以下所有AC题解程序来自“仙客传奇”团队。
AC的C++语言程序：
#include<iostream>
#include<string>
using namespace std;
string ans[]={"Typically Otaku",
"Eye-opener",
"Young Traveller",
"Excellent Traveller",
"Contemporary Xu Xiake"};
int main()
{
int T;
cin>>T;
while(T--){
int a[4],s=0;
for(int i=0;i<4;i++){
cin>>a[i];
if(a[i]) s++;
}
cout<<ans[s]<<endl;
}
}
AC的C++语言程序：
#include <iostream>
using namespace std;
int main(void) {
int cnt, x;
int t;
scanf("%d", &t);
while (t--) {
cnt = 0;
for (int i = 0; i < 4; i++) {
scanf("%d", &x);
if (x) cnt++;
}
if (cnt == 0) printf("Typically Otaku\n");
else if (cnt == 1) printf("Eye-opener\n");
else if (cnt == 2) printf("Young Traveller\n");
else if (cnt == 3) printf("Excellent Traveller\n");
else if (cnt == 4) printf("Contemporary Xu Xiake\n");
}
return 0;
}
B. Ultraman vs. Aodzilla and Bodzilla
题解链接：
【数学思维】【分类】codeforces102028B Ultraman vs. Aodzilla and Bodzilla
A-L CodeForces Gym 102028 简要题解
BCH 2018-2019 ACM-ICPC 焦作赛区 部分题解
题解链接：
Gym 102028C - Supreme Command - [思维题][2018-2019 ACM-ICPC Asia Jiaozuo Regional Contest Problem C]
A-L CodeForces Gym 102028 简要题解
BCH 2018-2019 ACM-ICPC 焦作赛区 部分题解
D. Keiichi Tsuchiya the Drift King
AC的C++语言程序：
#include<bits/stdc++.h>
using namespace std;
const double eps=1e-1;
const double Pi=acos(-1.0);
int main()
{
double a,b,r,d;
int T;
cin>>T;
while(T--){
cin>>a>>b>>r>>d;
double ang=atan((a+r)/b);
d=(d*Pi)/180;
if(2*(d+ang)>=Pi) {
printf("%.12f\n",sqrt(b*b+(a+r)*(a+r))-r);
}else
{
ang=sin(d+ang);
printf("%.12f\n",sqrt(b*b+(a+r)*(a+r))*ang-r);
}
}
}
AC的C++语言程序：
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#include <iostream>
#include <cmath>
using namespace std;
const double PI = acos(-1.0);
const double EPS = 1e-14;
inline int fixCmp(double a, double b) {
if (fabs(a - b) <= EPS) return 0;
else if (a > b) return 1;
return -1;
}
int main(void) {
int t;
double a, b, r, d;
scanf("%d", &t);
while (t--) {
scanf("%lf%lf%lf%lf", &a, &b, &r, &d);
double theta = d * PI / 180;
double bound = -atan((a + r) / b);
while (fixCmp(bound, 0) < 0) bound += PI / 2;
if (fixCmp(theta, bound) >= 0) {
printf("%.12f\n", sqrt((a + r) * (a + r) + b * b) - r);
} else {
printf("%.12f\n", b * sin(theta) + (a + r) * cos(theta) - r);
}
}
return 0;
}
AC的Java语言程序：
import java.util.Scanner;
import java.math.BigInteger;
public class Main{
static int prime[];
static boolean table[];
static final int maxn=1000;
public static void main(String []args){
Scanner in=new Scanner(System.in);
int cnt=0;
prime=new int[maxn];
table=new boolean[maxn];
for(int i=2;i<maxn;i++){
if(!table[i]){
prime[cnt++]=i;
for(int j=2;i*j<maxn;j++)
table[i*j]=true;
}
}
BigInteger n,a,b,g;
int T;
T=in.nextInt();
for(int s=0;s<T;s++){
n=in.nextBigInteger();
a=BigInteger.valueOf(1);
b=BigInteger.valueOf(1);
int num=0;
while(true){
a=a.multiply(BigInteger.valueOf(prime[num++]));
if(a.compareTo(n)>0) break;
}
a=a.divide(BigInteger.valueOf(prime[--num]));
for(int i=0;i<num;i++){
b=b.multiply(BigInteger.valueOf(prime[i]*prime[i]-1));
b=b.divide(BigInteger.valueOf(prime[i]-1));
}
g=a.gcd(b);
System.out.println(a.divide(g)+"/"+b.divide(g));
}
}
}
AC的Java语言程序：
import java.util.Scanner;
import java.math.BigInteger;
public class Main {
private static int prime[];
private static boolean notPrime[];
private final static int LIM = 10000;
private static int cur = 0;
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
// build prime table
notPrime = new boolean[LIM];
prime = new int[LIM];
for (int i = 2; i < LIM; i++) {
if (!notPrime[i]) prime[cur++] = i;
for (int j = 0; j < cur; j++) {
int tmp = prime[j] * i;
if (tmp >= LIM) break;
notPrime[tmp] = true;
if (i % prime[j] == 0) break;
}
}
// cnt
int t;
BigInteger n;
t = in.nextInt();
for (int k = 0; k < t; k++) {
n = in.nextBigInteger();
BigInteger mx = BigInteger.valueOf(1), crt = BigInteger.valueOf(1);
for (int i = 0; i < cur; i++) {
BigInteger tmp = mx.multiply(BigInteger.valueOf(prime[i]));
if (tmp.compareTo(n) > 0) break;
crt = crt.add(crt.multiply(BigInteger.valueOf(prime[i])));
mx = tmp;
}
BigInteger g = mx.gcd(crt);
System.out.println(mx.divide(g) + "/" + crt.divide(g));
}
}
}
AC的C++语言程序：
#include<bits/stdc++.h>
using namespace std;
const int maxn=10000+5;
int dx[]={-1,-1,1,1,-2,2};
int dy[]={-3,3,-3,3,0,0};
char gird[maxn][maxn];
int r,c;
bool vis[maxn][maxn];
struct Node
{
int x,y,step;
};
Node str,ter;
int bfs()
{
vis[str.x][str.y]=1;
Node first=str,u,v;
first.step=1;
queue<Node> q;
q.push(first);
while(!q.empty()){
u=q.front();q.pop();
if(u.x==ter.x&&u.y==ter.y) return u.step;
int dr,dc;
for(int i=0;i<6;i++){
dr=u.x+dx[i],dc=u.y+dy[i];
if(gird[dr][dc]=='/'||gird[dr][dc]=='\\'||gird[dr][dc]=='-'||vis[dr+dx[i]][dc+dy[i]]) continue;
v.x=dr+dx[i],v.y=dc+dy[i];
v.step=u.step+1;
q.push(v);
vis[v.x][v.y]=1;
}
}
return -1;
}
int main()
{
int T;
scanf("%d",&T);
while(T--){
scanf("%d%d",&r,&c);
r=4*r+3,c=6*c+3;
getchar();
for(int i=0;i<r;i++){
gets(gird[i]);
for(int j=0;j<c;j++){
if(gird[i][j]=='S') str.x=i,str.y=j;
if(gird[i][j]=='T') ter.x=i,ter.y=j;
vis[i][j]=false;
}
}
printf("%d\n",bfs());
}
}
AC的C++语言程序：
#include <iostream>
#include <queue>
using namespace std;
const int LIM = 1e3;
char dat[5 * LIM][7 * LIM];
int vis[5 * LIM][7 * LIM];
int rbound, down;
int t, r, c, stamp;
int sr, sc, tr, tc;
const int dr[] = {-2, 2, -1, -1, 1, 1};
const int dc[] = {0, 0, 3, -3, 3, -3};
struct Node {
int cnt, r, c;
Node(int nr = 0, int nc = 0, int ncnt = 0) {
cnt = ncnt, r = nr, c = nc;
}
};
int bfs() {
Node crt;
crt.cnt = 1;
crt.r = sr;
crt.c = sc;
queue<Node> q;
q.push(crt);
while (!q.empty()) {
crt = q.front();
q.pop();
if (dat[crt.r][crt.c] == 'T') return crt.cnt;
for (int i = 0; i < 6; i++) {
int nr = crt.r + dr[i];
int nc = crt.c + dc[i];
if (nr >= 0 && nr < down && nc >= 0 && nc < rbound
&& dat[nr][nc] == ' ' && vis[nr][nc] != stamp) {
vis[nr][nc] = stamp;
nr += dr[i], nc += dc[i];
if (nr >= 0 && nr < down && nc >= 0 && nc < rbound
&& vis[nr][nc] != stamp) {
vis[nr][nc] = stamp;
q.push(Node(nr, nc, crt.cnt + 1));
}
}
}
}
return -1;
}
inline void gl(char *cur) {
char ch;
while ((ch = getchar()) != '\n') *(cur++) = ch;
}
int main(void) {
scanf("%d", &t);
for (stamp = 1; stamp <= t; stamp++) {
scanf("%d%d", &r, &c);
gl(dat[0]);
for (int i = 0; i < r * 4 + 3; i++)
gl(dat[i]);
down = r * 4 + 3;
rbound = 6 * c + 3;
for (int i = 0; i < down; i++) {
for (int j = 0; j < rbound; j++) {
if (dat[i][j] == 'S')
sr = i, sc = j;
else if (dat[i][j] == 'T')
tr = i, tc = j;
}
}
printf("%d\n", bfs());
// destroy S and T
dat[sr][sc] = ' ';
dat[tr][tc] = ' ';
}
return 0;
}
G. Shortest Paths on Random Forests
题解链接：
A-L CodeForces Gym 102028 简要题解
H. Can You Solve the Harder Problem?
题解链接：
A-L CodeForces Gym 102028 简要题解
BCH 2018-2019 ACM-ICPC 焦作赛区 部分题解
AC的C++语言程序：
#include<iostream>
using namespace std;
typedef long long ll;
const int maxn = 1e5 + 10;
ll p[maxn];
int main()
{
ios::sync_with_stdio(false);
cin.tie(NULL);cout.tie(NULL);
int t,n,d;
ll ans,tmp;
cin>>t;
while(t--)
{
cin>>n;
for(int i = 2;i <= n;++i)
{
cin>>d;
p[i] = p[i - 1] + d;
}
int l = 1,r = n + 1;
ans = 0,tmp = 0;
cout<<"0";
for(int i = 1;i < n;++i)
{
if(i & 1)
{
tmp += p[--r] - p[l];
ans += tmp;
}
else
{
++l;
ans += tmp;
}
cout<<' '<<ans;
}
cout<<endl;
}
return 0;
}
AC的C++语言程序：
#include <iostream>
using namespace std;
typedef long long LL;
const LL LIM = 1e6;
LL dat[LIM];
int main(void) {
LL t, n, tmp, ans;
scanf("%lld", &t);
while (t--) {
scanf("%lld", &n);
LL cur = 0;
dat[cur++] = 0;
for (LL i = 1; i < n; i++) {
scanf("%lld", &tmp);
dat[cur] = dat[cur - 1] + tmp;
cur++;
}
LL left = 0, right = cur;
printf("0");
ans = tmp = 0;
for (LL i = 1; i < n; i++) {
if (i & 1) {
ans += tmp;
ans += dat[--right] - dat[left];
tmp += dat[right] - dat[left];
} else {
ans += tmp;
left++;
}
printf(" %lld", ans);
}
putchar('\n');
}
return 0;
}
AC的C++语言程序：
#include<bits/stdc++.h>
using namespace std;
const int maxn=1e5+5;
typedef long long ll;
ll a[maxn];
int main()
{
int T;
scanf("%d",&T);
while(T--){
int n;
ll sum=0;
scanf("%d",&n);
for(int i=1;i<n;i++){
scanf("%lld",&a[i]);
sum+=a[i];
}
ll tt=0,tmp=0;
int l=1,r=n-1;
for(int i=1;i<=n;i++){
if(i&1){
tmp+=tt;
}else{
tt+=sum;
sum-=a[l];
l++;
sum-=a[r];
r--;
tmp+=tt;
}
if(i==1){
printf("%lld",tmp);
}else{
printf(" %lld",tmp);
}
}
printf("\n");
}
return 0;
}
题解链接：
Gym102028J
A-L CodeForces Gym 102028 简要题解
K. Counting Failures on a Trie
题解链接：
A-L CodeForces Gym 102028 简要题解
题解链接：
A-L CodeForces Gym 102028 简要题解
题解连接：
ADEFI CodeForces GYM 102028 2018-2019 ACM-ICPC, Asia Jiaozuo Regional Contest
A-L CodeForces Gym 102028 简要题解
BCH 2018-2019 ACM-ICPC 焦作赛区 部分题解