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只会这些题了。。
A 恭喜小梁成为了宝可梦训练家~
#include <bits/stdc++.h>
#pragma GCC optimize(2)
#pragma GCC optimize(3)
typedef unsigned long long ull;
typedef long long ll;
ll a[1001],t,n;
using namespace std;
namespace IO{
char ibuf[1<<21],*ip=ibuf,*ip_=ibuf;
char obuf[1<<21],*op=obuf,*op_=obuf+(1<<21);
inline char gc(){
if(ip!=ip_)return *ip++;
ip=ibuf;ip_=ip+fread(ibuf,1,1<<21,stdin);
return ip==ip_?EOF:*ip++;
}
inline void pc(char c){
if(op==op_)fwrite(obuf,1,1<<21,stdout),op=obuf;
*op++=c;
}
inline ll read(){
register ll x=0,ch=gc(),w=1;
for(;ch<'0'||ch>'9';ch=gc())if(ch=='-')w=-1;
for(;ch>='0'&&ch<='9';ch=gc())x=x*10+ch-48;
return w*x;
}
template<class I>
inline void write(I x){
if(x<0)pc('-'),x=-x;
if(x>9)write(x/10);pc(x%10+'0');
}
class flusher_{
public:
~flusher_(){if(op!=obuf)fwrite(obuf,1,op-obuf,stdout);}
}IO_flusher;
}
using namespace IO;
int main()
{
t=read();
while(t--){
ll maxn=-0x3f3f3f;
ll minn=0x3f3f3f;
ll sum=0;
n=read();
for(int i=1;i<=n;i++)
{
a[i]=read();
maxn=max(maxn,a[i]);
minn=min(minn,a[i]);
sum+=a[i];
}
printf("MAX:%lld\n",maxn);
printf("MIN:%lld\n",minn);
printf("AVG:%.2lf\n",1.0*sum/n);
}
return 0;
}
B 皮(A)卡©皮(M)~
t=int(input())
while t>0:
t-=1
s=input().upper()
if 'A' not in s:
print("A")
elif 'C' not in s:
print("C")
elif 'M' not in s:
print("M")
else:
print("-1")
C 杰尼杰尼
水题没人做系列。
斜率相同就不算，hash一下就行了。
#include <bits/stdc++.h>
#pragma GCC optimize(2)
#pragma GCC optimize(3)
typedef unsigned long long ull;
typedef long long ll;
const ll maxn=1e3+10;
using namespace std;
namespace IO{
char ibuf[1<<21],*ip=ibuf,*ip_=ibuf;
char obuf[1<<21],*op=obuf,*op_=obuf+(1<<21);
inline char gc(){
if(ip!=ip_)return *ip++;
ip=ibuf;ip_=ip+fread(ibuf,1,1<<21,stdin);
return ip==ip_?EOF:*ip++;
}
inline void pc(char c){
if(op==op_)fwrite(obuf,1,1<<21,stdout),op=obuf;
*op++=c;
}
inline ll read(){
register ll x=0,ch=gc(),w=1;
for(;ch<'0'||ch>'9';ch=gc())if(ch=='-')w=-1;
for(;ch>='0'&&ch<='9';ch=gc())x=x*10+ch-48;
return w*x;
}
template<class I>
inline void write(I x){
if(x<0)pc('-'),x=-x;
if(x>9)write(x/10);pc(x%10+'0');
}
class flusher_{
public:
~flusher_(){if(op!=obuf)fwrite(obuf,1,op-obuf,stdout);}
}IO_flusher;
}
using namespace IO;
map< pair<double,double>, int> vis;
ll n,a[maxn],b[maxn],cnt;
int main()
{
n=read();
for(int i=1;i<=n;i++)
a[i]=read(),b[i]=read();
for(int i=1;i<=n;i++){
for(int j=i+1;j<=n;j++){
if(a[i]!=a[j]){
double x=(1.0*b[j]-1.0*b[i])/(1.0*a[i]-1.0*a[j]);
double y=x*a[i]+b[i];
if(!vis[{x,y}]){
cnt++;
vis[{x,y}]++;
}
}
}
}
if(cnt==0) printf("No Fire Point.");
else write(cnt);
}
G 遗迹逃亡
bfs板子题。
#include <bits/stdc++.h>
using namespace std;
int x2[4]={0,0,-1,1};
int y2[4]={1,-1,0,0};
struct sa
{
int x;
int y;
};
queue<sa>q;
char a[510][510];
int n,m,xt,yt,xg,yg;
int bfs(int x,int y)
{
q.push({x,y});
a[x][y]='#';
while(!q.empty()){
sa tmp;
tmp=q.front();
q.pop();
int x3=tmp.x;
int y3=tmp.y;
if(x3==xt&&y3==yt)
return 1;
for(int i=0;i<4;i++){
int x4=x3+x2[i];
int y4=y3+y2[i];
if(x4>=1&&x4<=n&&y4>=1&&y4<=m&&(a[x4][y4]=='.'||a[x4][y4]=='g')){
q.push({x4,y4});
a[x4][y4]='#';
}
}
}
return -1;
}
int main()
{
ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
cin>>n>>m;
for(int i=1;i<=n;i++){
for(int j=1;j<=m;j++){
cin>>a[i][j];
if(a[i][j]=='s') xg=i,yg=j;
if(a[i][j]=='g') xt=i,yt=j;
}
}
int cnt=bfs(xg,yg);
if(cnt==-1)
cout<<"No"<<endl;
else
cout<<"Yes"<<endl;
return 0;
}
L 小梁的道馆
赛后补题：并查集！
这里不搞快读什么的了，简单易懂 直接上代码。
会再写一篇简单说说菜鸡看代码想的并查集。
#include <bits/stdc++.h>
using namespace std;
int n, m, t, a, b, u, v;
int fa[1005];
int find(int x) {
if(fa[x] == x) return x;
return fa[x] = find(fa[x]);
}
int main() {
scanf("%d %d %d", &n, &m, &t);
for(int i = 1;i <= n; i++) fa[i] = i;
for(int i = 1;i <= m; i++) {
scanf("%d %d", &a, &b);
u = find(a); v = find(b);
fa[u] = v;
}
while(t--) {
scanf("%d %d", &a, &b);
u = find(a); v = find(b);
if(u == v) puts("YES");
else puts("NO");
}
}