有一个测试2（day3下）

t1 girl
这个题树一定是树的结点数
环是二，然后乘起来，边数大于点数就成零

#include<cstring>#include<cstdio>int n,m;int f[110000],cnt,t[110000],vis[110000];long long ans=1;int find (int x){    return f[x]=f[x]==x?x:find(f[x]);}int main(){    freopen("girl.in","r",stdin);    freopen("girl.out","w",stdout);    scanf("%d%d",&n,&m);    for(int i=1;i<=n;i++) f[i]=i,t[i]=1;    for(int i=1,x,y;i<=m;i++)    {        scanf("%d%d",&x,&y);        int fx=find(x);        int fy=find(y);        if(fx==fy&&!vis[fx]) t[fx]=2;        else if(fx==fy&&vis[fx]) t[fx]=0;        else if(fx!=fy){            f[fy]=fx;            t[fx]+=t[fy];            t[fy]=0;        }    }    for(int i=1;i<=n;i++)    if(t[i]){        ans=(1ll*ans*t[i])%1000000007;    }    printf("%d",ans);}

t2
自己写了80
用乘法原理做的，也没去掉一些比n大的数

好像正解是数位dp

#include<cstdio>#include<cmath> #include<iostream>#define ll long longll n,k;ll fastpow(ll a,ll w){    ll ans=1;    while(w){        if(w&1)        ans=ans*a;        a*=a;        w>>=1;    }    return ans;}ll ans;int main(){    freopen("endless.in","r",stdin);    freopen("endless.out","w",stdout);    scanf("%lld%lld",&n,&k);        ll t=log(n)/log(k)+0.00001;        //if(n==fastpow(k,t)) t--;        ll w=fastpow(k,(t+1)/2);        ll ww=fastpow(k,t);        ll i;        for( i=1;i<=k;i++)         if(i*ww>n) break;        ans=w;        if((t&1)==0)         ans*=i;        printf("%lld",ans);}

100
每一位都这么搞

#include<iostream>#include<cstdio>#include<cstring>#include<cmath>#include<queue>#include<algorithm>using namespace std;#define ll long longll n,k;ll ans;int sum;int a[100];ll fastpow(ll a,ll w){    ll ans=1;    while(w){        if(w&1)        ans=ans*a;        a*=a;        w>>=1;    }    return ans;}int main(){    scanf("%lld%lld",&n,&k);    int tot=0;    while(n)    a[++tot]=n%k,n/=k;    if(tot&1){        for(int i=tot;i>=1;i--)        if(a[i]){            if(i&1==0){                ans+=fastpow(k,i/2);                break;            }            ans+=1ll*a[i]*fastpow(k,i/2);            if(i==1) ans++;        }        cout<<ans;    }else        cout<<fastpow(k,tot/2);    return 0;}

t3 不会
求了个和交了上去，还有2个点。。。