Wannafly挑战赛17

A.走格子

按题意模拟即可

#include <bits/stdc++.h>
#define pb(x) push_back(x)
#define mem(W) memset(W,0,sizeof(W))
#define rep(i,a,b) for(int i=a;i<=b;++i)
inline int read() {
    char c=getchar();int x=0,f=1;
    while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}
    while(c>='0'&&c<='9'){x=x*10+c-'0';c=getchar();}
    return x*f;
}
typedef long long ll;
const int N = 1000 + 100;
using namespace std;
int n,a[N][N],m;
int dx[]={1,0,-1,0};
int dy[]={0,-1,0,1};
int ck(int x,int y){
    if(x<=0||x>n||y<=0||y>n)return 0;
    if(a[x][y])return 0;
    return 1;
}
int main() {
    scanf("%d%d",&n,&m);
    int ax=1,ay=1,f=0,stp=0,flag=0;
    a[ax][ay]=1;
    while(1) {
        while(ck(ax+dx[f%4],ay+dy[f%4])){
            ax+=dx[f%4],ay+=dy[f%4],++stp,a[ax][ay]=1;
            if(stp==m) {flag=1;break;}
        }
        if(flag)break;
        ++f;
    }
    printf("%d %d\n",ax,ay);
    return 0;
}

B.求值2

组合数第n层的平方和=C(2*n,n)，枚举求和即可

#include <bits/stdc++.h>
#define pb(x) push_back(x)
#define mem(W) memset(W,0,sizeof(W))
#define rep(i,a,b) for(int i=a;i<=b;++i)
inline int read() {
    char c=getchar();int x=0,f=1;
    while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}
    while(c>='0'&&c<='9'){x=x*10+c-'0';c=getchar();}
    return x*f;
}
typedef long long ll;
const int N = 2000000 + 100;
const ll mod = 998244353;
using namespace std;
int n;
ll ans=0;
ll fc[N],inv[N];
ll q_pow(ll a,ll b) {
    ll ans=1;
    while(b){
        if(b&1) ans=(ans*a)%mod;
        a=(a*a)%mod;
        b>>=1LL;
    }
    return ans;
}
ll C(ll n, ll m) {
    return (fc[n]*inv[m])%mod*inv[n-m]%mod;
}
int main() {
    n=read();
    fc[0]=1; rep(i,1,2*n)fc[i]=(fc[i-1]*i)%mod;
    inv[2*n]=q_pow(fc[2*n],mod-2);
    for(int i=2*n-1;i>=0;--i)inv[i]=(inv[i+1]*(i+1LL))%mod;
    ans=0;
    for(int i=1;i<=n;++i){
        ans=(ans+C(i<<1,i))%mod;
    }
    cout << ans << endl;
    return 0;
}

C.简单环

 dp[s][i]表示，以s最低位为起点，i为终点的简单路数目，转移dp[s|(1<<j)][j]+=dp[s][i]当i,j有边且j点未经过，转移的时候如果有以j点为起点，且i和j有边相连，就加到答案里。点数小于等于2不要计入答案。原题：codeforces 11D

#include <bits/stdc++.h>
#define pb(x) push_back(x)
#define mem(W) memset(W,0,sizeof(W))
#define rep(i,a,b) for(int i=a;i<=b;++i)
inline int read() {
    char c=getchar();int x=0,f=1;
    while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}
    while(c>='0'&&c<='9'){x=x*10+c-'0';c=getchar();}
    return x*f;
}
typedef long long ll;
const int N = 21;
const ll mod = 998244353;
using namespace std;
ll q_pow(ll a,ll b) {
    ll ans=1;
    while(b){
        if(b&1) ans=(ans*a)%mod;
        a=(a*a)%mod;
        b>>=1LL;
    }
    return ans;
}
int n,m,k,g[N][N],t,u,v;
ll dp[(1<<N)][N],ans[N],inv2;
int ct(int s){
    int ans=0;
    for(int i=s;i;i-=(i&(-i)))++ans;
    return ans;
}
int main() {
    inv2=q_pow(2LL,mod-2);
    n=read(),m=read(),k=read();
    rep(i,1,m)u=read(),v=read(),g[u-1][v-1]=g[v-1][u-1]=1;
    rep(i,0,n-1)dp[1<<i][i]=1;
    rep(s,1,(1<<n)){
        rep(i,0,n-1){
            if(!(s&(1<<i))||!dp[s][i])continue;
            for(t=0;!(s&(1<<t));++t);
            rep(j,t,n-1)if(g[i][j]){
                if(!(s&(1<<j)))dp[s|(1<<j)][j]+=dp[s][i];
                if((s&(1<<j))&&j==t&&ct(s)>2)ans[ct(s)%k]=(ans[ct(s)%k]+dp[s][i])%mod;
            }
        }
    }
    rep(i,0,k-1)printf("%lld\n",(ans[i]*inv2)%mod);
    return 0;
}