1.有两个用字符串表示的非常大的大整数,算出他们的乘积，也是用字符串表示。不能用系统自带的大整数类型。

解析：大数相乘，模拟一下。

package pdd.nt;import java.util.ArrayList;import java.util.List;import java.util.Scanner;/** * Created by cuijunyong on 2018/3/20. */public class Mul {  public static void main(String[] args) {    Scanner cin = new Scanner(System.in);    String a = cin.next();    String b = cin.next();    String c = mul(a, b);    System.out.println(c);  }  private static String mul(String x, String y) {    char[] a = x.toCharArray();    char[] b = y.toCharArray();    List
     
       c = new ArrayList();    for(int i = a.length - 1; i >= 0; i--){      long v = 0;      for(int j = b.length - 1; j >= 0; j--){        v += (a[i] - '0') * (b[j] - '0');        int pos = b.length - 1 - j + (a.length - 1 - i);        if(c.size() <= pos){          c.add(v % 10);        }else{          v += c.get(pos);          c.set(pos, v % 10);        }        v /= 10;      }      while (v > 0){        c.add(v % 10);        v /= 10;      }    }    String str = "";    for(int i = c.size() - 1; i >= 0; i-- ){      str += c.get(i);    }    return str;  }}
     

2.六一儿童节，老师带了很多好吃的巧克力到幼儿园。每块巧克力j的重量为w[j]，对于每个小朋友i，当他分到的巧克力大小达到h[i] (即w[j]>=h[i])，他才会上去表演节目。老师的目标是将巧克力分发给孩子们，使得最多的小孩上台表演。可以保证每个w[i]> 0且不能将多块巧克力分给一个孩子或将一块分给多个孩子。

解析：一看就先想到了二分匹配，看了解析看到他们用的排序模拟也做出来了。

package pdd.nt;import java.util.Scanner;/** * Created by cuijunyong on 2018/3/20. */public class day61 {  private static Scanner cin = new Scanner(System.in);  private static int n, m;  private static int[] h,w;  private static int[][] mp;  private static int[] d;  private static int[] v;  public static void main(String[] args) {    init();    int ans = 0;    for(int i = 0; i < n; i++){      for(int j = 0; j < m; j++){        v[j] = 0;      }      if(dfs(i)){        ans ++;      }    }    System.out.println(ans);  }  private static boolean dfs(int x){    for(int i = 0; i < m; i++){      if(v[i] == 0 && mp[x][i] == 1){        v[i] = 1;        if(d[i] == -1 || dfs(d[i])){          d[i] = x;          return true;        }      }    }    return false;  }  private static void init(){    n = cin.nextInt();    h = new int[n];    for(int i = 0; i < n; i++){      h[i] = cin.nextInt();    }    m = cin.nextInt();    w = new int[m];    for(int i = 0; i < m; i++){      w[i] = cin.nextInt();    }    d = new int[m];    for(int i = 0; i < m; i++){      d[i] = -1;    }    v = new int[m];    for(int i = 0; i < m; i++){      v[i] = 0;    }    mp = new int[n][m];    for(int i = 0; i < n; i++){      for(int j = 0; j < m; j++){        if(h[i] <= w[j]){          mp[i][j] = 1;        }      }    }  }}

 3.将1+3+3+4变成1+3*2+4

package pdd.cz;/** * Created by cuijunyong on 2018/3/20. */import java.util.*;public class A {  private static Map
     
       map = new HashMap
      
       ();  private static Set
       
         set = new HashSet
        
         ();  public static void main(String[] args) {    Scanner cin = new Scanner(System.in);    String str = cin.next();    init(str);  }  private static void init(String str) {    map.clear();    set.clear();    Long x = 0L;    Long sum = 0L;    str += "+";    for(int i = 0; i < str.length(); i++){      char c = str.charAt(i);      if(c >= '0' && c <= '9'){        x = x * 10 + c - '0';      }else{        Integer v = map.get(x);        if(v == null){          v = 0;        }        v++;        map.put(x, v);        set.add(x);        sum += x;        x = 0L;      }    }    Object[] objects = set.toArray();    long[] arr = new long[objects.length];    for(int i = 0; i < objects.length; i++){      arr[i] = (Long)objects[i];    }    Arrays.sort(arr);    String ans = "";    for(int i = 0; i < arr.length; i++){      Long next = arr[i];      if(i != 0){        ans += "+";      }      ans += next + "*" + map.get(next);    }    System.out.println(ans);    System.out.println(sum);  }}
        
       
      
     

4.坐出租车，每辆最多坐4个人，要求一队不能分开。

package pdd.cz;/** * Created by cuijunyong on 2018/3/20. *//*61 2 4 3 3 2*/import java.util.*;public class B {  public static void main(String[] args) {    Scanner cin = new Scanner(System.in);    int N;    N = cin.nextInt();    int[] p = new int[N];    int[] num = new int[5];    for(int i = 0; i < 5; i++){      num[i] = 0;    }    for(int i = 0; i < N; i++){      p[i] = cin.nextInt();      num[p[i]]++;    }    int ans;    ans = num[4] + (num[2] + 1)/2;    if(num[3] > num[1]){      ans += num[1] + num[3] - num[1];    }else{      ans += num[3] + (num[1] - num[3] + 3)/4;    }    if(num[2] % 2 == 1 && (num[1] - num[3])%4 <= 2 && (num[1] - num[3])%4 > 0){      ans --;    }    System.out.println(ans);  }}