今日SGU 5.18

SGU 125

题意：给你一个数组b[i][j],表示i,j的四周有多少个数字大于它的，问你能不能构造出一个a矩形

收获：dfs + 剪枝

一行一行的dfs，然后第一行去枚举0-9，下一行判断当前选择能否满足上一行对应列的情况，可以的话就继续dfs

#include<bits/stdc++.h>

#define de(x) cout<<#x<<"="<<x<<endl;

#define dd(x) cout<<#x<<"="<<x<<" ";

#define rep(i,a,b) for(int i=a;i<(b);++i)

#define repd(i,a,b) for(int i=a;i>=(b);--i)

#define repp(i,a,b,t) for(int i=a;i<(b);i+=t)

#define ll long long

#define mt(a,b) memset(a,b,sizeof(a))

#define fi first

#define se second

#define inf 0x3f3f3f3f

#define INF 0x3f3f3f3f3f3f3f3f

#define pii pair<int,int>

#define pdd pair<double,double>

#define pdi pair<double,int>

#define mp(u,v) make_pair(u,v)

#define sz(a) (int)a.size()

#define ull unsigned long long

#define ll long long

#define pb push_back

#define PI acos(-1.0)

#define qc std::ios::sync_with_stdio(false)

#define db double

#define all(a) a.begin(),a.end()

const int mod = 1e9+;

const int maxn = 1e5+;

const double eps = 1e-;

using namespace std;

bool eq(const db &a, const db &b) { return fabs(a - b) < eps; }

bool ls(const db &a, const db &b) { return a + eps < b; }

bool le(const db &a, const db &b) { return eq(a, b) || ls(a, b); }

ll gcd(ll a,ll b) { return a==?b:gcd(b%a,a); };

ll lcm(ll a,ll b) { return a/gcd(a,b)*b; }

ll kpow(ll a,ll b) {ll res=;a%=mod; if(b<) return ; for(;b;b>>=){if(b&)res=res*a%mod;a=a*a%mod;}return res;}

ll read(){

    ll x=,f=;char ch=getchar();

    while (ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}

    while (ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}

    return x*f;

}

//inv[1]=1;

//for(int i=2;i<=n;i++) inv[i]=(mod-mod/i)*inv[mod%i]%mod;

int a[][],n,b[][];

int dx[] = {,,,-};

int dy[] = {,-,,};

bool ok(int x,int y){

    int ret = ;

//    dd(x)dd(y)de(ret)

    rep(i,,) if(a[x+dx[i]][y+dy[i]] > a[x][y]) ret++;

    return ret==b[x][y];

}

bool dfs(int x,int y){

//    dd(n)dd(x)de(y)

    int tx = x,ty = y + ;

    if(x > n){

        rep(i,,n+) if(!ok(n,i)) return false;

        rep(i,,n+) rep(j,,n+) printf("%d%c",a[i][j]," \n"[j==n]);

        return true;

     }

    if(ty > n) ty = ,tx = x + ;

    rep(i,,){

        a[x][y] = i;

        if(x != ) if(!ok(x-,y)) continue;

        if(dfs(tx,ty)) return true;

    }

    return false;

}

int main(){

    scanf("%d",&n);

    rep(i,,n+) rep(j,,n+) scanf("%d",&b[i][j]);

    if(!dfs(,)) puts("NO SOLUTION");

    return ;

}

SGU 358

题意：求中位数的中位数

收获：无

#include<bits/stdc++.h>

#define de(x) cout<<#x<<"="<<x<<endl;

#define dd(x) cout<<#x<<"="<<x<<" ";

#define rep(i,a,b) for(int i=a;i<(b);++i)

#define repd(i,a,b) for(int i=a;i>=(b);--i)

#define repp(i,a,b,t) for(int i=a;i<(b);i+=t)

#define ll long long

#define mt(a,b) memset(a,b,sizeof(a))

#define fi first

#define se second

#define inf 0x3f3f3f3f

#define INF 0x3f3f3f3f3f3f3f3f

#define pii pair<int,int>

#define pdd pair<double,double>

#define pdi pair<double,int>

#define mp(u,v) make_pair(u,v)

#define sz(a) (int)a.size()

#define ull unsigned long long

#define ll long long

#define pb push_back

#define PI acos(-1.0)

#define qc std::ios::sync_with_stdio(false)

#define db double

#define all(a) a.begin(),a.end()

const int mod = 1e9+;

const int maxn = 1e5+;

const double eps = 1e-;

using namespace std;

bool eq(const db &a, const db &b) { return fabs(a - b) < eps; }

bool ls(const db &a, const db &b) { return a + eps < b; }

bool le(const db &a, const db &b) { return eq(a, b) || ls(a, b); }

ll gcd(ll a,ll b) { return a==?b:gcd(b%a,a); };

ll lcm(ll a,ll b) { return a/gcd(a,b)*b; }

ll kpow(ll a,ll b) {ll res=;a%=mod; if(b<) return ; for(;b;b>>=){if(b&)res=res*a%mod;a=a*a%mod;}return res;}

ll read(){

    ll x=,f=;char ch=getchar();

    while (ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}

    while (ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}

    return x*f;

}

//inv[1]=1;

//for(int i=2;i<=n;i++) inv[i]=(mod-mod/i)*inv[mod%i]%mod;

int a[maxn];

vector<int> ans;

int main(){

    rep(i,,){

        rep(j,,) scanf("%d",&a[j]);

        sort(a,a+);

        ans.pb(a[]);

    }

    sort(all(ans));

    printf("%d\n",ans[]);

    return ;

}

SGU 193

题意：给你一个N，让你求最大一个数字b，gcd(n,b)==1,(1<=b<n/2)

收获：不是奇数就是偶数，判断一下就行了

#include<bits/stdc++.h>

#define de(x) cout<<#x<<"="<<x<<endl;

#define dd(x) cout<<#x<<"="<<x<<" ";

#define rep(i,a,b) for(int i=a;i<(b);++i)

#define repd(i,a,b) for(int i=a;i>=(b);--i)

#define repp(i,a,b,t) for(int i=a;i<(b);i+=t)

#define ll long long

#define mt(a,b) memset(a,b,sizeof(a))

#define fi first

#define se second

#define inf 0x3f3f3f3f

#define INF 0x3f3f3f3f3f3f3f3f

#define pii pair<int,int>

#define pdd pair<double,double>

#define pdi pair<double,int>

#define mp(u,v) make_pair(u,v)

#define sz(a) (int)a.size()

#define ull unsigned long long

#define ll long long

#define pb push_back

#define PI acos(-1.0)

#define qc std::ios::sync_with_stdio(false)

#define db double

#define all(a) a.begin(),a.end()

const int mod = 1e9+;

const int maxn = 1e5+;

const double eps = 1e-;

using namespace std;

bool eq(const db &a, const db &b) { return fabs(a - b) < eps; }

bool ls(const db &a, const db &b) { return a + eps < b; }

bool le(const db &a, const db &b) { return eq(a, b) || ls(a, b); }

ll gcd(ll a,ll b) { return a==?b:gcd(b%a,a); };

ll lcm(ll a,ll b) { return a/gcd(a,b)*b; }

ll kpow(ll a,ll b) {ll res=;a%=mod; if(b<) return ; for(;b;b>>=){if(b&)res=res*a%mod;a=a*a%mod;}return res;}

ll read(){

    ll x=,f=;char ch=getchar();

    while (ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}

    while (ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}

    return x*f;

}

//inv[1]=1;

//for(int i=2;i<=n;i++) inv[i]=(mod-mod/i)*inv[mod%i]%mod;

int compare(string str1,string str2)

{

    if(str1.length()>str2.length()) return ;

    else if(str1.length()<str2.length())  return -;

    else return str1.compare(str2);

}

//高精度加法

//只能是两个正数相加

string add(string str1,string str2)//高精度加法

{

    string str;

    int len1=str1.length();

    int len2=str2.length();

    //前面补0，弄成长度相同

    if(len1<len2)

    {

        for(int i=;i<=len2-len1;i++)

           str1=""+str1;

    }

    else

    {

        for(int i=;i<=len1-len2;i++)

           str2=""+str2;

    }

    len1=str1.length();

    int cf=;

    int temp;

    for(int i=len1-;i>=;i--)

    {

        temp=str1[i]-''+str2[i]-''+cf;

        cf=temp/;

        temp%=;

        str=char(temp+'')+str;

    }

    if(cf!=)  str=char(cf+'')+str;

    return str;

}

//高精度减法

//只能是两个正数相减，而且要大减小

string sub(string str1,string str2)//高精度减法

{

    string str;

    int tmp=str1.length()-str2.length();

    int cf=;

    for(int i=str2.length()-;i>=;i--)

    {

        if(str1[tmp+i]<str2[i]+cf)

        {

            str=char(str1[tmp+i]-str2[i]-cf+''+)+str;

            cf=;

        }

        else

        {

            str=char(str1[tmp+i]-str2[i]-cf+'')+str;

            cf=;

        }

    }

    for(int i=tmp-;i>=;i--)

    {

        if(str1[i]-cf>='')

        {

            str=char(str1[i]-cf)+str;

            cf=;

        }

        else

        {

            str=char(str1[i]-cf+)+str;

            cf=;

        }

    }

    str.erase(,str.find_first_not_of(''));//去除结果中多余的前导0

    return str;

}

//高精度乘法

//只能是两个正数相乘

string mul(string str1,string str2)

{

    string str;

    int len1=str1.length();

    int len2=str2.length();

    string tempstr;

    for(int i=len2-;i>=;i--)

    {

        tempstr="";

        int temp=str2[i]-'';

        int t=;

        int cf=;

        if(temp!=)

        {

            for(int j=;j<=len2--i;j++)

              tempstr+="";

            for(int j=len1-;j>=;j--)

            {

                t=(temp*(str1[j]-'')+cf)%;

                cf=(temp*(str1[j]-'')+cf)/;

                tempstr=char(t+'')+tempstr;

            }

            if(cf!=) tempstr=char(cf+'')+tempstr;

        }

        str=add(str,tempstr);

    }

    str.erase(,str.find_first_not_of(''));

    return str;

}

//高精度除法

//两个正数相除，商为quotient,余数为residue

//需要高精度减法和乘法

void div(string str1,string str2,string &quotient,string &residue)

{

    quotient=residue="";//清空

    if(str2=="")//判断除数是否为0

    {

        quotient=residue="ERROR";

        return;

    }

    if(str1=="")//判断被除数是否为0

    {

        quotient=residue="";

        return;

    }

    int res=compare(str1,str2);

    if(res<)

    {

        quotient="";

        residue=str1;

        return;

    }

    else if(res==)

    {

        quotient="";

        residue="";

        return;

    }

    else

    {

        int len1=str1.length();

        int len2=str2.length();

        string tempstr;

        tempstr.append(str1,,len2-);

        for(int i=len2-;i<len1;i++)

        {

            tempstr=tempstr+str1[i];

            tempstr.erase(,tempstr.find_first_not_of(''));

            if(tempstr.empty())

              tempstr="";

            for(char ch='';ch>='';ch--)//试商

            {

                string str,tmp;

                str=str+ch;

                tmp=mul(str2,str);

                if(compare(tmp,tempstr)<=)//试商成功

                {

                    quotient=quotient+ch;

                    tempstr=sub(tempstr,tmp);

                    break;

                }

            }

        }

        residue=tempstr;

    }

    quotient.erase(,quotient.find_first_not_of(''));

    if(quotient.empty()) quotient="";

}

int main(){

    qc;

    string n,two="",one="",q,r,q1,r1;

//    cout<<sub(one,one)<<endl;

    cin>>n;

    div(n,two,q,r);

//    dd(r)de(q)

    if(r=="") cout<<q;

    else {

        div(q,two,q1,r1);

//        dd(r1)de(q1)

        if(r1=="") {

            if(q==one) cout<<one;

            else cout<<sub(q,two);

        }

        else {

            if(q==one) cout<<one;

            else cout<<sub(q,one);

        }

    }

    return ;

}

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