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 ;
}