【2016多校训练4】Multi-University Training Contest 4

//http://acm.hdu.edu.cn/showproblem.php?pid=5763
#include<iostream>
#include<string>
using namespace std;
#define M 1000000007
#define N 100005
int t,dp[N];
string A,B;
int main(){
    cin>>t;
    for(int c=1;c<=t;c++){
        cin>>A>>B;
        for(int i=0;i<A.size();i++){
            dp[i]=i?dp[i-1]:1;
            if(i+1>=B.size()&&A.substr(i-B.size()+1,B.size())==B){
                dp[i]=(dp[i]+dp[i-B.size()])%M;
                if(i+1==B.size())dp[i]=2;
            }
        }
        cout<<"Case #"<<c<<": "<<dp[A.size()-1]<<endl;
    }
    return 0;
}

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
typedef long long ll;
ll p[21],a[21];
ll p1[21],a1[21],num;
ll n,x,y;
ll ans;
ll exgcd(ll l,ll r,ll &x,ll &y)
{
    if(r==0){x=1;y=0;return l;}
    else
    {
        ll d=exgcd(r,l%r,y,x);
        y-=l/r*x;
        return d;
    }
}
ll qmul(ll a,ll b,ll m){//快速乘法
    ll ans=0;
    ll k=a;
    while(b){
        if(b&1)
            ans=(ans+k)%m;
        k=(k+k)%m;
        b>>=1;
    }
    return ans;
}
ll Remain(ll p[],ll a[],ll m){
    ll ans=0;
    for(int i=1;i<=num;i++)
    {
        ll x,y,Mk=m/p[i];
        exgcd(Mk,p[i],x,y);
        x=(x%p[i]+p[i])%p[i];
        ans=(ans+qmul(qmul(Mk,x,m),a[i],m))%m;//ans+=Mk*x*a[i]
    }
    return (ans+m)%m;
}
ll work()
{
    ll m=1;
    for(int i=1;i<=num;i++)
        m*=(ll)p1[i];
    ll sum=Remain(p1,a1,m);
    ll k1=(x-sum)/m;
    if(k1*m+sum<x)k1++;
    if(k1<0)k1=0;
    if(y<sum)return 0;
    ll k2=(y-sum)/m;
    return k2-k1+1;
}
void solve()
{
    ans=0;
    for(ll i=0;i<=(1<<n)-1;i++)
    {
        num=0;
        memset(p1,0,sizeof(p1));
        memset(a1,0,sizeof(a1));
        for(int j=0;j<n;j++)
            if((i&(1<<j)))
            {
                p1[++num]=p[j+1];
                a1[num]=a[j+1];
            }
        p1[++num]=7;a1[num]=0;
        if(num%2==1)ans+=work();
        else ans-=work();
    }
    return;
}
int main()
{
    int t;
    cin>>t;
    for(int l=1;l<=t;l++)
    {
        cin>>n>>x>>y;
        for(int i=1;i<=n;i++)cin>>p[i]>>a[i];
        solve();
        cout<<"Case #"<<l<<": "<<ans<<endl;
    }
    return 0;
}

#include<iostream>
#include<cstdio>
#include<cstring>
#define N 100005
using namespace std;
int t,n,p[N],sum[N],ans[N];
inline int lowbit(int x){return x&(-x);}
int getsum(int x){
    int s=0;
    for(;x;x-=lowbit(x))s+=sum[x];
    return s;
}
void update(int v,int x){
    for(;x<=n;x+=lowbit(x))sum[x]+=v;
}
int main(){
    cin>>t;
    for(int c=1;c<=t;c++){
        memset(sum,0,sizeof sum);
        cout<<"Case #"<<c<<":";
        cin>>n;
        for(int i=1;i<=n;i++)
        {   
            cin>>p[i];
            update(1,p[i]);
            int r=getsum(p[i]);
            ans[p[i]]=max(i-r,p[i]-r);
        }
        for(int i=1;i<=n;i++)
            cout<<" "<<ans[i];
        cout<<endl;
    }
    return 0;
}