OAuth2實戰QQ,微信第三方登入
阿新 • • 發佈:2020-07-25
#include<iostream> #include<vector> #include<string.h> #include<map> #include<string> #include<algorithm> #include<cstdio> using namespace std; #define sz(v) (int)(v.size()) #define rep(i,a,n) for(int i=a;i<n;++i) #define per(i,a,n) for(int i=n-1;i>=a;--i) #definemem(a,t) memset(a,t,sizeof(a)) #define pb push_back #define mp make_pair #define fi first #define se second #define N 25 #define MAXN 9999 #define MAXSIZE 1010 #define DLEN 4 class BigNum { private: int a[500]; //可以控制大數的位數 int len; public: BigNum(){len=1;memset(a,0,sizeof(a));} //建構函式 BigNum(constint); //將一個int型別的變數轉化成大數 BigNum(const __int64); //將一個__int64型別的變數轉化成大數 BigNum(const char*); //將一個字串型別的變數轉化為大數 BigNum(const BigNum &); //拷貝建構函式 BigNum &operator=(const BigNum &); //過載賦值運算子,大數之間進行賦值運算 friend istream& operator>>(istream&,BigNum&); //過載輸入運算子friend ostream& operator<<(ostream&,BigNum&); //過載輸出運算子 BigNum operator+(const BigNum &)const; //過載加法運算子,兩個大數之間的相加運算 BigNum operator-(const BigNum &)const; //過載減法運算子,兩個大數之間的相減運算 BigNum operator*(const BigNum &)const; //過載乘法運算子,兩個大數之間的相乘運算 BigNum operator/(const int &)const; //過載除法運算子,大數對一個整數進行相除運算 BigNum operator/(const __int64 &)const; //過載除法運算子,大數對一個64位整數進行相除運算 BigNum operator^(const int &)const; //大數的n次方運算 int operator%(const int &)const; //大數對一個int型別的變數進行取模運算 __int64 operator%(const __int64 &)const; //大數對一個__int64型別的變數進行取模運算 bool operator>(const BigNum &T)const; //大數和另一個大數的大小比較 bool operator>(const int &t)const; //大數和一個int型別的變數的大小比較 bool operator>(const __int64 &t)const; //大數和一個__int64型別的變數的大小比較 void print(); //輸出大數 }; BigNum::BigNum(const int b) //將一個int型別的變數轉化為大數 { int c,d=b; len=0; memset(a,0,sizeof(a)); while(d>MAXN) //每位存四位數字,MAXN=9999; { c=d-(d/(MAXN+1))*(MAXN+1); d=d/(MAXN+1); a[len++]=c; } a[len++]=d; } BigNum::BigNum(const __int64 b) //將一個int型別的變數轉化為大數 { __int64 c,d=b; len=0; memset(a,0,sizeof(a)); while(d>MAXN) //每位存四位數字,MAXN=9999; { c=d-(d/(MAXN+1))*(MAXN+1); d=d/(MAXN+1); a[len++]=c; } a[len++]=d; } BigNum::BigNum(const char *s) //將一個字串型別的變數轉化為大數 { 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) //拷貝建構函式 { int i; memset(a,0,sizeof(a)); for(i=0;i<len;i++) a[i]=T.a[i]; } BigNum & BigNum::operator=(const BigNum &n) //過載賦值運算子,大數之間賦值運算 { 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) { 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) //過載輸出運算子 { int i; cout<<b.a[b.len-1]; for(i=b.len-2;i>=0;i--) { printf("%04d",b.a[i]); } return out; } BigNum BigNum::operator+(const BigNum &T)const //兩個大數之間的相加運算 { BigNum t(*this); int i,big; 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 //兩個大數之間的相減運算 { 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 //兩個大數之間的相乘 { 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 //大數對一個整數進行相除運算 { 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; } BigNum BigNum::operator/(const __int64 &b)const //大數對一個整數進行相除運算 { BigNum ret; __int64 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 //大數對一個 int型別的變數進行取模 { int i,d=0; for(i=len-1;i>=0;i--) d=((d*(MAXN+1))%b+a[i])%b; return d; } __int64 BigNum::operator%(const __int64 &b)const //大數對一個 64位整數型別的變數進行取模 { __int64 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 //大數的n次方運算 { 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 //大數和另一個大數的大小比較 { 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 //大數和一個int型別的變數的大小比較 { BigNum b(t); return *this>b; } void BigNum::print() //輸出大數 { int i; printf("%d",a[len-1]); for(i=len-2;i>=0;i--) printf("%04d",a[i]); printf("\n"); } int main() { //freopen("in.txt","r",stdin); //freopen("out.txt","w",stdout); int m,n; int T; scanf("%d",&T); while(T--) { scanf("%d%d",&m,&n); BigNum tt = 1; //cin>>t; //大整數讀入 for(int i = 1;i < n;i++) tt = tt*m; int tmp = n; for(int i = 2;i <= n;i++) { while( tmp%i == 0 && (tt%i == 0) ) { tmp /= i; tt = tt/i; } } printf("%d/",tmp); tt.print(); //大整數輸出 } return 0; }
Cut the Cake
Time Limit: 2000/1000 MS (Java/Others)Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 821Accepted Submission(s): 396
For each case, two integers M, N indicate the number of her friends and the number of strawberry.
(2 < M, N <= 20, T <= 400) Output As the probability could be very small, you should output the probability in the form of a fraction in lowest terms. For each case, output the probability in a single line. Please see the sample for more details. Sample Input 2 3 3 3 4 Sample Output 1/3 4/27