戰績:
A-恭喜小梁成爲了寶可夢訓練家~
簽到題
#pragma GCC optimize(2)
#include<bits/stdc++.h>
#define ll long long
#define maxn 1005
#define inf 1e9
#define pb push_back
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define per(i,a,b) for(int i=a;i>=b;i--)
using namespace std;
inline ll read()
{
ll x=0,w=1; char c=getchar();
while(c<'0'||c>'9') {if(c=='-') w=-1; c=getchar();}
while(c<='9'&&c>='0') {x=(x<<1)+(x<<3)+c-'0'; c=getchar();}
return w==1?x:-x;
}
const int N=1e2+10;
int a[N],n;
int main()
{
int _=read();while(_--)
{
n=read();
int mx,mi;
double sum=0;
rep(i,1,n)
{
a[i]=read();
}
mx=mi=a[1];
rep(i,1,n){
mx=max(mx,a[i]);
mi=min(mi,a[i]);
sum+=a[i];
}
printf("MAX:%d\n",mx);
printf("MIN:%d\n",mi);
printf("AVG:%.2f\n",sum/n);
}
}
B-皮(A)卡(C)皮(M)~
簽到題
#pragma GCC optimize(2)
#include<bits/stdc++.h>
#define ll long long
#define maxn 1005
#define inf 1e9
#define pb push_back
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define per(i,a,b) for(int i=a;i>=b;i--)
using namespace std;
inline ll read()
{
ll x=0,w=1; char c=getchar();
while(c<'0'||c>'9') {if(c=='-') w=-1; c=getchar();}
while(c<='9'&&c>='0') {x=(x<<1)+(x<<3)+c-'0'; c=getchar();}
return w==1?x:-x;
}
const int N=1e2+10;
int a[N],n;
int main()
{
int _=read();while(_--)
{
string s;
cin>>s;
int vis[26];
memset(vis,0,sizeof(vis));
for(int i=0;i<s.size();++i){
vis[s[i]-'A']++;
}
if(vis['A'-'A']==0) puts("A");
else if(vis['C'-'A']==0) puts("C");
else if(vis['M'-'A']==0) puts("M");
else puts("-1");
}
}
C-傑尼傑尼
做法:n比較小 n^2枚舉 計算交點,map去重即可。
#include <bits/stdc++.h>
using namespace std;
#define ll long long
ll input(){
ll x=0,f=0;char ch=getchar();
while(ch<'0'||ch>'9') f|=ch=='-',ch=getchar();
while(ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar();
return f? -x:x;
}
#define PII pair <double,double>
#define fr first
#define sc second
#define mp make_pair
map <PII,int> p;
double k[107],b[107];
int main(){
int n=input();
for(int i=1;i<=n;i++) k[i]=input(),b[i]=input();
for(int i=1;i<=n;i++){
for(int j=i+1;j<=n;j++){
if(k[i]==k[j]) continue;
double x=(b[j]-b[i])/(k[i]-k[j]);
double y=k[i]*x+b[i];
//printf("%f %f\n",x,y);
p[mp(x,y)]=1;
}
}
if(p.size()==0) printf("No Fire Point.\n");
else printf("%d\n",p.size());
}
/*
2
1 2
1 1
*/
D-古代遺蹟:字符王國
做法:統計S串前綴各個字符的個數即可。然後枚舉t串長度的區間,判斷區間內 各個字符 不同的數量數是否在1以內,且只有兩個字符數量不同,或者字符數量都相同也是對答案產生了貢獻
#pragma GCC optimize(2)
#include<bits/stdc++.h>
#define ll long long
#define maxn 1005
#define inf 1e9
#define pb push_back
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define per(i,a,b) for(int i=a;i>=b;i--)
using namespace std;
inline ll read()
{
ll x=0,w=1; char c=getchar();
while(c<'0'||c>'9') {if(c=='-') w=-1; c=getchar();}
while(c<='9'&&c>='0') {x=(x<<1)+(x<<3)+c-'0'; c=getchar();}
return w==1?x:-x;
}
const int N=2e5+10;
char s[N],t[N];
int vs[30],vis[N][26],tmp[26];
int main()
{
int _=read();while(_--)
{
scanf("%s%s",s+1,t+1);
int ls=strlen(s+1);
int lt=strlen(t+1);
for(int i=0;i<=ls;++i){
for(int j=0;j<26;++j) vis[i][j]=vs[j]=0;
}
for(int i=1;i<=lt;++i) vs[t[i]-'a']++;
for(int i=1;i<=ls;++i){
for(int j=0;j<26;++j) vis[i][j]=vis[i-1][j];
vis[i][s[i]-'a']++;
}
int ans=0;
for(int i=1;i+lt-1<=ls;++i){
int l=i,r=i+lt-1;
for(int j=0;j<26;++j) tmp[j]=vis[r][j]-vis[l-1][j];
int s1=0,s2=0,flag=1;
for(int j=0;j<26&&flag;++j){
if(tmp[j]>vs[j]){
if(tmp[j]-vs[j]>1) flag=0;
else s1++;
}
else if(tmp[j]<vs[j]){
if(vs[j]-tmp[j]>1) flag=0;
else s2++;
}
}
//printf("i:%d s1:%d s2:%d\n",i,s1,s2);
if(flag&&(s1==1&&s2==1||s1==0&&s2==0))ans++;
}
printf("%d\n",ans);
}
}
E-皮卡丘這麼可愛,當然要.....
做法:二進制優化多重揹包,詳細博客:博客
F-Doctor Rabbit 的統計
做法:簡單模擬沒人寫系列,按照題意模擬即可。
#pragma GCC optimize(2)
#include<bits/stdc++.h>
#define ll long long
#define maxn 1005
#define inf 1e9
#define pb push_back
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define per(i,a,b) for(int i=a;i>=b;i--)
using namespace std;
inline ll read()
{
ll x=0,w=1; char c=getchar();
while(c<'0'||c>'9') {if(c=='-') w=-1; c=getchar();}
while(c<='9'&&c>='0') {x=(x<<1)+(x<<3)+c-'0'; c=getchar();}
return w==1?x:-x;
}
const int N=1e2+10,M=2000;
char s[N];
struct node{
int sex,t,ty;
}a[N];
int na[M],nv[M],n;
void print(int x)
{
int h=x/60;
int m=x%60;
printf("%02d:%02d\n",h,m);
}
int main()
{
n=read();
int num=0;
int start=8*60+10,en=19*60;
rep(i,1,n)
{
int h,m;
scanf("%s %d %d:%d %d",s,&a[i].sex,&h,&m,&a[i].ty);
if(a[i].ty>37) continue;
a[i].t=h*60+m+2;
if(a[i].t>en) continue;
if(a[i].sex==0) nv[a[i].t]++;
else na[a[i].t]++;
num++;
}
int en1=21*60;
int mx=0,mi=0;
for(int i=start;i<=en1;i+=10){
//printf("i: ");
//print(i);
int flag=0;
for(int j=i;j>=i-9;--j){
if(na[j]){
mi=i+3;
flag=1;
break;
}
}
if(flag) break;
}
for(int i=start;i<=en1;i+=10){
for(int j=i;j>=i-9;--j){
if(nv[j]){
mx=max(mx,i+5);
}
}
}
if(!num) puts("-1");
else{
printf("%d\n",num);
if(mi) print(mi); else puts("-1");
if(mx) print(mx); else puts("-1");
}
}
G-遺蹟逃亡
做法:bfs經典入門水題
#pragma GCC optimize(2)
#include<bits/stdc++.h>
#define maxn 1005
#define inf 1e9
#define pb push_back
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define per(i,a,b) for(int i=a;i>=b;i--)
using namespace std;
typedef long long ll;
inline ll read()
{
ll x=0,w=1; char c=getchar();
while(c<'0'||c>'9') {if(c=='-') w=-1; c=getchar();}
while(c<='9'&&c>='0') {x=(x<<1)+(x<<3)+c-'0'; c=getchar();}
return w==1?x:-x;
}
const int N=5e2+10;
int n,m,si,sj,ei,ej;
char s[N][N];
int dir[4][2]={1,0,0,1,-1,0,0,-1},vis[N][N];
void bfs()
{
queue<pair<int,int> >que;
que.push({si,sj});
vis[si][sj]=1;
while(que.size()){
auto now=que.front();que.pop();
if(now.first==ei&&now.second==ej){
puts("Yes");return ;
}
for(int i=0;i<4;++i){
int x=now.first+dir[i][0];
int y=now.second+dir[i][1];
if(x<=0||y<=0||x>n||y>m||s[x][y]=='#') continue;
if(vis[x][y]) continue;
vis[x][y]=1;
que.push({x,y});
}
}
puts("No");
}
int main()
{
n=read(),m=read();
rep(i,1,n) {
scanf("%s",s[i]+1);
for(int j=1;j<=m;++j){
if(s[i][j]=='s') si=i,sj=j;
if(s[i][j]=='g') ei=i,ej=j;
}
}
bfs();
}
H-阿羅拉聯盟賽
做法:又一個題意看起來複雜,做起來很水,但是沒人寫的一個題。
按照題意模擬即可。
#pragma GCC optimize(2)
#include<bits/stdc++.h>
#define ll long long
#define maxn 1005
#define inf 1e9
#define pb push_back
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define per(i,a,b) for(int i=a;i>=b;i--)
using namespace std;
inline ll read()
{
ll x=0,w=1; char c=getchar();
while(c<'0'||c>'9') {if(c=='-') w=-1; c=getchar();}
while(c<='9'&&c>='0') {x=(x<<1)+(x<<3)+c-'0'; c=getchar();}
return w==1?x:-x;
}
const int N=1e5+10;
char X;
int n,A[N],B[N];
struct node
{
int l,r,x,id;
}a[N],b[N];
bool cmp1(node a,node b)
{
if(a.x!=b.x) return a.x>b.x;
return a.id<b.id;
}
bool cmp2(node a,node b)
{
if(a.x!=b.x) return a.x<b.x;
return a.id<b.id;
}
int main()
{
cin>>n>>X;
rep(i,1,n) a[i].l=read();
rep(i,1,n) a[i].r=read();
rep(i,1,n) b[i].l=read();
rep(i,1,n) b[i].r=read();
rep(i,1,n) A[i]=read();
rep(i,1,n) B[i]=read();
rep(i,1,n) a[i].x=a[i].l+a[i].r,a[i].id=i;
rep(i,1,n) b[i].x=b[i].l+b[i].r,b[i].id=i;
sort(a+1,a+1+n,cmp1);
sort(b+1,b+1+n,cmp2);
int id1=1,id2=1;
int sum1=0,sum2=0,ans1=0,ans2=0;
if(X=='A') {
while(id1<=n&&id2<=n){
//printf("id1:%d id2:%d\n",id1,id2);
int mi=a[A[id1]].l;
sum1+=mi;
b[B[id2]].r-=mi;
if(b[B[id2]].r<=0) id2++,ans2++;
if(id2>n) break;
mi=b[B[id2]].l;
sum2+=mi;
a[A[id1]].r-=mi;
if(a[A[id1]].r<=0) id1++,ans1++;
}
}
else{
while(id1<=n&&id2<=n){
int mi=b[B[id2]].l;
sum2+=mi;
a[A[id1]].r-=mi;
if(a[A[id1]].r<=0) id1++,ans1++;
if(id1>n) break;
mi=a[A[id1]].l;
sum1+=mi;
b[B[id2]].r-=mi;
if(b[B[id2]].r<=0) id2++,ans2++;
}
}
printf("%d %d %d %d\n",sum1,sum2,ans1,ans2);
}
I-雪拉比的求救
做法:由於兩個目標是相向而行的,那麼對s、t分別跑dij 並保存方案數
方案數怎麼求?兩種情況:
1、在點上 (s到點i方案數)*(t到點i方案數)* s到點i方案數)*(t到點i方案數) 爲什麼乘了兩邊?因爲當從t到達i 從i 到達s 時又是一次方案數
2、在邊上 判斷能否在這邊上相遇即可。
#include<bits/stdc++.h>
using namespace std;
const int N=1e5+10;
typedef long long ll;
const ll inf=0x3f3f3f3f,mod=1e9+7;
vector<pair<int,int> >G[N];
int n,m,s,t;
int vis[N];
ll dis[2][N],dp[2][N];
void dij(int s,ll d[],ll dp[])
{
for(int i=1;i<=n;++i) d[i]=1e18,vis[i]=0;//初始化
priority_queue<pair<ll,int> >que;
que.push({0,s});
d[s]=0;
dp[s]=1;
while(que.size()){
auto now=que.top();que.pop();
int u=now.second;
if(vis[u]) continue;
vis[u]=1;
for(auto it:G[u]){
int v=it.first;
if(d[v]>d[u]+it.second){
d[v]=d[u]+it.second;
que.push({-d[v],v});
dp[v]=dp[u];
//printf("v:%d u:%d dp[u]:%lld\n",v,u,dp[u]);
}
else if(d[v]==d[u]+it.second) dp[v]=(dp[v]+dp[u])%mod;
}
}
}
int main()
{
scanf("%d%d",&n,&m);
scanf("%d%d",&s,&t);
for(int i=1;i<=m;++i){
int u,v,w;
scanf("%d%d%d",&u,&v,&w);
G[u].push_back({v,w});
G[v].push_back({u,w});
}
dij(s,dis[0],dp[0]);
dij(t,dis[1],dp[1]);
ll L=dis[0][t];
ll ans=((dp[0][t]%mod)*(dp[1][s]%mod))%mod;
for(int i=1;i<=n;++i){
if(dis[1][i]==dis[0][i]&&dis[1][i]*2==L){
ans=(ans-dp[0][i]*dp[0][i]%mod*dp[1][i]%mod*dp[1][i]%mod+mod)%mod;
}
for(auto now:G[i]){
int v=now.first,w=now.second;
if(dis[0][i]+w+dis[1][v]==L&&dis[0][i]*2<L&&dis[1][v]*2<L){
ans=(ans-dp[0][i]*dp[0][i]%mod*dp[1][v]%mod*dp[1][v]%mod+mod)%mod;
}
}
}
printf("%lld\n",ans);
}
J-小梁的揹包
做法:經典01揹包,但是他這數據 的理論時間複雜度 不支持他標程能過,可能是t個數據的n之和小於1e4
#pragma GCC optimize(2)
#include<bits/stdc++.h>
#define ll long long
#define maxn 1005
#define inf 1e9
#define pb push_back
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define per(i,a,b) for(int i=a;i>=b;i--)
using namespace std;
inline ll read()
{
ll x=0,w=1; char c=getchar();
while(c<'0'||c>'9') {if(c=='-') w=-1; c=getchar();}
while(c<='9'&&c>='0') {x=(x<<1)+(x<<3)+c-'0'; c=getchar();}
return w==1?x:-x;
}
const int N=1e4+10;
int n,s;
ll w[N],v[N];
pair<ll,ll>ans[N];
int main()
{
int _=read();while(_--)
{
n=read(),s=read();
rep(i,1,n) w[i]=read(),v[i]=read();
rep(i,0,s) ans[i].first=0,ans[i].second=0;
rep(i,1,n){
for(int j=s;j>=w[i];--j){
if(ans[j-w[i]].first+v[i]>ans[j].first){
ans[j].first=ans[j-w[i]].first+v[i];
ans[j].second=ans[j-w[i]].second+1;
}
}
}
printf("%lld %lld\n",ans[s].first,ans[s].second);
}
}
K-訓練師的變強祕訣:時間管理
做法:有點需要思維思考,主要是沒想到 排序後 還可以動態的更新後一個的遊戲結束時間
//#pragma GCC optimize(2)
#include <bits/stdc++.h>
using namespace std;
const int maxn=1e5+5;
typedef long long ll;
int n;
struct node{
ll s,e,ti,en;
}a[maxn];
bool cmp(node x,node y){
if(x.en!=y.en) return x.en<y.en;
return x.s<y.s;
}
int main(){
scanf("%d",&n);
int f=0;
for(int i=1;i<=n;i++){
scanf("%d%d",&a[i].s,&a[i].e);
a[i].ti=(a[i].e-a[i].s+2)/2;//至少玩的時長
a[i].en=a[i].s+a[i].ti;//結束時間
}
sort(a+1,a+1+n,cmp);
for(int i=1;i<n;i++){
if(a[i].en+a[i+1].ti>a[i+1].e){//當前結束時間+下一個需要玩的時長大於結束時間,無解
f=1;
break;
}
else if(a[i].en>a[i+1].s) a[i+1].en=a[i].en+a[i+1].ti;//更新下一個結束的時間
}
if(f==0) printf("YES\n");
else printf("NO\n");
return 0;
}
L-小梁的道館
做法:連通塊、並查集水題。
#pragma GCC optimize(2)
#include<bits/stdc++.h>
#define ll long long
#define maxn 1005
#define inf 1e9
#define pb push_back
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define per(i,a,b) for(int i=a;i>=b;i--)
using namespace std;
inline ll read()
{
ll x=0,w=1; char c=getchar();
while(c<'0'||c>'9') {if(c=='-') w=-1; c=getchar();}
while(c<='9'&&c>='0') {x=(x<<1)+(x<<3)+c-'0'; c=getchar();}
return w==1?x:-x;
}
const int N=1e3+10;
int n,m,t,sz[N],cnt;
vector<int>G[N];
void bfs(int u,int cnt)
{
queue<int>que;
que.push(u);
while(que.size()){
int now=que.front();que.pop();
sz[now]=cnt;
for(int v:G[now]){
if(!sz[v]){
que.push(v);
}
}
}
}
int main()
{
n=read(),m=read(),t=read();
for(int i=1;i<=m;++i){
int u=read(),v=read();
G[u].push_back(v);
G[v].push_back(u);
}
for(int i=1;i<=n;++i){
if(!sz[i]) bfs(i,++cnt);
}
while(t--)
{
int u=read(),v=read();
if(sz[u]==sz[v])puts("YES");
else puts("NO");
}
}