今日 SGU 5.6

SGU 106

题意：问你有多少个<x,y>，满足ax+by+c=0，x1<=x<=x2,y1<=y<=y2

收货：拓展欧几里得求解的是这种方程，ax+by=1,gcd(a,b)=1

如果gcd(a,b)不等于1的话，那么你直接传进egcd函数里求出的x，y还是a1x+b1y=1的解,a1=a/gcd(a,b),b1=b/gcd(a,b)

还有注意y1，x0，y0会和系统的里面变量冲突

#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;
void egcd(ll a,ll b,ll &d,ll &x,ll &y){
    if(!b) d=a ,x= ,y= ;
    else{
        egcd(b,a%b,d,y,x);
        y-=a/b*x;
    }
}
ll x,y,d;
ll a,b,c,x1,x2,yy1,yy2;
int main(){
     scanf("%lld%lld%lld%lld%lld%lld%lld",&a,&b,&c,&x1,&x2,&yy1,&yy2);
    c=-c;
    if(c<) a=-a,b=-b,c=-c;
    if(a<) a=-a,swap(x1,x2),x1=-x1,x2=-x2;
    if(b<) b=-b,swap(yy1,yy2),yy1=-yy1,yy2=-yy2;
    if(a==||b==){
        if(a==&&b==){
            if(c!=) puts("");
            else printf("%lld\n",(x2-x1+)*(yy2-yy1+));
        }else if(a==){
            if(c%b==&&c/b>=yy1&&c/b<=yy2) printf("%lld\n",x2-x1+);
            else return puts(""),;
        }else {
            if(c%a==&&c/a>=x1&&c/a<=x2) printf("%lld\n",yy2-yy1+);
            else puts("");
        }
        return ;
    }
    egcd(a,b,d,x,y);//这边求得是a/d*x+b/d*y=1(d=gcd(a,b))的解，那么乘以c/gcd(a,b)就得到a/d*x+b/d*y=c/d的一个解了
    if(c%d) return puts(""),;
    double aa = a/d,bb = b/d;
    ll cc = c/d;
    x*=cc,y*=cc;
    ll r=min(floor((x2-x)/bb),floor((y-yy1)/aa)) ,l=max(ceil((x1-x)/bb),ceil((y-yy2)/aa));
    if(r>=l) printf("%lld\n",r-l+);
    else puts("");
    return ;
}

SGU 111

题意：求一个大整数的开方

收获：大整数开方模板

#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 l;
int work(int o,char *O,int I){//求大整数开根号
    char c, *D=O ;
    if(o>){
        for(l=;D[l];D[l++]-=)
        {
            D[l++]-=;
            D[l]-=;
            while(!work(,O,l))
                D[l]+=;
            putchar((D[l]+)/);
        }
        putchar();
    }
    else{
        c=o+(D[I]+)%-(I>l/)*(D[I-l+I]+)/-;
        D[I]+=I< ?  : !(o=work(c/,O,I-))*((c+)%-(D[I]+)%);
    }
    return o;
}

int main(){
    char s[maxn];s[]='';
    scanf("%s",s+);
    if(strlen(s)%==) work(,s+,);
    else work(,s,);
    return ;
}

//转载别人的手算开方方法，防止自己以后忘记

SGU 181

题意求：xk，然后给你个x的递推公式

收获：找循环节，你要记录循环节的头

#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 maxn = 1e3+;
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 mod;
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;
}
ll x[maxn];
map<ll,int> m;
int main(){
    qc;
    ll a,alpha,beta,gamma,k;
    cin>>a>>alpha>>beta>>gamma>>mod>>k;
    x[]=a;
    if(k==) {
        cout<<a<<endl;
        return ;
    }
    int t=mod,be=;
    rep(i,,k+) {
        x[i]=(x[i-]*x[i-]*alpha+beta*x[i-]+gamma)%mod;
        if(m.count(x[i])) {
            t = i - m[x[i]];
            be = m[x[i]];
            break;
        }
        m[x[i]] = i;
    }
    cout<<x[be+(k-be)%t];
    return ;
}