欧拉降幂定理
#include <iostream> #include<cstdio> #include<cmath> #include<algorithm> typedef long long ll; using namespace std; ll a,b,c,p; ll qpow(ll a,ll b,ll mod){ ll ret=1; while(b){ if(b&1) ret=ret*a%mod; a=a*a%mod; b>>=1; } return ret; } ll getphi(ll x){ ll ret=x; for(ll i=2;i*i<=x;i++) if(x%i==0){ ret/=i,ret*=(i-1); while(x%i==0)x/=i; } if(x>1)ret/=x,ret*=x-1; return ret; } int main() { scanf("%lld%lld%lld%lld",&a,&b,&c,&p); ll phi=getphi(p); ll phi_=getphi(phi); ll tmp=qpow(b,c%phi_+phi_,phi); ll ans=qpow(a,tmp+phi,p); printf("%lld\n",ans); return 0; }
C.约数的平方和
考虑每个约数对答案的贡献
#include <iostream> #include<cstdio> #include<cmath> typedef long long ll; using namespace std; ll n,p; int main() { scanf("%lld%lld",&n,&p); ll ans=0; for(ll i=1;i<=1000005;i++){ ans=(ans+i*i%p*(n/i)%p)%p; } printf("%lld\n",ans); return 0; }
D.随机配对
H.幻想机器人
模拟
#include <iostream> #include<cstdio> #include<cstring> #include<cmath> #include<map> #include<algorithm> typedef long long ll; using namespace std; char s[100005]; struct Node{ ll x,y; }now; map<pair<ll,ll>,ll> mp; int main() { ll n,m,k,len,x,y; scanf("%lld%lld%lld%lld%lld%lld",&n,&m,&k,&len,&x,&y); scanf("%s",s); ll cnt=0; mp.clear(); now.x=x,now.y=y; mp[make_pair(now.x,now.y)]=1,cnt++; for(ll i=0;i<len;i++){ if(s[i]=='U'){ if(now.x-1>=1){ now.x--; if((i+1)%k==0){ if(mp[make_pair(now.x,now.y)]) printf("AKTang!\n"); else mp[make_pair(now.x,now.y)]=1,cnt++; } } else{ printf("AWaDa!\n"); if((i+1)%k==0){ if(mp[make_pair(now.x,now.y)]) printf("AKTang!\n"); else mp[make_pair(now.x,now.y)]=1,cnt++; } } } else if(s[i]=='D'){ if(now.x+1<=n){ now.x++; if((i+1)%k==0){ if(mp[make_pair(now.x,now.y)]) printf("AKTang!\n"); else mp[make_pair(now.x,now.y)]=1,cnt++; } } else{ printf("AWaDa!\n"); if((i+1)%k==0){ if(mp[make_pair(now.x,now.y)]) printf("AKTang!\n"); else mp[make_pair(now.x,now.y)]=1,cnt++; } } } else if(s[i]=='L'){ if(now.y-1>=1){ now.y--; if((i+1)%k==0){ if(mp[make_pair(now.x,now.y)]) printf("AKTang!\n"); else mp[make_pair(now.x,now.y)]=1,cnt++; } } else{ printf("AWaDa!\n"); if((i+1)%k==0){ if(mp[make_pair(now.x,now.y)]) printf("AKTang!\n"); else mp[make_pair(now.x,now.y)]=1,cnt++; } } } if(s[i]=='R'){ if(now.y+1<=m){ now.y++; if((i+1)%k==0){ if(mp[make_pair(now.x,now.y)]) printf("AKTang!\n"); else mp[make_pair(now.x,now.y)]=1,cnt++; } } else{ printf("AWaDa!\n"); if((i+1)%k==0){ if(mp[make_pair(now.x,now.y)]) printf("AKTang!\n"); else mp[make_pair(now.x,now.y)]=1,cnt++; } } } } printf("%lld\n",n*m-cnt); return 0; }
B.阿卡吃馅饼
大数+枚举
#include<iostream> #include<cstdio> #include<cstring> #include<string> #include<iomanip> #include<algorithm> using namespace std; #define MAXN 9999 #define MAXSIZE 10 #define DLEN 4 class BigNum { private: int a[1000]; //可以控制大数的位数 int len; //大数长度 public: BigNum(){ len = 1;memset(a,0,sizeof(a)); } //构造函数 BigNum(const int); //将一个int类型的变量转化为大数 BigNum(const char*); //将一个字符串类型的变量转化为大数 BigNum(const BigNum &); //拷贝构造函数 BigNum &operator=(const BigNum &); //重载赋值运算符，大数之间进行赋值运算 friend istream& operator>>(istream&, BigNum&); //重载输入运算符 friend ostream& operator<<(ostream&, BigNum&); //重载输出运算符 BigNum operator+(const BigNum &) const; //重载加法运算符，两个大数之间的相加运算 BigNum operator-(const BigNum &) const; //重载减法运算符，两个大数之间的相减运算 BigNum operator*(const BigNum &) const; //重载乘法运算符，两个大数之间的相乘运算 BigNum operator/(const int &) const; //重载除法运算符，大数对一个整数进行相除运算 BigNum operator^(const int &) const; //大数的n次方运算 int operator%(const int &) const; //大数对一个int类型的变量进行取模运算 bool operator>(const BigNum & T)const; //大数和另一个大数的大小比较 bool operator>(const int & t)const; //大数和一个int类型的变量的大小比较 void print(); //输出大数 }; BigNum::BigNum(const int b) //将一个int类型的变量转化为大数 { int c,d = b; len = 0; memset(a,0,sizeof(a)); while(d > MAXN) { c = d - (d / (MAXN + 1)) * (MAXN + 1); d = d / (MAXN + 1); a[len++] = c; } a[len++] = d; } BigNum::BigNum(const char*s) //将一个字符串类型的变量转化为大数 { int t,k,index,l,i; memset(a,0,sizeof(a)); l=strlen(s); len=l/DLEN; if(l%DLEN) len++; index=0; for(i=l-1;i>=0;i-=DLEN) { t=0; k=i-DLEN+1; if(k<0) k=0; for(int j=k;j<=i;j++) t=t*10+s[j]-'0'; a[index++]=t; } } BigNum::BigNum(const BigNum & T) : len(T.len) //拷贝构造函数 { int i; memset(a,0,sizeof(a)); for(i = 0 ; i < len ; i++) a[i] = T.a[i]; } BigNum & BigNum::operator=(const BigNum & n) //重载赋值运算符，大数之间进行赋值运算 { int i; len = n.len; memset(a,0,sizeof(a)); for(i = 0 ; i < len ; i++) a[i] = n.a[i]; return *this; } istream& operator>>(istream & in, BigNum & b) //重载输入运算符 { char ch[MAXSIZE*4]; int i = -1; in>>ch; int l=strlen(ch); int count=0,sum=0; for(i=l-1;i>=0;) { sum = 0; int t=1; for(int j=0;j<4&&i>=0;j++,i--,t*=10) { sum+=(ch[i]-'0')*t; } b.a[count]=sum; count++; } b.len =count++; return in; } ostream& operator<<(ostream& out, BigNum& b) //重载输出运算符 { int i; cout << b.a[b.len - 1]; for(i = b.len - 2 ; i >= 0 ; i--) { cout.width(DLEN); cout.fill('0'); cout << b.a[i]; } return out; } BigNum BigNum::operator+(const BigNum & T) const //两个大数之间的相加运算 { BigNum t(*this); int i,big; //位数 big = T.len > len ? T.len : len; for(i = 0 ; i < big ; i++) { t.a[i] +=T.a[i]; if(t.a[i] > MAXN) { t.a[i + 1]++; t.a[i] -=MAXN+1; } } if(t.a[big] != 0) t.len = big + 1; else t.len = big; return t; } BigNum BigNum::operator-(const BigNum & T) const //两个大数之间的相减运算 { int i,j,big; bool flag; BigNum t1,t2; if(*this>T) { t1=*this; t2=T; flag=0; } else { t1=T; t2=*this; flag=1; } big=t1.len; for(i = 0 ; i < big ; i++) { if(t1.a[i] < t2.a[i]) { j = i + 1; while(t1.a[j] == 0) j++; t1.a[j--]--; while(j > i) t1.a[j--] += MAXN; t1.a[i] += MAXN + 1 - t2.a[i]; } else t1.a[i] -= t2.a[i]; } t1.len = big; while(t1.a[len - 1] == 0 && t1.len > 1) { t1.len--; big--; } if(flag) t1.a[big-1]=0-t1.a[big-1]; return t1; } BigNum BigNum::operator*(const BigNum & T) const //两个大数之间的相乘运算 { BigNum ret; int i,j,up; int temp,temp1; for(i = 0 ; i < len ; i++) { up = 0; for(j = 0 ; j < T.len ; j++) { temp = a[i] * T.a[j] + ret.a[i + j] + up; if(temp > MAXN) { temp1 = temp - temp / (MAXN + 1) * (MAXN + 1); up = temp / (MAXN + 1); ret.a[i + j] = temp1; } else { up = 0; ret.a[i + j] = temp; } } if(up != 0) ret.a[i + j] = up; } ret.len = i + j; while(ret.a[ret.len - 1] == 0 && ret.len > 1) ret.len--; return ret; } BigNum BigNum::operator/(const int & b) const //大数对一个整数进行相除运算 { BigNum ret; int i,down = 0; for(i = len - 1 ; i >= 0 ; i--) { ret.a[i] = (a[i] + down * (MAXN + 1)) / b; down = a[i] + down * (MAXN + 1) - ret.a[i] * b; } ret.len = len; while(ret.a[ret.len - 1] == 0 && ret.len > 1) ret.len--; return ret; } int BigNum::operator %(const int & b) const //大数对一个int类型的变量进行取模运算 { int i,d=0; for (i = len-1; i>=0; i--) { d = ((d * (MAXN+1))% b + a[i])% b; } return d; } BigNum BigNum::operator^(const int & n) const //大数的n次方运算 { BigNum t,ret(1); int i; if(n<0) exit(-1); if(n==0) return 1; if(n==1) return *this; int m=n; while(m>1) { t=*this; for( i=1;i<<1<=m;i<<=1) { t=t*t; } m-=i; ret=ret*t; if(m==1) ret=ret*(*this); } return ret; } bool BigNum::operator>(const BigNum & T) const //大数和另一个大数的大小比较 { int ln; if(len > T.len) return true; else if(len == T.len) { ln = len - 1; while(a[ln] == T.a[ln] && ln >= 0) ln--; if(ln >= 0 && a[ln] > T.a[ln]) return true; else return false; } else return false; } bool BigNum::operator >(const int & t) const //大数和一个int类型的变量的大小比较 { BigNum b(t); return *this>b; } void BigNum::print() //输出大数 { int i; cout << a[len - 1]; for(i = len - 2 ; i >= 0 ; i--) { cout.width(DLEN); cout.fill('0'); cout << a[i]; } cout << endl; } BigNum qpow(BigNum a,int b){ BigNum ret=1; while(b){ if(b&1) ret=ret*a; a=a*a; b>>=1; } return ret; } int main() { int T; scanf("%d",&T); while(T--){ int p,r;scanf("%d%d",&p,&r); int x; BigNum now(1),tot(0); for(x=0;x<=1000;x++){ tot=tot+now*p; //tot.print(); if(tot%r==0) break; now=now*10; } if(x>1000) puts("GG"); else printf("%d\n",x); } }
E.大战幻想珠
暴力
#include <iostream> #include<cstdio> #include<cmath> #include<cstring> #include<algorithm> using namespace std; char s[3005]; bool check(char a,char b){ if(a==b||a=='?'||b=='?') return 1; return 0; } int main() { int n;scanf("%d",&n); scanf("%s",s); int ans=0; for(int i=0;i<n;i++){ for(int j=0;i-j>=0&&i+j<n;j++){ if(check(s[i-j],s[i+j])) ans++; else break; } for(int j=0;i+j<n&&i-j-1>=0;j++){ if(check(s[i-j-1],s[i+j])) ans++; else break; } } printf("%d\n",ans); return 0; }
C.烤乐滋打虎
可以证明，不拐弯最优
#include <iostream> #include<cstdio> #include<cstring> #define inf 0x3f3f3f3f using namespace std; int a[105],d[105]; int vis[105]; int main() { int n;scanf("%d",&n); for(int i=1;i<=n;i++){ scanf("%d",&a[i]); } for(int i=1;i<=n-1;i++){ scanf("%d",&d[i]); } int ans=inf; for(int i=2;i<=n-1;i++){ memset(vis,0,sizeof(vis));vis[i]=1; int p=i,dire=-1,cnt=1,tot=a[p],now=0; while(cnt<n){ if(dire==-1) p--,now+=d[p]; else now+=d[p],p++; if(!vis[p]) vis[p]=1,cnt++,tot+=(now+a[p]); if(p==1) dire=1; if(p==n) dire=-1; } ans=min(ans,tot); memset(vis,0,sizeof(vis));vis[i]=1; p=i,dire=1,cnt=1,tot=a[p],now=0; while(cnt<n){ if(dire==1) now+=d[p],p++; else p--,now+=d[p]; if(!vis[p]) vis[p]=1,cnt++,tot+=(now+a[p]); if(p==1) dire=1; if(p==n) dire=-1; } ans=min(ans,tot); } int tot=0,now=0; for(int i=1;i<=n;i++){ tot+=(now+a[i]); now+=d[i]; } ans=min(ans,tot); tot=0,now=0; for(int i=n;i>=1;i--){ tot+=(now+a[i]); now+=d[i-1]; } ans=min(ans,tot); printf("%d\n",ans); return 0; }
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