http://codeforces.com/gym/101933/problem/I
I. Intergalactic Bidding
time limit per test
2.0 s
memory limit per test
256 MB
input
standard input
output
standard output
Today the Intergalactic Council of Pebble Coins (ICPC) conducted an intergalactic auction of the Neutronium Chaos Pebble Coin (NCPC). This coin, which was forged in the Ancient Coin Machine (ACM), is rumored to be the key to ruling the universe.
Due to the extremely competitive nature of the auction, as well as the odd mechanics of the intergalactic currency used (far too advanced for mere mortals to understand), the auction was conducted with the following rules:
- only one participant was allowed to make a bid at a time,
- each participant was only allowed to make one bid, and
- a participant making a bid had to bid at least twice the amount of the highest bid at the time.
The first participant making a bid was allowed to make a bid of any positive amount.
After the auction there were a lot of sore losers – understandably, having just lost their chance at world domination. To make the losers feel a little better and prevent possible rioting, the ICPC has decided to hold a lottery for the participants. The winners of the lottery are determined as follows. The ICPC picks a random number ss. A group of participants is called winning if the sum of their bets from the auction is equal to ss. A participant wins the lottery and receives a prize – a shiny Pebble Coin – if they belong to any winning group of participants.
Given the names of the participants, the bets that they made, and the random number ss chosen by the ICPC, help them determine which participants won the lottery.
Input
The first line of input contains two integers nn and ss, where 1≤n≤10001≤n≤1000 is the number of participants, and 1≤s<1010001≤s<101000 is the random number chosen by the ICPC.
Then follow nn lines describing the participants. Each line contains a string tt and an integer bb, where tt is the name of a participant, and 1≤b<1010001≤b<101000 is the amount of his bet. The name of each participant is unique and consists of between 11 and 2020 letters from the English alphabet.
Output
Output an integer kk denoting the number of participants that won the lottery. Then output kk lines containing the names of the participants that won the lottery, one per line, in any order.
Examples
input
Copy
5 63 Vader 3 Voldemort 7 BorgQueen 20 Terminator 40 Megatron 101
output
Copy
3 Terminator BorgQueen Vader
input
Copy
4 1112 Blorg 10 Glorg 1000 Klorg 1 Zlorg 100
output
Copy
0
把所有的數從大到小排序，因爲每一個數是前一個的數的兩倍，所以從大到小要一直加
如果沒有大於等於隨機數，就永遠不會等於隨機數了，因爲數是前面的二分之一
套一個大整數模板
#include<bits/stdc++.h>
#define MAXN 9999
#define M 1000//
#define DLEN 4
using namespace std;
class Bint
{
private:
int a[M];
int len;
public:
Bint()//Construction function
{
len=1;
memset(a,0,sizeof(a));
}
Bint(const int);
Bint(const char*);
Bint(const Bint &);
Bint(const std::string s);
friend istream & operator>>(istream&, Bint&);//Overloaded IO stream
friend ostream & operator<<(ostream&, Bint&);
Bint &operator=(const Bint &);//Overloaded operator
Bint operator+(const Bint &) const;
Bint operator-(const Bint &) const;
Bint operator*(const Bint &) const;
Bint operator/(const int &) const;
Bint operator^(const int &) const;
int operator%(const int &) const;
bool operator<(const Bint& b)const
{
if(len!=b.len)return len<b.len;
for (int i=len-1; i>=0; --i)if(a[i]!=b.a[i])return a[i]<b.a[i];
return false;
}
bool operator>(const Bint& b)const
{
return b<*this;
}
bool operator<=(const Bint& b)const
{
return !(b<*this);
}
bool operator>=(const Bint& b)const
{
return !(*this<b);
}
bool operator!=(const Bint& b)const
{
return b<*this||*this<b;
}
bool operator==(const Bint& b)const
{
return !(b<*this)&&!(*this<b);
}
void print()//output
{
printf("%d",a[len-1]);
for(int i=len-2; i>=0; --i)printf("%04d",a[i]);
putchar('\n');
};
};
Bint::Bint(const int b)
{
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;
}
Bint::Bint(const char*s)
{
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;
}
}
Bint::Bint(const Bint & T):len(T.len) //拷貝構造函數
{
memset(a,0,sizeof(a));
for(int i=0; i<len; ++i) a[i]=T.a[i];
}
Bint::Bint(const std::string s)
{
int t,k,index,l,i;
memset(a,0,sizeof(a));
l=s.length();
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;
}
}
Bint & Bint::operator=(const Bint & n) //重載賦值運算符，大數之間進行賦值運算
{
len=n.len;
memset(a,0,sizeof(a));
for(int i=0; i<len; ++i)a[i]=n.a[i];
return *this;
}
istream& operator>>(istream & in,Bint & b)
{
char ch[M*4];
in>>ch;
int l=strlen(ch),count=0,sum=0,i=-1;
for(int 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,Bint& b) //重載輸出運算符
{
cout<<b.a[b.len-1];
for(int i=b.len-2; i>=0; --i)
{
cout.width(DLEN);
cout.fill('0');
cout<<b.a[i];
}
return out;
}
Bint Bint::operator+(const Bint & T) const
{
Bint t(*this);
int i,big; //位數
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;
}
Bint Bint::operator-(const Bint & T) const
{
int i,j,big;
bool flag;
Bint 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[t1.len-1]==0&&t1.len>1)--t1.len,--big;
if(flag)t1.a[big-1]=0-t1.a[big-1];
return t1;
}
Bint Bint::operator*(const Bint & T) const
{
Bint 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;
}
Bint Bint::operator/(const int & b) const
{
Bint 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 Bint::operator %(const int & b) const
{
int i,d=0;
for (i=len-1; i>=0; --i)d=((d*(MAXN+1))%b+a[i])%b;
return d;
}
Bint Bint::operator^(const int & n) const
{
Bint t,ret(1);
if(n<0)exit(-1);
if(n==0)return 1;
if(n==1)return *this;
int m=n,i;
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;
}
struct node
{
string name;
Bint coin;
bool operator < (const node &t)
{
return t.coin < coin;
}
}a[1005];
int ans[1005], n, k;
Bint zero,sum;
int main()
{
cin >> n >> sum;
for(int i = 1; i <= n; i++)
cin >> a[i].name >> a[i].coin;
sort(a+1, a+n+1);
for(int i = 1; i <= n ; i++)
if(a[i].coin <= sum)
{
sum = sum - a[i].coin;
ans[k++] = i;
}
if(sum == zero)
{
cout << k << endl;
for(int i = 0; i < k; i++)
cout << a[ans[i]].name << endl;
}
else
cout << '0' << endl;
return 0;
}