今日SGU 5.29

sgu 299

题意：给你n个线段，然后问你能不能选出其中三个组成一个三角形，数字很大

收获：另一个大整数模板

那么考虑下为什么如果连续三个不可以的话，一定是不存在呢？

连续上个不合法的话，一定是 ai-1 + ai-2 < = ai;

那么如果我们取右边的数，那是不是aj ,那么aj >= ai就更不可能成立了，

取左边的一样可以证明出不可以

#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 = 1e3+;
const double eps = 1e-;
using namespace std;
class bign
{
public:
    int len, s[maxn];//数的长度，记录数组
//构造函数
    bign();
    bign(const char*);
    bign(int);
    bool sign;//符号 1正数 0负数
    string toStr() const;//转化为字符串，主要是便于输出
    friend istream& operator>>(istream &,bign &);//重载输入流
    friend ostream& operator<<(ostream &,bign &);//重载输出流
//重载复制
    bign operator=(const char*);
    bign operator=(int);
    bign operator=(const string);
//重载各种比较
    bool operator>(const bign &) const;
    bool operator>=(const bign &) const;
    bool operator<(const bign &) const;
    bool operator<=(const bign &) const;
    bool operator==(const bign &) const;
    bool operator!=(const bign &) const;
//重载四则运算
    bign operator+(const bign &) const;
    bign operator++();
    bign operator++(int);
    bign operator+=(const bign&);
    bign operator-(const bign &) const;
    bign operator--();
    bign operator--(int);
    bign operator-=(const bign&);
    bign operator*(const bign &)const;
    bign operator*(const int num)const;
    bign operator*=(const bign&);
    bign operator/(const bign&)const;
    bign operator/=(const bign&);
//四则运算的衍生运算
    bign operator%(const bign&)const;//取模（余数）
    bign factorial()const;//阶乘
    bign Sqrt()const;//整数开根（向下取整）
    bign pow(const bign&)const;//次方
//一些乱乱的函数
    void clean();
    ~bign();
};
#define max(a,b) a>b ? a : b
#define min(a,b) a<b ? a : b  

bign::bign()
{
    memset(s, , sizeof(s));
    len = ;
    sign = ;
}  

bign::bign(const char *num)
{
    *this = num;
}  

bign::bign(int num)
{
    *this = num;
}  

string bign::toStr() const
{
    string res;
    res = "";
    for (int i = ; i < len; i++)
        res = (char)(s[i] + '') + res;
    if (res == "")
        res = "";
    if (!sign&&res != "")
        res = "-" + res;
    return res;
}  

istream &operator>>(istream &in, bign &num)
{
    string str;
    in>>str;
    num=str;
    return in;
}  

ostream &operator<<(ostream &out, bign &num)
{
    out<<num.toStr();
    return out;
}  

bign bign::operator=(const char *num)
{
    memset(s, , sizeof(s));
    char a[maxn] = "";
    if (num[] != '-')
        strcpy(a, num);
    else
        for (int i = ; i < strlen(num); i++)
            a[i - ] = num[i];
    sign = !(num[] == '-');
    len = strlen(a);
    for (int i = ; i < strlen(a); i++)
        s[i] = a[len - i - ] - ;
    return *this;
}  

bign bign::operator=(int num)
{
    if (num < )
        sign = , num = -num;
    else
        sign = ;
    char temp[maxn];
    sprintf(temp, "%d", num);
    *this = temp;
    return *this;
}  

bign bign::operator=(const string num)
{
    const char *tmp;
    tmp = num.c_str();
    *this = tmp;
    return *this;
}  

bool bign::operator<(const bign &num) const
{
    if (sign^num.sign)
        return num.sign;
    if (len != num.len)
        return len < num.len;
    for (int i = len - ; i >= ; i--)
        if (s[i] != num.s[i])
            return sign ? (s[i] < num.s[i]) : (!(s[i] < num.s[i]));
    return !sign;
}  

bool bign::operator>(const bign&num)const
{
    return num < *this;
}  

bool bign::operator<=(const bign&num)const
{
    return !(*this>num);
}  

bool bign::operator>=(const bign&num)const
{
    return !(*this<num);
}  

bool bign::operator!=(const bign&num)const
{
    return *this > num || *this < num;
}  

bool bign::operator==(const bign&num)const
{
    return !(num != *this);
}  

bign bign::operator+(const bign &num) const
{
    if (sign^num.sign)
    {
        bign tmp = sign ? num : *this;
        tmp.sign = ;
        return sign ? *this - tmp : num - tmp;
    }
    bign result;
    result.len = ;
    int temp = ;
    for (int i = ; temp || i < (max(len, num.len)); i++)
    {
        int t = s[i] + num.s[i] + temp;
        result.s[result.len++] = t % ;
        temp = t / ;
    }
    result.sign = sign;
    return result;
}  

bign bign::operator++()
{
    *this = *this + ;
    return *this;
}  

bign bign::operator++(int)
{
    bign old = *this;
    ++(*this);
    return old;
}  

bign bign::operator+=(const bign &num)
{
    *this = *this + num;
    return *this;
}  

bign bign::operator-(const bign &num) const
{
    bign b=num,a=*this;
    if (!num.sign && !sign)
    {
        b.sign=;
        a.sign=;
        return b-a;
    }
    if (!b.sign)
    {
        b.sign=;
        return a+b;
    }
    if (!a.sign)
    {
        a.sign=;
        b=bign()-(a+b);
        return b;
    }
    if (a<b)
    {
        bign c=(b-a);
        c.sign=false;
        return c;
    }
    bign result;
    result.len = ;
    for (int i = , g = ; i < a.len; i++)
    {
        int x = a.s[i] - g;
        if (i < b.len) x -= b.s[i];
        if (x >= ) g = ;
        else
        {
            g = ;
            x += ;
        }
        result.s[result.len++] = x;
    }
    result.clean();
    return result;
}  

bign bign::operator * (const bign &num)const
{
    bign result;
    result.len = len + num.len;  

    for (int i = ; i < len; i++)
        for (int j = ; j < num.len; j++)
            result.s[i + j] += s[i] * num.s[j];  

    for (int i = ; i < result.len; i++)
    {
        result.s[i + ] += result.s[i] / ;
        result.s[i] %= ;
    }
    result.clean();
    result.sign = !(sign^num.sign);
    return result;
}
bign bign::operator*(const int num)const
{
    bign x = num;
    bign z = *this;
    return x*z;
}
void bign::clean()
{
    if (len == ) len++;
    while (len >  && s[len - ] == '\0')
        len--;
}
bign::~bign()
{
}
bign a[maxn];
int main(){
    int n;
    scanf("%d",&n);
    rep(i,,n+) cin>>a[i];
    sort(a+,a+n+);
    rep(i,,n+) if(a[i] < a[i-] + a[i-]) {
        cout<<a[i]<<" "<<a[i-]<<" "<<a[i-]<<endl;
        return ;
    }
    puts("0 0 0");
    return ;
}

sgu 230

题意：告诉你硬币1 到 n，然后重量依次递增，然后给你n个盒子，然后让你用n个盒子装这n个硬币，然后重量大的

要装重量大的硬币

收获：好好看题意，拓扑

#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 = 2e2+;
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=; if(b<) return ; for(;b;b>>=){if(b&)res=res*a;a=a*a;}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;
}
int n,m;
int in[maxn];
vector<int> G[maxn],ans;
bool topo(){
    queue<int> q; while(sz(q)) q.pop();
    rep(i,,n+) if(!in[i]) q.push(i);
    while(sz(q)){
        int u = q.front(); q.pop();
        ans.pb(u);
        rep(i,,sz(G[u])){
            int v = G[u][i];
            in[v]--;
            if(!in[v]) q.push(v);
        }
    }
    return sz(ans) == n;
}
int a[maxn];
int main(){
    scanf("%d%d",&n,&m);
    rep(i,,m) {
        int u,v;
        scanf("%d%d",&u,&v);
        G[u].pb(v);
        in[v]++;
    }
    if(!topo()) return puts("No solution"),;
    rep(i,,sz(ans)) a[ans[i]]=i+;//printf("%d%c",ans[i]," \n"[i+1==sz(ans)]);
    rep(i,,n+) printf("%d%c",a[i]," \n"[i==n]);
    return ;
}

sgu 249

题意：给你0到2^(n+m)-1这些数字，要你放到2^n行，2^m列的矩形里，要求相邻两个数字的二进制只差一位

收获：格雷码

#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;
//格雷码的相邻数，二进制只差一位
int main(){
    int n,m;
    scanf("%d%d",&n,&m);
    rep(i,,<<n) rep(j,,<<m) printf("%d%c",(((i^(i>>))<<m)) | (j^(j>>))," \n"[j+==(<<m)]);
    return ;
}