不会
#include <bits/stdc++.h>
using namespace std;
const double eps = 1e-8;
int cmp(double x) {
if (fabs(x) < eps) return 0;
if (x > 0) return 1;
return -1;
}
const double pi = acos(-1.0);
inline double sqr(double x) {
return x * x;
}
struct point {
double x, y;
point() {}
point(double a, double b) : x(a), y(b) {}
void input() {
scanf("%lf%lf", &x, &y);
}
friend point operator + (const point &a, const point &b) {
return point(a.x + b.x, a.y + b.y);
}
friend point operator - (const point &a, const point &b) {
return point(a.x - b.x, a.y - b.y);
}
friend bool operator == (const point &a, const point &b) {
return cmp(a.x - b.x) == 0 && cmp(a.y - b.y) == 0;
}
friend point operator * (const point &a, const double &b) {
return point(a.x * b, a.y * b);
}
friend point operator * (const double &a, const point &b) {
return point(a * b.x, a * b.y);
}
friend point operator / (const point &a, const double &b) {
return point(a.x / b, a.y / b);
}
double norm(){
return sqrt(sqr(x) + sqr(y));
}
};
double det(const point &a, const point &b) {
return a.x * b.y - a.y * b.x;
}
double dot(const point &a, const point &b) {
return a.x * b.x + a.y * b.y;
}
double dist(const point &a, const point &b) {
return (a - b).norm();
}
point rotate_point(const point &p, double A){
double tx = p.x, ty = p.y;
return point(tx * cos(A) - ty * sin(A), tx * sin(A) + ty * cos(A));
}
struct line {
point a, b;
line() {}
line(point x, point y) : a(x), b(y) {}
};
line point_make_line(const point a, const point b) {
return line(a, b);
}
double dis_point_segment(const point p, const point s, const point t) {
if(cmp(dot(p - s, t - s)) < 0) return (p - s).norm();
if(cmp(dot(p - t, s - t)) < 0) return (p - t).norm();
return fabs(det(s - p, t - p) / dist(s, t));
}
void PointProjLine(const point p, const point s, const point t, point &cp){
double r = dot((t - s), (p - s)) / dot(t - s, t - s);
cp = s + r * (t - s);
}
bool PointOnSegment(point p, point s, point t){
return cmp(det(p - s, t - s)) == 0 && cmp(dot(p - s, p - t)) <= 0;
}
bool parallel(line a, line b){
return !cmp(det(a.a - a.b, b.a - b.b));
}
bool line_make_point(line a, line b, point &res){
if(parallel(a, b)) return false;
double s1 = det(a.a - b.a, b.b - b.a);
double s2 = det(a.b - b.a, b.b - b.a);
res = (s1 * a.b - s2 * a.a) / (s1 - s2);
return true;
}
line move_d(line a, const double &len){
point d = a.b - a.a;
d = d / d.norm();
d = rotate_point(d, pi / 2);
return line(a.a + d * len, a.b + d * len);
}
bool SegmentSamePoints(line i, line j){
return max(i.a.x, i.b.x) >= min(j.a.x, j.b.x) &&
max(j.a.x, j.b.x) >= min(i.a.x, i.b.x) &&
max(i.a.y, i.b.y) >= min(j.a.y, j.b.y) &&
max(j.a.y, j.b.y) >= min(i.a.y, i.b.y) &&
cmp(det(j.a - i.a, i.b - i.a) * det(j.b - i.a, i.b - i.a)) <= 0 &&
cmp(det(i.a - j.a, j.b - j.a) * det(i.b - j.a, j.b - j.a)) <= 0;
}
int T;
int n;
point a[1005];
int main(){
scanf("%d", &T);
while(T--){
scanf("%d", &n);
for(int i = 0; i < n; i++){
a[i].input();
}
if(n < 5){
puts("Y");
continue;
}
bool ans = false;
// 关于某条边的中垂线对称
for(int i = 0; i < n; i++){
point fir = a[i];
point sec = a[(i + 1) % n];
point mid_p = (fir + sec) / 2;
point ss1 = sec - fir;
point mid_q = rotate_point(ss1, pi / 2) * (2500.0) + mid_p;
point mid_r = rotate_point(ss1, -pi / 2) * (2500.0) + mid_p;
int cnt = 0;
for(int j = 1; j <= (n + 1) / 2 - 1; j++){
fir = a[(i + 1 + j) % n];
sec = a[(i - j + n) % n];
if(!(fir == sec) && cmp(dis_point_segment(fir, mid_q, mid_r)) == 0 && cmp(dis_point_segment(sec, mid_q, mid_r)) == 0){
cnt = 20;
break;
}
point mid_c = (fir + sec) / 2;
if(cmp(dis_point_segment(mid_c, mid_q, mid_r)) != 0){
cnt++;
}
point cp1, cp2;
PointProjLine(fir, mid_q, mid_r, cp1);
PointProjLine(sec, mid_q, mid_r, cp2);
if(!(cp1 == cp2)){
cnt++;
}
if(n % 2 == 0 && j == (n + 1) / 2 - 1){
point ss1 = a[(i + j) % n] - mid_q;
point ss2 = a[(i + j + 1) % n] - mid_q;
point ss3 = a[(i + j + 2) % n] - mid_q;
point ss4 = a[(i + j + 3) % n] - mid_q;
point ss5 = mid_r - mid_q;
int bb1 = cmp(det(ss5, ss2));
int bb2 = cmp(det(ss5, ss1));
int bb3 = cmp(det(ss5, ss3));
int bb4 = cmp(det(ss5, ss4));
int bb5 = bb1 * bb2;
int bb6 = bb4 * bb3;
if(bb5 > 0 || bb6 > 0){
;
}
else{
cnt += 20;
}
}
if(cnt > 1){
break;
}
}
if(cnt <= 1){
ans = true;
break;
}
}
if(ans){
puts("Y");
continue;
}
// 偶数点时：关于对边连线对称
if(n % 2 == 0){
for(int i = 0; i < n / 2; i++){
point fir = a[i];
point sec = a[i + n / 2];
point ss1 = sec - fir;
point mid_q = ss1 * 2500.0 + fir;
point mid_r = ss1 * (-2500.0) + fir;
int cnt = 0;
for(int j = 1; j < n / 2; j++){
fir = a[(i + j) % n];
sec = a[(i - j + n) % n];
if(!(fir == sec) && cmp(dis_point_segment(fir, mid_q, mid_r)) == 0 && cmp(dis_point_segment(sec, mid_q, mid_r)) == 0){
cnt = 20;
break;
}
point mid_c = (fir + sec) / 2;
if(cmp(dis_point_segment(mid_c, mid_q, mid_r)) != 0){
cnt++;
continue;
}
point cp1, cp2;
PointProjLine(fir, mid_q, mid_r, cp1);
PointProjLine(sec, mid_q, mid_r, cp2);
if(!(cp1 == cp2)){
cnt++;
}
if(cnt > 1){
break;
}
}
if(cnt <= 1){
ans = true;
break;
}
}
if(ans){
puts("Y");
continue;
}
}
// 奇数点时：检查能不能调整正对的两个点。
else{
for(int i = 0; i < n; i++){
point fir = a[(i + 1) % n];
point sec = a[(i - 1 + n) % n];
point mid_p = (fir + sec) / 2;
point ss1 = mid_p - a[i];
point mid_q = ss1 * 2500.0 + a[i];
point mid_r = ss1 * (-2500.0) + a[i];
int cnt = 0;
for(int j = 1; j <= n / 2; j++){
// printf("%d
", j);
fir = a[(i + j) % n];
sec = a[(i - j + n) % n];
if(!(fir == sec) && cmp(dis_point_segment(fir, mid_q, mid_r)) == 0 && cmp(dis_point_segment(sec, mid_q, mid_r)) == 0){
cnt = 20;
break;
}
point mid_c = (fir + sec) / 2;
if(cmp(dis_point_segment(mid_c, mid_q, mid_r)) != 0){
// printf("%d: mid wrong
", j);
cnt++;
}
point cp1, cp2;
PointProjLine(fir, mid_q, mid_r, cp1);
PointProjLine(sec, mid_q, mid_r, cp2);
if(!(cp1 == cp2)){
// printf("%d: cp wrong
", j);
cnt++;
}
point ss1 = a[(i + j - 1 + n) % n] - mid_q;
point ss2 = a[(i + j) % n] - mid_q;
point ss3 = a[(i + j + 1) % n] - mid_q;
point ss4 = a[(i + j + 2) % n] - mid_q;
point ss5 = mid_r - mid_q;
int bb1 = cmp(det(ss5, ss2));
int bb2 = cmp(det(ss5, ss1));
int bb3 = cmp(det(ss5, ss3));
int bb4 = cmp(det(ss5, ss4));
int bb5 = bb1 * bb2;
int bb6 = bb4 * bb3;
if(bb5 > 0 || bb6 > 0){
;
}
else{
cnt += 20;
}
if(cnt > 1){
break;
}
}
if(cnt <= 1){
// printf("%d
", i);
ans = true;
break;
}
}
if(ans){
puts("Y");
continue;
}
}
// 偶数时：还要检查是不是正对面的点是偏的
if(n % 2 == 0){
for(int i = 0; i < n; i++){
point fir = a[(i + 1) % n];
point sec = a[(i - 1 + n) % n];
point mid_p = (fir + sec) / 2;
point ss1 = mid_p - a[i];
point mid_q = ss1 * 2500.0 + a[i];
point mid_r = ss1 * (-2500.0) + a[i];
int cnt = 0;
for(int j = 1; j <= n / 2; j++){
fir = a[(i + j) % n];
sec = a[(i - j + n) % n];
if(!(fir == sec) && cmp(dis_point_segment(fir, mid_q, mid_r)) == 0 && cmp(dis_point_segment(sec, mid_q, mid_r)) == 0){
cnt = 20;
break;
}
point mid_c = (fir + sec) / 2;
if(cmp(dis_point_segment(mid_c, mid_q, mid_r)) != 0){
cnt++;
continue;
}
point cp1, cp2;
PointProjLine(fir, mid_q, mid_r, cp1);
PointProjLine(sec, mid_q, mid_r, cp2);
if(!(cp1 == cp2)){
cnt++;
}
if(cnt > 1){
break;
}
}
if(cnt <= 1){
ans = true;
break;
}
}
if(ans){
puts("Y");
continue;
}
}
if(ans){
puts("Y");
}
else{
puts("N");
}
}
return 0;
}