只A两题,技不如人~
题解待补完
#include
#include
#include
#include
#include
#include
#include
const int INF=1<<30;
using namespace std;
int main(){
int t;
scanf("%d",&t);
while(t--){
int n,k;
scanf("%d %d",&n,&k);
n=n-1;
n%2?puts("LYF"):puts("BH");
}
return 0;
}
#include
#include
#include
#include
#include
#include
#include
const int INF=1<<30;
using namespace std;
const int N=1010;
int a[N];
int main(){
int t;
scanf("%d",&t);
while(t--){
int n;
scanf("%d",&n);
for(int i=1;i<=n;i++) scanf("%d",&a[i]);
sort(a+1,a+n+1);
printf("%d\n",a[n]-a[1]);
}
return 0;
}
#include
#include
#include
#include
#include
#include
#include
const int INF=1<<30;
using namespace std;
bool leap(int i){
if(i%4==0&&i%100) return 1;
else if(i%4==0&&i%100==0&&i%400==0) return 1;
else return 0;
}
int main(){
int mon[2018][13]={0,31,28,31,30,31,30,31,31,30,31,30,31};
for(int i=1;i<=2017;i++){
mon[i][1]=31;
mon[i][2]=28;
mon[i][3]=31;
mon[i][4]=30;
mon[i][5]=31;
mon[i][6]=30;
mon[i][7]=31;
mon[i][8]=31;
mon[i][9]=30;
mon[i][10]=31;
mon[i][11]=30;
mon[i][12]=31;
if(leap(i)) mon[i][2]=29;
}
mon[2017][10]=28;
int t;
scanf("%d",&t);
while(t--){
int y,m,d;
scanf("%d %d %d",&y,&m,&d);
int ans=0;
int y1=2017-y,m1=10-m,d1=28-d;
if(y1==0){
if(m1==0){
ans=d1;
}
else {
ans=mon[y][m]-d;
for(int i=m+1;i<=10;i++)
ans+=mon[y][i];
}
}
else if(y1){
ans=mon[y][m]-d;
for(int i=m+1;i<=12;i++){
ans+=mon[y][i];
}
for(int i=1;i<=10;i++){
ans+=mon[2017][i];
}
for(int i=y+1;i<2017;i++){
for(int j=1;j<=12;j++){
ans+=mon[i][j];
}
}
}
printf("%d\n",ans);
}
return 0;
}
#include
#include
#include
#include
#include
#include
#include
#define S(l,r,d,a,b,s) l*l*(a-sin(a))+r*r*(b-sin(b))-s
const double eps=1e-8;
const double PI=3.14159265357;
using namespace std;
double sqr(double x){
return x*x;
}
int main(){
int t;
scanf("%d",&t);
while(t--){
double x0,y0,x1,y1,r;
scanf("%lf %lf %lf %lf %lf",&x0,&y0,&x1,&y1,&r);
double a=0.0,b=0.0;
double d=sqrt(sqr(x0-x1)+sqr(y0-y1));
double s=sqr(r)*PI;
double lf=d,rt=sqrt(sqr(r)+sqr(d)),mid=0.0;
if(d+r/sqrt(2)-rprintf("%.4f\n",r/sqrt(2));
continue;
}
/*
2
0 0 10 10 2
*/
while(rt-lf>eps){
mid=(rt+lf)/2.0;
double a=2*acos((sqr(mid)+sqr(d)-sqr(r))/(2.0*mid*d));
double b=2*acos((sqr(d)+sqr(r)-sqr(mid))/(2.0*d*r));
if(S(mid,r,d,a,b,s)>eps){
rt=mid-eps;
}
else{
lf=mid+eps;
}
}
printf("%.4f\n",mid);
}
}
#include
#include
#include
#include
#include
#include
#include
const int INF=1<<30;
using namespace std;
const int N=1010;
int a[N];
int vis[N];
int vis2[N];
int main(){
int t;
scanf("%d",&t);
while(t--){
vector<int> v;
memset(vis,0,sizeof(vis));
memset(vis2,0,sizeof(vis2));
int n;
scanf("%d",&n);
for(int i=1;i<=n;i++){
scanf("%d",&a[i]);
vis[a[i]]+=1;
vis2[a[i]]+=1;
}
sort(vis+1,vis+1001);
int cnt1=0;
int cnt=vis[1000];
for(int i=1;i<=1000;i++){
if(vis2[i]==cnt){
v.push_back(i);
cnt1++;
}
}
for(int i=0;iif(i==cnt1-1) printf("%d\n",v[i]);
else printf("%d ",v[i]);
}
}
return 0;
}
#include
#include
#include
#include
#include
#include
typedef long long ll;
using namespace std;
const int N=100;
ll f[N];
void fi(){
f[0]=1;f[1]=1;
for(int i=2;i<=90;i++){
f[i]=f[i-1]+f[i-2];
}
}
int f1(ll n,ll m){
if(n==1) return 1;
if(n==2) return m==1?1:0;
if(m>f[n-1]) return f1(n-2,m-f[n-1]);
else return f1(n-1,m);
}
int main(){
int t;
scanf("%d",&t);
fi();
while(t--){
ll n,m;
scanf("%lld %lld",&n,&m);
f1(n,m)?puts("a"):puts("b");
}
}
#include
#include
#include
#include
#include
#include
#include
const int INF=1<<30;
typedef long long ll;
using namespace std;
char s[100010];
int main(){
int t;
scanf("%d",&t);
while(t--){
int n,k;
scanf("%d %d",&n,&k);
scanf("%s",s);
int ans=n;
int cnta=0,cntsum=0;
for(int i=0;s[i];i++){
if(s[i]=='a'){
cnta++;
cntsum++;
}
else if(s[i]=='b'){
cntsum++;
if(cnta){
ans++;
cnta=0;
cntsum=1;
}
}
else if(s[i]=='c'){
if(cntsum>=k){
ans++;
cntsum=1;
}
else cntsum++;
cnta=0;
}
}
printf("%d\n",ans);
}
return 0;
}