SGU123

题意：求和

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;}

int read(){

    int 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;

ll f[];

int main(){

    f[]=;f[]=;

    rep(i,,) f[i]=f[i-]+f[i-];

    int k = read();

    ll sum = ;

    rep(i,,k+) sum+=f[i];

    printf("%lld\n",sum);

    return ;

}

SGU115

题意：求2001某月某日是星期几

收获：注意无解的情况

#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 month[]={,,,,,,,,,,,,};

int main(){

    int n,m;

    scanf("%d%d",&n,&m);

    if(m>) return puts("Impossible"),;

    if(n>month[m]) return puts("Impossible"),;

    rep(i,,m) n+=month[i];

    n%=;if(!n)n=;

    printf("%d\n",n);

    return ;

}

SGU105

题意：1,12,123，...，1....N,求这n个数字中被3整除的个数

收获：打表

#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 bit(int x){

    int ret = ;

    while(x) x/=,ret++;

    return ret;

}

void dabiao(){

    ll sum = ;

    rep(i,,){

        sum = sum * kpow(,bit(i)) + i;

        if(sum%==) de(sum)

    }

}

int main(){

//    dabiao();

    int n,now=;

    scanf("%d",&n);

    n--;

    int t = n/;

    int d = n%;

    printf("%d\n",t*+d);

    return ;

}

SGU135

题意：问你画n个线段，最多把一个无穷大的矩形分成几个区域

收获：打表，找规律，或者可以这么想，你新加入第k条直线，最多与k-1一条直线同时相交，那么最多就会比上一次多弄出k个区间

#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 = ;

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;

ll f[maxn];

int main(){

    f[]=;f[]=;

    int n;

    scanf("%d",&n);

    rep(i,,n+) f[i]=f[i-]+i;

    printf("%lld\n",f[n]);

    return ;

}

SGU184

题意：让你做饼，要求最大数量

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 main(){

    int p,m,c,k,r,v;

    scanf("%d%d%d%d%d%d",&p,&m,&c,&k,&r,&v);

//    de(c)de(v)

    printf("%d\n",min(p/k,min(m/r,c/v)));

    return ;

}

SGU113

题意：求一个数能不能分解成两个素数相乘

收获：素数打表

#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 N = 1e5+;

bool isPrime[N];

int prim[];

void prime(){

    int num = ;

    memset(isPrime,true,sizeof(isPrime));

    isPrime[] = isPrime[] = false;

    for(int i= ; i<=N ; i++){

        if( isPrime[i] ) prim[num++] = i;

        for(int j= ; j<num ; j++){

            if( i*prim[j]>N ) break;

            isPrime[ i*prim[j] ] = false;

            if( i%prim[j] ==  ) break;

        }

    }

}

bool isprime(int x){

    for(int i=;i*i<=x;++i) if(x%i==) return false;

    return true;

}

bool ok(int x){

    for(int i=;i*i<=x;++i){

        if(x%i==&&isPrime[i]){

            if(isprime(x/i)) return true;

        }

    }

    return false;

}

int main(){

    prime();

    int n,x;

    scanf("%d",&n);

    rep(i,,n){

        scanf("%d",&x);

        puts(ok(x)?"Yes":"No");

    }

    return ;

}

SGU112

题意：求a^b-b^a

收获：kuangbin的string高精度板子，用了std::ios::sync_with_stdio(false)，不能再用printf和scanf了，会出现奇怪的错误，会wa

#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;

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;

}

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;

}

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;

}

int bit(int x){

    int ret = ;

    while(x) x/=,ret++;

    return ret;

}

string change(int x){

    char s[];

    int len = bit(x) - ;

    s[len+] = '\0';

    while(x){

        s[len--]=(x%+'');

        x/=;

    }

    string ss = s;

    return ss;

}

bool big(string a,string b){

    if(sz(a)<sz(b)) return false;

    if(sz(a)>sz(b)) return true;

    return a>b;

}

int main(){

    qc;

    bool fg=false;

    int ta,tb;

    cin>>ta>>tb;

    string a,b;

    a=change(ta),b=change(tb);

//    de(a)de(b)

    string ansa=a,ansb=b;

    rep(i,,tb) ansa=mul(ansa,a);

    rep(i,,ta) ansb=mul(ansb,b);

    if(big(ansb,ansa)) fg=true,swap(ansa,ansb);

    if(fg) cout<<'-';

    string ans=sub(ansa,ansb);

    cout<<ans;

    return ;

}

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