2020牛客多校第六场
B.Binary Vector
看样例，求逆元异或前缀和
#include
using namespace std;
typedef unsigned long long ll;
const int mod=1e9+7;
const int maxn=2e7+10;
ll ksm(ll x)
{
ll ans=1,n=mod-2;
while(n)
{
if(n%2==1)
ans*=x,ans%=mod;
x*=x;
x%=mod;
n/=2;
}
return ans;
}
ll a[maxn],b[maxn],c[maxn];
int main()
{
ll x=1,y=2;
a[0]=1;
ll p=ksm(2);
b[0]=1;
y=p;
for(int i=1;i<=maxn-10;i++)
{
a[i]=x*a[i-1];
x=x*2+1;
a[i]%=mod;
x%=mod;
b[i]=b[i-1]*y;
y*=p;
y%=mod;
b[i]%=mod;
//y*=p;
c[i]=b[i]*a[i];
//y%mod;
c[i]%=mod;
c[i]^=c[i-1];
}
int t,n;
cin>>t;
while(t--)
{
scanf("%d",&n);
//x=ksm(b[n])*ksm(a[n]);
//x%=mod;
printf("%lld\n",c[n]);
}
return 0;
} C.Combination of Physics and Maths
队友过的签到题
代码：
#include
using namespace std;
typedef unsigned long long ll;
#define sc(a) scanf("%d",&a)
#define pii pair
#define pb push_back
#define pf printf
const int mod=1e9+9;
const int N=2e2+10;
int n,m;
int a[N][N];
int s[N];
int main()
{
int t;sc(t);
while(t--)
{
cin>>n>>m;
for(int i=1;i<=n;i++)
for(int j=1;j<=m;j++)
sc(a[i][j]);
double ans=0;
for(int i=1;i<=m;i++)
{
int s=0;
for(int j=1;j<=n;j++)
{
s+=a[j][i];
ans=max(ans,1.0*s/a[j][i]);
}
}pf("%.8f\n",ans);
}
return 0;
}
E.Easy ConstructionEasy Construction
#include
using namespace std;
typedef unsigned long long ll;
const int mod=1e9+9;
const int maxn=1e5+10;
int f[maxn];
int main()
{
int n,k,s=0,x;
cin>>n>>k;
if(n%2==0)
{
if(k!=n/2)
{
printf("-1");
return 0;
}
printf("%d %d",n,n/2);
for(int i=3;i<=n;i+=2)
{
printf(" %d %d",i/2,n-i/2);
}
}
else
{
if(k!=0)
{
printf("-1");
return 0;
}
printf("%d",n);
for(int i=1;i<=n/2;i++)
{
printf(" %d %d",i,n-i);
}
}
return 0;
} K.K-Bag
构造模拟
代码：
#include
using namespace std;
typedef unsigned long long ll;
const int mod=1e9+7;
const int maxn=5e5+10;
int a[maxn],f[maxn],b[maxn],lt[maxn];
struct cc{
int x,id;
}d[maxn];
bool cmp(cc x,cc y)
{
return x.x>t;
while(t--)
{
scanf("%d %d",&n,&k);
int pp=0;
for(int i=1;i<=n;i++)
{
scanf("%d",&a[i]);
b[i]=0;
lt[i]=0;
if(a[i]>k)
pp=1;
}
if(pp)
{
printf("NO\n");
continue;
}
if(k>n) //
{
for(int i=1;i<=n;i++)
d[i].id=i,d[i].x=a[i];
sort(d+1,d+1+n,cmp);
int flag=1,cnt=1;
a[d[1].id]=cnt;
for(int i=2;i<=n;i++)
{
if(d[i].x!=d[i-1].x)
cnt++;
a[d[i].id]=cnt;
}
for(int p=1;p<=cnt;p++)
f[p]=0;
int i;
for(i=1;i<=n;i++)
{
f[a[i]]++;
if(f[a[i]]==2)
break;
}
for(int p=1;p<=cnt;p++)
f[p]=0;
for(;i<=n;i++)
{
f[a[i]]++;
if(f[a[i]]==2)
break;
}
if(i>n)
printf("YES\n");
else
printf("NO\n");
continue;
}
for(int i=1;i<=n;i++)
b[min(a[i],k+1)]++;
m=1;
int mmin=n;
for(int i=1;i<=k;i++)
if(b[i]n)
break;
//减去a[j]
f[a[j]]--;
if(f[a[j]]==0)
s--;
//加上a[j+k]
if(f[a[j+k]]==0)
s++;
f[a[j+k]]++;
if(s==k)
{
lt[j+1]++;
//cout<0)
flag=0;
f[a[j]]++;
}
for(int p=1;p<=k;p++)
f[p]=0;
for(j=1;j0)
flag=0;
f[a[j]]++;
}
s+=flag;
}
}
//
for(int p=1;p<=k;p++)
f[p]=0;
int i;
for(i=1;i<=n;i++)
{
f[a[i]]++;
if(f[a[i]]==2)
break;
}
for(int p=1;p<=k;p++)
f[p]=0;
for(;i<=n;i++)
{
f[a[i]]++;
if(f[a[i]]==2)
break;
}
if(i>n)
s++;
//
if(s>0)
printf("YES\n");
else
printf("NO\n");
}
return 0;
}
/*
17 3
1 2 3 4 3 7 6 5 4 2 1 3 4 5 3 2 1
*/