220803 总结
220803 总结
总之就是挂大分
顺序开题
T1 解方程：意料之中，没想到可以边读入边取模。
T2 三角形：明明暴力能拿 \(40\) 分，非要人类智慧乱搞，痛失 \(20\) 分
T3 游戏（屠龙勇士）：屠龙勇士反被龙屠考场上推出来式子了但是忘了 \(x\) 前面有系数该怎么解于是去打暴力，盯着 \(p=1\) 的 \(30\) 分去的，又由于没想到剑的攻击力可以重复，开的 set 没开 multiset 痛失 \(10\) 分
T4 找朋友：不知道怎么回事反正输出全是零（
T1 解方程
因为 \(N,M\) 范围都很小，所以完全可以在 \([1,M]\) 的区间内将所有的正整数 \(x\) 代入方程看结果是否为 \(0\)
问题出在离谱的 \(|a_i|\leq10^{10000}\) 上，但是因为最后只需要判断结果是否为 \(0\) 即可，所以可以在模意义下做这题
#include<bits/stdc++.h>
using namespace std;
#define FO(name) freopen(name".in","r",stdin);freopen(name".out","w",stdout)
const int maxN = 1e6+10;
typedef long long ll;
int N,M;
ll A[maxN];
int Ans[maxN];
const int MOD = 998244353;
inline bool clc(ll x){
int ans = 0;
for(int i = N;i>=1;i--){
ans = ((ans + A[i]) * x) % MOD;
}
ans = (ans + A[0]) % MOD;
return ans == 0;
}
inline ll read(){
ll x=0;int f=1;char c=getchar();
while(!isdigit(c)){if(c=='-')f=-1;c=getchar();}
while(isdigit(c)){x=(x<<3)+(x<<1)+(c^48);x=x%MOD;c=getchar();}
return x*f;
}
int cnt;
int main(){
scanf("%d%d",&N,&M);
for(int i = 0;i<=N;i++) A[i] = read();
for(int i = 1;i<=M;i++){
if(clc(i)) Ans[++cnt] = i;
}
printf("%d\n",cnt);
for(int i = 1;i<=cnt;i++) printf("%d\n",Ans[i]);
return 0;
}
T2 三角形
平面分治，可惜我不会写
人类智慧，玄学旋转，转两次更保险。
#include<bits/stdc++.h>
using namespace std;
#define FO(name) freopen(name".in","r",stdin);freopen(name".out","w",stdout)
//#define double long double
typedef long long ll;
const int maxN = 1e6+10;
struct Point{
double x,y;
Point(){};
Point(int x,int y):x((double)x),y((double)y){};
friend bool operator < (Point A,Point B){
if(A.x != B.x) return A.x < B.x;
return A.y < B.y;
}
}P[maxN];
inline double Q2p(double x){
return x*x;
}
inline double dis(Point A,Point B){
double t = sqrt(Q2p(A.x-B.x)+Q2p(A.y-B.y));
return t;
}
inline double C(int a,int b,int c){
return dis(P[a],P[b])+dis(P[a],P[c])+dis(P[b],P[c]);
}
double ans = 1e18;
int N;
void Turn(int D){
for(int i = 1;i<=N;i++){
double x = P[i].x;
double y = P[i].y;
P[i].x = cos(D)*x-sin(D)*y;
P[i].y = sin(D)*x+cos(D)*y;
}
sort(P+1,P+1+N);
for(int i = 1;i<N-1;i++){
for(int j = i+1;j<min(i+10,N);j++){
for(int k = j+1;k<=min(j + 10,N);k++){
ans = min(ans,C(i,j,k));
}
}
}
}
int main(){
srand(time(0));
scanf("%d",&N);
for(int i = 1;i<=N;i++){
scanf("%lf%lf",&P[i].x,&P[i].y);
}
Turn(rand()%360);
Turn(rand()%360);
printf("%.6lf\n",ans);
return 0;
}
T3 游戏
用 multiset 存剑的攻击力，可以得到杀每条龙用的剑的攻击力序列 \(S\)。
解一个同余方程组，类似
\(\left\{\begin{matrix} xS_1 \equiv A_1 \pmod {p_1} \\ xS_2 \equiv A_2 \pmod {p_2} \\ xS_3 \equiv A_3 \pmod {p_3} \\ \vdots \\ xS_n \equiv A_n \pmod {p_n} \end{matrix}\right.\)
如果 \(S_i=1\) 的话，可以直接用 exgcd 求解。
设前 \(i-1\) 组方程的解为 \(ans\) ，\(M=\operatorname{lcm}(p_1,p_2,\dots,p_n)\) 。
那么前 \(i\) 组同余方程的解需要满足 \(S_i(ans+Mx)\equiv A_i \pmod{p_i}\)
移项得 \(S_iMx\equiv A_i-S_i\cdot ans \pmod {p_i}\)
用 exgcd 求解 \((S_iM)x+(p_i)y=(A_i-S_i\cdot ans)\)
#include<bits/stdc++.h>
using namespace std;
#define int long long
const int maxN = 1e5 + 10;
#define FO(name) freopen(name".in","r",stdin);freopen(name".out","w",stdout)
inline int read() {
int x = 0, f = 1;
char c = getchar();
while (!isdigit(c)) {
if (c == '-')f = -1;
c = getchar();
}
while (isdigit(c)) {
x = (x << 3) + (x << 1) + (c ^ 48);
c = getchar();
}
return x * f;
}
multiset<int> Sw;
int N, M;
int A[maxN];
int P[maxN];
int S[maxN];
int T[maxN];
inline void input() {
Sw.clear();
scanf("%lld%lld", &N, &M);
for (int i = 1; i <= N; i++) A[i] = read();
for (int i = 1; i <= N; i++) P[i] = read();
for (int i = 1; i <= N; i++) T[i] = read();
for (int i = 1; i <= M; i++) {
int t = read();
Sw.insert(t);
}
for (int i = 1; i <= N; i++) {
auto p = Sw.upper_bound(A[i]);
p--;
if (p == Sw.begin()) {
p = Sw.lower_bound(0);
S[i] = *p;
Sw.erase(p);
} else {
S[i] = *p;
Sw.erase(p++);
}
Sw.insert(T[i]);
}
// for(int i = 1;i<=N;i++) printf("%d ",S[i]);
// printf("\n");
}
inline void exgcd(int a, int b, int &g, int &x, int &y) {
if (b == 0) g = a, x = 1, y = 0;
else {
exgcd(b, a % b, g, x, y);
int t = x;
x = y;
y = t - (a / b) * y;
}
}
int ma = -1;
int exCRT() {
int ans = 0, lcm = 1;
int x, y, g, ATK, B, C;
for (int i = 1; i <= N; i++) {
ATK = (__int128)S[i] * lcm % P[i];
B = P[i];
C = (A[i] - S[i] * ans % P[i] + P[i]) % P[i];
exgcd(ATK, B, g, x, y);
x = (x % B + B) % B;
if (C % g) return -1;
ans += (__int128)(C / g) * x % (B / g) * lcm % (lcm *= (B / g));
ans %= lcm;
}
if (ans < ma) ans += ((ma - ans - 1) / lcm + 1) / lcm;
return ans;
}
inline bool TP1() {
for (int i = 1; i <= N; i++) if (P[i] != 1) return false;
return true;
}
inline int quickPow(int a, int b, int p) {
int ans = 1;
while (b) {
if (b & 1) ans = ans * a % p;
a = a * a % p;
b >>= 1;
}
return ans;
}
inline int CRT() {
int M = 1;
for (int i = 1; i <= N; i++) {
M *= P[i];
}
int ans = 0;
for (int i = 1; i <= N; i++) {
int M1 = quickPow(M, P[i] - 2, P[i]);
ans = (ans + A[i] * M * M1) % P[i];
}
return 0;
// return ans * ;
}
inline void solve() {
input();
ma = -1;
for (int i = 1; i <= N; i++) {
ma = max(ma, (int)(ceil(1.0 * A[i] / S[i])));
}
if(TP1()){
printf("%lld\n",ma);
return;
}
printf("%lld\n", exCRT());
}
signed main() {
int T;
scanf("%lld", &T);
while (T--) {
solve();
}
return 0;
}
T4 找朋友
跟输入杠了一个多点最后发现把 char 当 int 搁那移位，乐死我了
一旦输入写对了基本就能对了，但是会 TLE
暴力思路就是对每个新生算一遍所有人与他的友好度最后输出。
但是由于 \(k \leq 15\) ，如果我们用 \(16\) 个 \(16\) 位整数来存储 DNA 的话，至少会有一段 DNA 是完全相同的。
用一个 vector 表示第 \(i\) 组 DNA 片段 为 \(j\) 的学生的编号。
对于每个转校生，先看 \(16\) 组片段分别与哪些学生相同，再去计算友好值
然后就是卡常乐
函数前 inline ，快读，循环展开，优先使用位运算，用 __builtin_popcount(a^b) 计算相似度，const 定义常量等
#include<bits/stdc++.h>
using namespace std;
typedef unsigned long long ull;
const int maxN = 400000;
bool s[maxN + 1][256];
#define FO(name) freopen(name".in","r",stdin);freopen(name".out","w",stdout)
int lastans = 0;
inline ull myRand(ull &k1, ull &k2) {
ull k3 = k1, k4 = k2;
k1 = k4;
k3 ^= (k3 << 23);
k2 = k3 ^ k4 ^ (k3 >> 17) ^ (k4 >> 26);
return k2 + k4;
}
inline void gen(int n, ull a1, ull a2) {
for (int i = 1; i <= n; i++)
for (int j = 0; j < 256; j++)
s[i][j] = (myRand(a1, a2) & (1ull << 32)) ? 1 : 0;
}
bool s2[maxN+2];
int N, Q;
ull a1, a2;
int k;
int M[maxN+2][20];
int T[maxN+2];
const int TSY = 256*256+10;
vector<int> V[17][TSY];
inline int toInt(char c) {
return isdigit(c) ? c - '0' : c - 'A' + 10;
}
inline int read() {
char c = getchar();
while (!isdigit(c)) {
if (c >= 'A' && c <= 'Z') return c - 'A' + 10;
c = getchar();
}
while (isdigit(c)) return c - '0';
}
inline void printInBin(int x, char endline = '\n') {
for (int i = 15; i >= 0; i--) {
putchar((x & (1 << i)) ? '1' : '0');
}
if (endline) putchar(endline);
}
int popcount(int x) {
int cnt = 0;
for (int i = 0; i < 16; i++) {
cnt = cnt + (((1 << i) & x) ? 1 : 0);
}
return cnt;
}
int a, b, c, d;
int main() {
// getM();
scanf("%d%d%llu%llu", &N, &Q, &a1, &a2);
gen(N, a1, a2);
// for(int i = 0;i<16;i++){
// for(int j = 0;j<(1<<16);j++){
// V[i][j].resize(32);
// }
// }
int p = 0;
for (register int i = 1; i <= N; i++) {
p = 0;
for (int j = 0; j < 16; j++) {
for (int k = 15; k >= 0; k--) {
M[i][j] |= s[i][p++] ? (1 << k) : 0;
}
V[j][M[i][j]].push_back(i);
}
}
while (Q--) {
for (int i = 0; i < 16; i += 8) {
T[i] = 0;
a = read();
b = read();
c = read();
d = read();
T[i] = (a << 12) + (b << 8) + (c << 4) + d;
if (lastans)T[i] = ~(T[i] | (~((1 << 16) - 1)));
T[i + 1] = 0;
a = read();
b = read();
c = read();
d = read();
T[i + 1] = (a << 12) + (b << 8) + (c << 4) + d;
if (lastans)T[i + 1] = ~(T[i + 1] | (~((1 << 16) - 1)));
T[i + 2] = 0;
a = read();
b = read();
c = read();
d = read();
T[i + 2] = (a << 12) + (b << 8) + (c << 4) + d;
if (lastans)T[i + 2] = ~(T[i + 2] | (~((1 << 16) - 1)));
T[i + 3] = 0;
a = read();
b = read();
c = read();
d = read();
T[i + 3] = (a << 12) + (b << 8) + (c << 4) + d;
if (lastans)T[i + 3] = ~(T[i + 3] | (~((1 << 16) - 1)));
T[i + 4] = 0;
a = read();
b = read();
c = read();
d = read();
T[i + 4] = (a << 12) + (b << 8) + (c << 4) + d;
if (lastans)T[i + 4] = ~(T[i + 4] | (~((1 << 16) - 1)));
T[i + 5] = 0;
a = read();
b = read();
c = read();
d = read();
T[i + 5] = (a << 12) + (b << 8) + (c << 4) + d;
if (lastans)T[i + 5] = ~(T[i + 5] | (~((1 << 16) - 1)));
T[i + 6] = 0;
a = read();
b = read();
c = read();
d = read();
T[i + 6] = (a << 12) + (b << 8) + (c << 4) + d;
if (lastans)T[i + 6] = ~(T[i + 6] | (~((1 << 16) - 1)));
T[i + 7] = 0;
a = read();
b = read();
c = read();
d = read();
T[i + 7] = (a << 12) + (b << 8) + (c << 4) + d;
if (lastans)T[i + 7] = ~(T[i + 7] | (~((1 << 16) - 1)));
}
int k;
scanf("%d", &k);
bool flg = false;
// for (int i = 0; i < 16; i++) {
// if (V[i][T[i]].size()) {
// AT = i;
// siz = V[i][T[i]].size();
// break;
// }
// }
// if(AT == -1) goto awa;
for (int i = 0; i < 15; i++) {
for (int t : V[i][T[i]]) {
int cnt = 0;
for (register int j = 0; j < 16; j++) {
cnt += __builtin_popcount(M[t][j] ^ T[j]);
if(cnt > k) break;
}
if (cnt <= k) {
printf("1\n");
flg = true;
lastans = 1;
goto qwq;
}
}
}qwq:{};
if (flg) continue;
lastans = 0;
printf("0\n");
}
return 0;
}
