标签:期望 str import highlight -- continue while sub top
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; } }
标签:期望 str import highlight -- continue while sub top
原文地址:http://www.cnblogs.com/jianglangcaijin/p/7436366.html