欧拉函数模板
//直接求解欧拉函数
int euler(int n){ //返回euler(n)
int res=n,a=n;
for(int i=2;i*i<=a;i++){
if(a%i==0){
res=res/i*(i-1);//先进行除法是为了防止中间数据的溢出
while(a%i==0) a/=i;
}
}
if(a>1) res=res/a*(a-1);
return res;
}
//筛选法打欧拉函数表
#define Max 1000001
int euler[Max];
void Init(){
euler[1]=1;
for(int i=2;i<Max;i++)
euler[i]=i;
for(int i=2;i<Max;i++)
if(euler[i]==i)
for(int j=i;j<Max;j+=i)
euler[j]=euler[j]/i*(i-1);//先进行除法是为了防止中间数据的溢出
}
扩展欧几里得
long long ex_gcd(long long a, long long b, long long &x, long long &y){
if(b == 0){
x = 1, y= 0;
return a;
}else{
long long r = ex_gcd(b, a% b, y, x);
y -= x*(a/b);
return r;
}
}
dinic网络流
src：源点
sink：汇点
#include<queue>
#include<iostream>
#include<string.h>
#include<stdio.h>
using namespace std;
const int inf = 1000000000;
const int maxn = 20000, maxm = 500000;
struct Edge{
int v, f, nxt;
};
int src, sink;
int g[maxn + 10];
int nume;
Edge e[maxm*2+10];
void addedge(int u, int v, int c){
e[++nume].v = v;
e[nume].f = c;
e[nume].nxt = g[u];
g[u] = nume;
e[++nume].v = u;
e[nume].f = 0;
e[nume].nxt = g[v];
g[v] = nume;
}
void init(){
memset(g, 0, sizeof(g));
nume = 1;
}
queue<int> que;
bool vis[maxn +10];
int dist[maxn + 10];
void bfs(){
memset(dist, 0, sizeof(dist));
while(!que.empty()) que.pop();
vis[src] = 1;
que.push(src);
while(!que.empty()){
int u = que.front();
que.pop();
for(int i = g[u]; i; i = e[i].nxt)
if(e[i].f && !vis[e[i].v]){
que.push(e[i].v);
dist[e[i].v] = dist[u] + 1;
vis[e[i].v] = true;
}
}
}
int dfs(int u, int delta){
if(u == sink){
return delta;
}else{
int ret = 0;
for(int i = g[u]; delta && i; i = e[i].nxt){
if(e[i].f && dist[e[i].v] == dist[u] +1){
int dd = dfs(e[i].v, min(e[i].f, delta));
e[i].f -= dd;
e[i^1].f += dd;
delta -= dd;
ret += dd;
}
}
return ret;
}
}
int maxflow(){
int ret = 0;
while(true){
memset(vis, 0, sizeof(vis));
bfs();
if(!vis[sink])return ret;
ret += dfs(src, inf);
}
return ret;
}
int main(){
int n, m;
while(scanf("%d%d", &n, &m)!=EOF){
init();
src = 1;
sink = m;
for(int i = 0; i < n; i++){
int x, y, z;
scanf("%d%d%d", &x, &y, &z);
addedge(x, y, z);
}
printf("%d\n", maxflow());
}
}
ac自动机
#include<iostream>
#include<string.h>
#include<stdio.h>
#include<algorithm>
#include<queue>
using namespace std;
const int maxa = 500000;
const int cha = 26;
int n, m, k;
struct Tire{
int next[maxa][cha], fail[maxa], end[maxa];
int root, L;
int newnode(){
for(int i = 0; i < cha; i++){
next[L][i] = -1;
}
end[L++] = 0;
return L-1;
}
void init(){
L = 0;
root = newnode();
}
void insert(char buf[]){
int len = strlen(buf);
int now = root;
for(int i = 0; i < len; i++){
if(next[now][buf[i] - ‘a‘] == -1)
next[now][buf[i]-‘a‘] = newnode();
now = next[now][buf[i]-‘a‘];
//printf("%d ", now);
}//puts("");
end[now] ++;
}
void build(){
queue<int>Q;
fail[root] = root;
for(int i = 0; i < cha; i++){
if(next[root][i] == -1)
next[root][i] = root;
else{
fail[next[root][i]] = root;
Q.push(next[root][i]);
}
}
while(!Q.empty()){
int now = Q.front();
Q.pop();
// end[now] |= end[fail[now]];
for(int i = 0; i < cha; i++){
if(next[now][i] == -1)
next[now][i] = next[fail[now]][i];
else{
fail[next[now][i]] = next[fail[now]][i];
Q.push(next[now][i]);
// printf("**%d %d\n",next[now][i],next[fail[now]][i]);
}
}
}
}
int solve(char *s){
int ans = 0, k = 0;
for(int i = 0; s[i]; i++){
int t = s[i] - ‘a‘;
k = next[k][t];
int j = k;
while(j){
ans += end[j];
//if(end[j]) printf("%d ",j);
end[j] = 0;
j = fail[j];
}//puts("");
}
return ans;
}
};
char buf[1000005];
Tire ac;
int main(){
int t, n;
scanf("%d", &t);
while(t--){
scanf("%d", &n);
ac.init();
//memset(ac.end, 0, sizeof(ac.end));
for(int i = 0; i < n; i++){
scanf("%s", buf);
ac.insert(buf);
}
ac.build();
scanf("%s", buf);
printf("%d\n", ac.solve(buf));
}
}
/*
abcdefg
bcdefg
cdef
de
e
ssaabcdefg
*/
凸包
#include<iostream>
#include<stdio.h>
#include<string.h>
#include<cmath>
#include<algorithm>
using namespace std;
const int maxa = 1005;
struct edge{
int x, y;
}e[maxa], q[maxa];
int cmp(edge a, edge b){
if(a.x != b.x)
return a.x < b.x;
return a.y < b.y;
}
int det(int x1, int y1, int x2, int y2){
return x1*y2 - x2*y1;
}
int cross(edge a, edge b, edge c, edge d){
return det(b.x - a.x, b.y -a.y, d.x - c.x, d.y - c.y);
}
int make_tubao(int n){
sort(e, e+n, cmp);
int m = 0;
for(int i = 0; i < n; i++){
while(m >= 2 && cross(q[m-2], q[m-1], q[m-1], e[i])>= 0){
m--;
}
q[m++] = e[i];
}
int k = m;
for(int i = n-2; i >= 0; i--){
while(m > k && cross(q[m-2], q[m-1], q[m-1], e[i])>= 0){
m--;
}
q[m++] = e[i];
}
return m;
}
double dis(edge a, edge b){
return sqrt((b.x - a.x)*(b.x - a.x) + (b.y-a.y)*(b.y-a.y));
}
int print(int n, int m){
n = make_tubao(n);
double ans = 0;
for(int i = 0;i < n-1; i++){
// printf("%d %d\n", q[i].x, q[i].y);
ans += dis(q[i], q[i+1]);
}
printf("%.0f\n", ans + 3.1415926*2*m);
}
int main(){
int n, m;
while(scanf("%d%d", &n, &m)!=EOF){
for(int i = 0 ; i < n; i++){
scanf("%d%d", &e[i].x, &e[i].y);
}
print(n, m);
}
}
rmq
#include<iostream>
#include<string.h>
#include<stdio.h>
using namespace std;
const int maxa = 50005;
int rmp_max[maxa][100];
int rmp_min[maxa][100];
int log(int n){
int cnt = 0;
while(n){
cnt ++;
n /= 2;
}
return cnt - 1;
}
int main(){
int n, m;
while(scanf("%d%d", &n ,&m)!=EOF){
for(int i = 0;i < n; i++){
scanf("%d", &rmp_max[i][0]);
rmp_min[i][0] = rmp_max[i][0];
}
int l = log(n);
for(int i = 1; i <= l; i++){
for(int j = 0; j+(1<<(i-1)) < n; j++){
rmp_max[j][i] = max(rmp_max[j][i-1], rmp_max[j+(1<<(i-1))][i-1]);
rmp_min[j][i] = min(rmp_min[j][i-1], rmp_min[j+(1<<(i-1))][i-1]);
}
}
while(m--){
int a, b;
scanf("%d%d", &a, &b);
a--, b--;
int j = log(b-a+1);
int maxn = max(rmp_max[a][j], rmp_max[b-(1<<j)+1][j]);
int minn = min(rmp_min[a][j], rmp_min[b-(1<<j)+1][j]);
printf("%d\n", maxn-minn);
}
}
}
时间： 2024-10-26 12:45:20