# Monthly Archives: June 2012

## Poj Solution 2195

http://poj.org/problem?id=2195 #include <vector> using namespace std; typedef int type; //�������� const int size = 210; //ͼ�Ķ���ģ const type MAX_FEE = 10000000; //�������� const int MAX = 1000; //������� class minfee_flow { public: //ɾ�����б� void clear(); //���һ���from��to ����Ϊc ��λ������Ϊw �ı� void … Continue reading

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## Poj Solution 2193

http://poj.org/problem?id=2193 //* @author: import java.util.*; public class Main { public static void main(String[] args){ Scanner sc=new Scanner(System.in); long dp[][]=new long; int nn=sc.nextInt(); for (int i=1;i<=2000;i++) dp[i]=1; for (int i=2;i<=10;i++) for (int j=i;j<=2000;j++) for (int k=1;k<=j/2;k++) dp[i][j]+=dp[i-1][k]; for (int ii=1;ii<=nn;ii++) { … Continue reading

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## Poj Solution 2192

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## Poj Solution 2191

http://poj.org/problem?id=2191 //* @author popop0p0popo import java.util.*; import java.io.*; public class Main{ public static void main(String[] args){ String[] store={"23 * 89 = 2047 = ( 2 ^ 11 ) – 1", "47 * 178481 = 8388607 = ( 2 ^ 23 … Continue reading

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## Poj Solution 2190

http://poj.org/problem?id=2190 //* @author: 82638882@163.com import java.util.Scanner; public class Main { public static void main(String[] args) { Scanner in=new Scanner(System.in); String a=in.next(); int t=0; int n=-1; for(int i=0;i< 10;i++) { char c=a.charAt(i); if(c!=’?’&&c!=’X’){ t+=(c-‘0′)*(10-i); } else if(c==’X’) t+=10; else if(c==’?’) n=10-i; … Continue reading

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## Poj Solution 2189

http://poj.org/problem?id=2189 #include <stdio.h> #include <algorithm> #include <memory.h> using namespace std; int n,p,c; int s; int main() { int i, j, k; scanf( "%d %d %d", &n, &p, &c ); memset( s, 0, sizeof( s ) ); for( i=0; i<n; i++ … Continue reading

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## Poj Solution 2188

http://poj.org/problem?id=2188 //* @author: 82638882@163.com import java.io.*; public class Main{ static long ans=0; public static void main(String[] args) throws NumberFormatException, IOException { InputStreamReader is=new InputStreamReader(System.in); BufferedReader in=new BufferedReader(is); int a=Integer.parseInt(in.readLine()); ans=0; int[] arr=new int[a]; int[] aww=new int[a]; for(int i=0;i< a;i++) { … Continue reading

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## Poj Solution 2187

http://poj.org/problem?id=2187 #include"stdio.h" #include"math.h" #include<vector> #include<algorithm> using namespace std; //////////////////////////////// typedef int Type;/*????????*/ //////////////////////////////// //???????? struct point { Type x,y; point(){x=y=0;} point(Type x,Type y):x(x),y(y){;} bool operator==(point &a){return x==a.x&&y==a.y;} }; //???? inline Type cheng(point a,point b,point c) {return (b.x-a.x)*(c.y-a.y)-(c.x-a.x)*(b.y-a.y);} inline Type cheng(point … Continue reading

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## Poj Solution 2186

http://poj.org/problem?id=2186 //PKU_2186_Popular Cows_ǿlͨ��_���(kosaraju) #include <iostream> using namespace std; struct Point { int Index; Point *Next; Point(int index = 0) { Index = index; Next = NULL; } }; struct HeadPoint { Point *List; Point *LastPoint; HeadPoint() { List = NULL; … Continue reading

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## Poj Solution 2185

http://poj.org/problem?id=2185 #include <stdio.h> char map; int n,m; bool sign; int main() { int i, j, n, m, k, h, l; bool key; scanf( "%d %d", &n, &m ); for( i=0; i<n; i++ ) scanf( "%s", map[i] ); for( k=1; k<m; … Continue reading

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## Poj Solution 2184

http://poj.org/problem?id=2184 #include <stdio.h> #include <memory.h> #include <algorithm> #include <functional> using namespace std; bool mem,*sign = mem+100000; int mem2,*ans = mem2+100000; typedef pair<int,int> node; node c; int n; node mem3,*s = mem3, *t = mem3; int sn,tn; int main() { int … Continue reading

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## Poj Solution 2183

http://poj.org/problem?id=2183 //* @author: <strong> import java.util.*; public class Main { private int t; private int b[]; public Main(int t){ this.t=t; b=new int; } private void doIt(){ int st=0; while(true){ t/=10; t%=10000;//ȡt���м���λ t*=t;//ƽ�� t%=1000000;//ȡt�ĺ���λ if (b[t]!=0){//���t���ֹ�,����ѭ���� st++; System.out.println(t+" "+(st-b[t])+" "+st);//ѭ���ڵĵ�һ����ѭ���ڴ�С����Ҫ���η���ѭ���� break; … Continue reading

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