topcoder srm 335 div1

problem1 link

直接模拟即可。

import java.util.*;
import java.math.*;
import static java.lang.Math.*;

public class Multifactorial {
	
	public String calcMultiFact(int n, int k) {
		long result=1;
		final long nlimit=1000000000000000000l;
		while(true) {

			if(result>nlimit/n) {
				return "overflow";
			}
			result*=n;
			if(n<=k) {
				break;
			}
			n-=k;
		}
		return Long.toString(result);
	}
}

problem2 link

记录到达$(x,y)$的步数以及当前新一步的和，dp即可。

import java.util.*;
import java.math.*;
import static java.lang.Math.*;

public class ExpensiveTravel {

	static class Fraction {
		public int a,b;
		public Fraction() {
			a=1;
			b=1;
		}
		public Fraction(int a,int b) {
			this.a=a;
			this.b=b;
		}


		public static int gcd(int x,int y) {
			if(y==0) {
				return x;
			}
			return gcd(y,x%y);
		}

		private Fraction simple() {
			int t=gcd(a,b);
			a/=t;
			b/=t;
			return this;
		}

		public Fraction add(Fraction p) {
			int bb=p.b*b;
			int aa=a*p.b+p.a*b;
			return new Fraction(aa,bb).simple();
		}
		public boolean ok() {
			return a>b;
		}

		public boolean less(Fraction p) {
			return a*p.b<p.a*b;
		}
	}

	static class Node {
		public Fraction pref;
		public Fraction f;
		public int cost;
		public boolean inq;

		public Node() {
			f=new Fraction(0,1);
			cost=0;
			inq=false;
		}

		Node add(int t) {
			Node p=new Node();
			p.f=new Fraction(f.a,f.b).add(new Fraction(1,t));
			p.cost=cost;
			p.inq=inq;

			if(p.f.ok()) {
				++p.cost;
				p.f=pref.add(new Fraction(1,t));
			}
			p.pref=new Fraction(1,t);
			return p;
		}

		public boolean less(Node p) {
			return cost<p.cost||cost==p.cost&&f.less(p.f);
		}

		public int result() {
			if(cost==-1) {
				return -1;
			}
			return cost+1;
		}

	}
	
	public int minTime(String[] m, int startRow, int startCol, int endRow, int endCol) {
		final int N=m.length;
		final int M=m[0].length();
		int[][] g=new int[N][M];
		for(int i=0;i<N;++i) {
			for(int j=0;j<M;++j) {
				char c=m[i].charAt(j);
				g[i][j]=c-'0';
			}
		}
		--startRow;
		--startCol;
		--endRow;
		--endCol;
		if(g[startRow][startCol]==1||g[endRow][endCol]==1) {
			return -1;
		}

		Node[][] f=new Node[N][];
		for(int i=0;i<N;++i) {
			f[i]=new Node[M];
			for(int j=0;j<M;++j) {
				f[i][j]=new Node();
				f[i][j].cost=-1;
			}
		}

		Queue<Integer> queue=new LinkedList<>();
		f[startRow][startCol].f=new Fraction(1,g[startRow][startCol]);
		f[startRow][startCol].pref=new Fraction(1,g[startRow][startCol]);
		f[startRow][startCol].inq=true;
		f[startRow][startCol].cost=0;
		queue.offer(startRow*100+startCol);

		final int[] dx={0,0,1,-1};
		final int[] dy={1,-1,0,0};

		while(!queue.isEmpty()) {
			final int x=queue.peek()/100;
			final int y=queue.peek()%100;
			queue.poll();
			f[x][y].inq=false;
			if(x==endRow&&y==endCol) {
				continue;
			}
			for(int i=0;i<4;++i) {
				final int xx=x+dx[i];
				final int yy=y+dy[i];
				if(xx<0||xx>=N||yy<0||yy>=M) {
					continue;
				}
				if(g[xx][yy]==1) {
					continue;
				}
				Node t=f[x][y].add(g[xx][yy]);
				if(f[xx][yy].cost==-1||t.less(f[xx][yy])) {
					f[xx][yy]=t;
					if(!f[xx][yy].inq) {
						f[xx][yy].inq=true;
						queue.offer(xx*100+yy);
					}
				}
			}
		}
		return f[endRow][endCol].result();
	}
}

problem3 link

根据期望的可加性，A组中每个数$x$比B组中每个小于$x$的值$y$的贡献值$\frac{(x-y)^{2}}{n}$为正，对于每个大于$x$的值$z$的贡献值$\frac{(x-z)^{2}}{n}$为负。

import java.util.*;
import java.math.*;
import static java.lang.Math.*;

public class RandomFights {


	int[] get(int[] X,int n) {
		final int m=X.length;
		int j=0;
		int[] R=new int[n];
		for(int i=0;i<n;++i) {
			R[i]=X[j];
			int s=(j+1)%m;
			X[j]=((X[j]^X[s])+13)%49999;
			j=s;
		}
		return R;
	}


	BigInteger int2big(long x) {
		return new BigInteger(Long.toString(x));
	}


	public double expectedNrOfPoints(int[] A,int[] B,int n) {
		int[] a=get(A,n);
		int[] b=get(B,n);

		Arrays.sort(a);
		Arrays.sort(b);


		BigInteger nxt=BigInteger.ZERO,nxt2=BigInteger.ZERO;
		for(int i=0;i<n;++i) {
			nxt=nxt.add(int2big(b[i]));
			nxt2=nxt2.add(int2big((long)b[i]*b[i]));
		}

		BigInteger result=BigInteger.ZERO;
		BigInteger pre=BigInteger.ZERO,pre2=BigInteger.ZERO;
		int k=0;
		for(int i=0;i<n;++i) {
			while(k<n&&b[k]<=a[i]) {
				pre=pre.add(int2big(b[k]));
				pre2=pre2.add(int2big((long)b[k]*b[k]));
				nxt=nxt.subtract(int2big(b[k]));
				nxt2=nxt2.subtract(int2big((long)b[k]*b[k]));
				++k;
			}

			BigInteger tmp=int2big((long)k*a[i]*a[i]).subtract(pre.multiply(int2big(a[i]*2))).add(pre2);
			result=result.add(tmp);
			tmp=int2big((long)(n-k)*a[i]*a[i]).subtract(nxt.multiply(int2big(a[i]*2))).add(nxt2);
			result=result.subtract(tmp);

		}
		BigInteger[] last=result.divideAndRemainder(int2big(n));
		return Double.valueOf(last[0].toString())+Double.valueOf(last[1].toString())/n;
	}
}