T1 打气球
题解:
期望递推,不关心具体位置,所以\(F[i][j]\)表示还有\(i\),\(j\)行,列未完;
\(code\):
#include<cstdio>
#include<algorithm>
#include<ctype.h>
#include<vector>
#include<queue>
#include<cstring>
#define lowbit(x) (x&-x)
#define ll long long
#define ld double
#include<map>
#include<stdlib.h>
#include<ctime>
#define mod 19260817
using namespace std;
char buf[1<<20],*p1,*p2;
inline char gc()
{
// return getchar();
return p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<20,stdin))==p1?0:*p1++;
}
template<typename T>
inline void read(T &x)
{
char tt;
bool flag=0;
while(!isdigit(tt=gc())&&tt!='-');
tt=='-'?(flag=1,x=0):(x=tt-'0');
while(isdigit(tt=gc())) x=x*10+tt-'0';
if(flag) x=-x;
}
int n,m;
int hang,lie;
ld f[2005][2005];
bool heng[2005],zong[2005];
ld dfs(int x,int y)
{
if(f[x][y]!=-1) return f[x][y];
if(!x&&!y) return f[x][y]=0.0;f[x][y]=0.0;
f[x][y]+=(x)?(dfs(x-1,y)*1.0*x*(n-y)):0.0;
// printf("%.2lf\n",f[x][y]);
f[x][y]+=(y)?(dfs(x,y-1)*1.0*y*(n-x)):0.0;
f[x][y]+=(x&&y)?(dfs(x-1,y-1)*1.0*x*y):0.0;
return f[x][y]=(f[x][y]+n*n*1.0)/((x+y)*n-x*y);
}
int main()
{
read(n),read(m);
for(int i=1;i<=m;i++)
{
int x,y;
read(x),read(y);
heng[x]=zong[y]=1;
}
for(int i=1;i<=n;i++) hang+=!heng[i],lie+=!zong[i];
for(int i=0;i<=n;i++)
for(int j=0;j<=n;j++)
f[i][j]=-1;
printf("%.2lf",dfs(hang,lie));
return 0;
}
T2 有趣的数列
题解:
卡特兰数+质因数分解
\(code:\)
#include<cstdio>
#include<algorithm>
#include<ctype.h>
#include<vector>
#include<queue>
#include<cstring>
#define lowbit(x) (x&-x)
#define ll long long
#define ld double
#include<map>
#include<stdlib.h>
#include<ctime>
using namespace std;
char buf[1<<20],*p1,*p2;
inline char gc()
{
// return getchar();
return p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<20,stdin))==p1?0:*p1++;
}
template<typename T>
inline void read(T &x)
{
char tt;
bool flag=0;
while(!isdigit(tt=gc())&&tt!='-');
tt=='-'?(flag=1,x=0):(x=tt-'0');
while(isdigit(tt=gc())) x=x*10+tt-'0';
if(flag) x=-x;
}
const int maxn=1000002;
ll pri[maxn<<1],phi[maxn<<1],s[maxn<<1],ans=1;
ll n,mod,tot;
bool book[maxn<<1];
void modify(int x,int d){
while(1){
if(pri[phi[x]]==0) break;
s[phi[x]]+=d;
x/=pri[phi[x]];
}
}
void sent(){
for(int i=2;i<=n<<1;i++){
if(!book[i]) pri[++tot]=i,phi[i]=tot;
for (int j=1;pri[j]*i<=(n<<1)&&j<=tot;j++){
book[pri[j]*i]=1,phi[pri[j]*i]=j;
if(i%pri[j]==0) break;
}
}
}
int main(){
// 1 1 5 14 42
// freopen("interesting.in","r",stdin);
// freopen("interesting.out","w",stdout);
read(n),read(mod);sent();
for (int i=n<<1;i>n;i--) modify(i,1);
for (int i=1;i<=n;i++) modify(i,-1);
modify(n+1,-1);
for(int i=1;i<=tot;i++)
while(s[i]--) ans=(ans*pri[i])%mod;
printf("%lld",ans);
return 0;
}
T3 距离
题解:
待补
\(code:\)
待补
原文地址:https://www.cnblogs.com/KatouKatou/p/9813947.html
时间: 2024-11-10 00:24:55