[ABC266] AtCoder Beginner Contest 266
先贴代码，题解有空再补。
Tasks
| Task Name | Time Limit | Memory Limit | ||
|---|---|---|---|---|
| A | Middle Letter | 2 sec | 1024 MB | Submit |
| B | Modulo Number | 2 sec | 1024 MB | Submit |
| C | Convex Quadrilateral | 2 sec | 1024 MB | Submit |
| D | Snuke Panic (1D) | 2 sec | 1024 MB | Submit |
| E | Throwing the Die | 2 sec | 1024 MB | Submit |
| F | Well-defined Path Queries on a Namori | 3 sec | 1024 MB | Submit |
| G | Yet Another RGB Sequence | 2 sec | 1024 MB | Submit |
| Ex | Snuke Panic (2D) | 5 sec | 1024 MB | Submit |
Solutions
A
#include<set>
#include<map>
#include<queue>
#include<stack>
#include<cmath>
#include<ctime>
#include<bitset>
#include<vector>
#include<cstdio>
#include<complex>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
typedef long long LL;
typedef long double LD;
typedef unsigned long long ULL;
int main()
{
// freopen("1.in","r",stdin);
string str;
cin>>str;
printf("%c\n",str[str.length()/2]);
return 0;
}
B
#include<set>
#include<map>
#include<queue>
#include<stack>
#include<cmath>
#include<ctime>
#include<bitset>
#include<vector>
#include<cstdio>
#include<complex>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
typedef long long LL;
typedef long double LD;
typedef unsigned long long ULL;
int main()
{
// freopen("1.in","r",stdin);
LL x;
cin>>x;
const LL MOD=998244353;
printf("%lld\n",(x%MOD+MOD)%MOD);
return 0;
}
C
#include<set>
#include<map>
#include<queue>
#include<stack>
#include<cmath>
#include<ctime>
#include<bitset>
#include<vector>
#include<cstdio>
#include<complex>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
typedef long long LL;
typedef long double LD;
typedef pair<int,int> PII;
typedef unsigned long long ULL;
const int N=1000;
PII a[N];
int S(int x,int y,int z)
{
int x_1=a[y].first-a[x].first;
int y_1=a[y].second-a[x].second;
int x_2=a[z].first-a[x].first;
int y_2=a[z].second-a[x].second;
return abs(x_1*y_2-x_2*y_1);
}
int main()
{
for(int i=1;i<=4;i++) {
cin>>a[i].first>>a[i].second;
a[i+4]=a[i];
a[i+8]=a[i];
}
int ans=S(1,2,3)+S(1,3,4);
for(int i=2;i<=9;i++) {
int siz=S(i-1,i,i+1)+S(i-1,i+1,i+2);
if(siz!=ans) return puts("No")&0;
}
puts("Yes");
return 0;
}
D
#include<set>
#include<map>
#include<queue>
#include<stack>
#include<cmath>
#include<ctime>
#include<bitset>
#include<vector>
#include<cstdio>
#include<complex>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
typedef long long LL;
typedef long double LD;
typedef unsigned long long ULL;
const int N=1e5+5;
int n,m;
LL f[N][5],a[N][5];
int main()
{
// freopen("1.in","r",stdin);
scanf("%d",&n);
for(int i=1,t,x,z;i<=n;i++) {
scanf("%d%d%d",&t,&x,&z);
a[t][x]+=z;
m=max(m,t);
}
memset(f,-0x3f,sizeof f);
f[0][0]=0;
for(int i=1;i<=m;i++)
for(int j=0;j<=4;j++) {
f[i][j]=f[i-1][j]+a[i][j];
if(j-1>=0) f[i][j]=max(f[i][j],f[i-1][j-1]+a[i][j]);
if(j+1<=4) f[i][j]=max(f[i][j],f[i-1][j+1]+a[i][j]);
}
LL ans=0;
for(int i=0;i<=4;i++) ans=max(ans,f[m][i]);
printf("%lld\n",ans);
return 0;
}
E
一道期望题，重点在于读懂题目什么意思。
#include<set>
#include<map>
#include<queue>
#include<stack>
#include<cmath>
#include<ctime>
#include<bitset>
#include<vector>
#include<cstdio>
#include<complex>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
typedef long long LL;
typedef long double LD;
typedef unsigned long long ULL;
const int N=1000+5;
double f[N]; // f[i][j] 表示策略下在前 i 轮拿到 j 的概率
int n;
int main()
{
scanf("%d",&n);
f[1]=3.5;
for(int i=2;i<=n;i++) {
for(int j=1;j<=6;j++) {
if(j<f[i-1]) f[i]+=f[i-1];
else f[i]+=j;
}
f[i]/=6;
}
printf("%.10lf\n",f[n]);
return 0;
}
F
拓扑排序
#include<set>
#include<map>
#include<queue>
#include<stack>
#include<cmath>
#include<ctime>
#include<bitset>
#include<vector>
#include<cstdio>
#include<complex>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
typedef long long LL;
typedef long double LD;
typedef unsigned long long ULL;
const int N=4e5+5;
int one[N],idx;
int ver[N],Next[N];
void AddEdge(int a,int b)
{
Next[idx]=one[a]; ver[idx]=b; one[a]=idx++;
}
int n,m;
queue<int> q;
int deg[N];
bool vis[N];
void topsort()
{
while(q.size()) q.pop();
memset(vis,false,sizeof vis);
for(int i=1;i<=n;i++) {
if(deg[i]==1) {
q.push(i);
vis[i]=true;
}
}
while(q.size()) {
int x=q.front(); q.pop();
for(int i=one[x],y;~i;i=Next[i]) {
y=ver[i];
if(vis[y]) continue;
if((--deg[y])==1) q.push(y),vis[y]=true;
}
}
}
int col[N];
void color(int x,int fa,int c)
{
col[x]=c;
for(int i=one[x];~i;i=Next[i]) {
int y=ver[i];
if(y==fa || !vis[y]) continue;
color(y,x,c);
}
}
int main()
{
// freopen("1.in","r",stdin);
scanf("%d",&n);
memset(one,-1,sizeof one); idx=0;
for(int i=1,x,y;i<=n;i++) {
scanf("%d%d",&x,&y);
AddEdge(x,y); AddEdge(y,x);
deg[x]++; deg[y]++;
}
topsort();
for(int i=1;i<=n;i++) {
if(vis[i]) continue;
color(i,-1,i);
}
int x,y;
scanf("%d",&m);
while(m--) {
scanf("%d%d",&x,&y);
if(col[x]==col[y]) puts("Yes");
else puts("No");
}
return 0;
}
G
计数问题
#include<set>
#include<map>
#include<queue>
#include<stack>
#include<cmath>
#include<ctime>
#include<bitset>
#include<vector>
#include<cstdio>
#include<complex>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
typedef long long LL;
typedef long double LD;
typedef unsigned long long ULL;
const LL P=998244353;
const int N=3e6+5;
LL fact[N];
LL power(LL x,LL k)
{
LL res=1;
while(k) {
if(k&1) res=res*x%P;
x=x*x%P; k>>=1;
}
return res%P;
}
LL inv(LL x) { return power(x,P-2); }
LL C(LL n,LL m)
{
if(m>n || m<0) return 0;
else return fact[n]*inv(fact[m])%P*inv(fact[n-m])%P;
}
LL R,G,B,k;
LL n;
int main()
{
// freopen("1.in","r",stdin);
fact[0]=1;
for(int i=1;i<N;i++)
fact[i]=fact[i-1]*i%P;
scanf("%lld%lld%lld%lld",&R,&G,&B,&k);
n=R+G+B-k;
LL ans=C(n-(G-k),k)*C(n-(G-k)-k,R-k)%P; // 排 RG,R,B
ans=ans*(C(B+k+G-k,G-k)%P)%P;
printf("%lld\n",ans);
return 0;
}
