Xingkong vs C++

SecretSum 是本次 google 竞赛中第二轮淘汰赛的一道分值为 500 分竞赛题。事实上，这道题目反而比同轮比赛中的那道 1000 分值的RecurringNumbers 难(RecurringNumbers 的难度水准充其量不过是道初一学生奥数竞赛题)。好了，闲话少叙，来看 SecretSum 的题目吧:

一、竞赛题目

Problem Statement

We can substitute each digit of a number with a unique letter from ''A'' to ''Z''. This way we can write groups of numbers in a single representation. For example "ABA" can represent any of the following: 101, 151, 343, 767, 929. However, "ABA" cannot mean 555 because each letter must stand for a distinct digit. Furthermore, numbers cannot begin with 0 and thus ''A'' cannot be replaced by 0. Given two such representations num1 and num2 and the result of their summation return the total number of possible combinations of numbers for which the equation holds. If no combinations are possible then return 0.

Definition

Class: SecretSum
Method: countPossible
Parameters: String, String, String
Returns: int
Method signature: int countPossible(String num1, String num2, String result)
(be sure your method is public)

Constraints
- num1, num2, and result will each contain exactly 3 uppercase letters (''A'' - ''Z'').

Examples

0)

"AAA"
"BBB"
"CCC"
Returns: 32

1)

"ABB"
"DEE"
"TTT"
Returns: 112

2)

"ABC"
"ABA"
"ACC"
Returns: 0
Leading zeroes are not allowed.

3)

"AAA"
"CDD"
"BAA"
Returns: 32

4)

"TEF"
"FET"
"AAA"
Returns: 12

5)

"ABC"
"ABC"
"BCE"
Returns: 5
We can have the following 5 sums:
124 + 124 = 248
125 + 125 = 250
249 + 249 = 498
374 + 374 = 748
375 + 375 = 750


6)


"AAA"
"AAA"
"BBB"
Returns: 4
We can have the following 4 sums:
111 + 111 = 222
222 + 222 = 444
333 + 333 = 666
444 + 444 = 888


This problem statement is the exclusive and proprietary property of TopCoder, Inc. Any unauthorized use or reproduction of this information without the prior written consent of TopCoder, Inc. is strictly prohibited. (c)2003, TopCoder, Inc. All rights reserved.

题目来源:SRM250 DIV2
级别:500Points
题目:
Problem Statement:
Problem Statement
    
For computers it can be hard to determine in which language a given text is written. A simple way to try to determine the language is the following: for the given text and for some sample texts, for which we know the languages, we determine the letter frequencies and compare these.
The frequency of a letter is the total number of occurrences of that letter divided by the total number of letters in the text. To determine this, we ignore case and non-letter characters.
Once the letter frequencies of the text and of a language are known, we can calculate the difference between the two. This difference we define by the sum of the squared differences of the frequencies:
 
The lesser this value, the closer text resembles that language. Compare text with each element of languages and return the (0-based) index of the language that has the smallest difference with text. In case of a tie, return the smallest index.
Definition
    
Class:
LanguageRecognition
Method:
whichLanguage
Parameters:
vector <string>, string
Returns:
int
Method signature:
int whichLanguage(vector <string> languages, string text)
(be sure your method is public)
    

Constraints
-
languages contains between 1 and 50 elements, inclusive.
-
Each element of languages has length between 1 and 50, inclusive.
-
text has length between 1 and 50, inclusive.
-
Each element of languages and text consists only of characters with ASCII value between 32 and 127, inclusive.
-
Each element of languages and text contains at least one letter ('A'-'Z' and 'a'-'z').
Examples
0)

    
{"This is an English sentence.",
 "Dieser ist ein Deutscher Satz.",
 "C'est une phrase Francaise.",
 "Dit is een Nederlandse zin."
}
"In welke taal is deze zin geschreven?"
Returns: 3
The differences are 0.0385, 0.0377, 0.0430 and 0.0276, so the sentence is written in language 3, Dutch. Note that Dutch is somewhat similar to German, somewhat less similar to English and not similar to French.
1)

    
{"aaaaa","bbbb","ccc","dd","e"}
"xxx"
Returns: 0
In case of a tie, return the language with the smallest index.
2)

    
{"AABB","AaBb","A? B!","ab!@#$%"}
"ab"
Returns: 0
Ignore case and the non-letter characters.

//优化的冒泡排序
//编译成功

#include<iostream.h>
void swap(int a,int b);
int main()
{
 
 int a[10],i,j;
 for( i=0;i<10;i++)
 {
  cout<<"enter";
  cin>>a[i];
 }
 cout<<"数组的原次序为：";
 for( i=0;i<10;i++)
  cout<<a[i];
 for( i=0;i<9;i++)
 {
  int temp=0;
  for( j=8;j>i;j--)
  {
   if(a[j]<a[j-1])
   {
    swap(a[j],a[j-1]);
    temp=1;
   }
  }
  if(temp==0)
   break;
 }
cout<<"排序后次序为";
for(i=0;i<10;i++)
cout<<a[i];
 return 0;
}
void swap(int a,int b)
{
 int temp;
 temp=a;
 a=b;
 b=temp;
}

//给出kruskal球联通网络的最小生成树的算法实现
//调试成功
#include<iostream>
#include<stack>
using namespace std;
int main()
{
 stack<int> travstack;
 int i,j,k,n=5;
 cout<<"以矩阵形式表示"<<endl;
 int a[5][5],b[5][5],c[5][5],d[5];
 cout<<"请输入各路径权值";
 for( i=0;i<n;i++)
    for(j=i+1;j<n;j++)
 {
        cout<<"a[i][j]="<<endl;
        cin>>a[i][j];
 }
 int biggest=0,big=1000;
 for(k=0;k<n*n-n;k++)
     for(i=0;i<n;i++)
        for(j=i+1;j<n;j++)
  {
            if(a[i][j]>biggest&&a[i][j]<big)
   {
                 biggest=a[i][j];
                 travstack.push(i);
                 travstack.push(j);
   }
           big=a[i][j];
  }
 for(i=0;i<n;i++)
     for(j=0;j<n;j++)
  {
         int p1,p2;
         p1=travstack.top();//p1=j
         travstack.pop();
         p2=travstack.top();//p2=i
         travstack.pop();
         if(b[p1][p2]==0)
   {
            b[p1][p2]=1;
            b[p2][p1]=1;
            c[p2][p1]=a[p2][p1];
            if(d[p1]==0&&d[p2]==0)
              d[p1]=d[p2]=1;
  
            else if(d[p1]==0)
   {
              d[p1]=1;
              for( i=0;i<n;i++)
     {
                 if(b[p2][i]==1)
     {
                    b[p1][i]=1;
                    b[i][p1]=1;
     }
     }
   }
           else if(d[p2]==0)
     {
              d[p2]=1;
              for(i=0;i<n;i++)
     {
                 if(b[p1][i]==1)
     {
                    b[p2][i]=1;
                    b[i][p2]=1;
     }
     }
     }
   }
  }
  for(i=0;i<n;i++)
  {
      for(j=i+1;j<n;j++)
      cout<<"c[i][j]="<<c[i][j];
      cout<<endl;
      for(int k=0;k<n;k++)
      cout<<"*";
 }
 return 0;
}