算法题3

题目：

Problem Statement

 

***Note:  Please keep programs under 7000 characters in length.  Thank youClass Name: SquareDigitsMethod Name: smallestResultParameters: intReturns: intDefine the function S(x) as the sum of the squares of the digits of x.   For example: S(3)=3*3=9 and S(230)=2*2+3*3+0*0=13.Define the set T(x) to be the set of unique numbers that are produced byrepeatedly applying S to x.  That is: S(x), S(S(x)), S(S(S(x))), etc... For example, repeatedly applying S to 37:S(37)=3*3+7*7=58.  S(58)=5*5+8*8=89.S(89)=145.S(145)=42. S(42)=20.S(20)=4.S(4)=16.S(16)=37. Note this sequence will repeat so we can stop calculating now and: T(37)={58,89,145,42,20,4,16,37}.However, note T(x) may not necessarily contain x.  Implement a class SquareDigits, which contains a method smallestResult.  Themethod takes an int, n, as a parameter and returns the smallest int, x, suchthat T(x) contains n.The method signature is (be sure your method is public): int smallestResult(int n); TopCoder will ensure n is non-negative and is between 0 and 199 inclusive.Examples:If n=0: S(0) = 0, so T(0)={0}, so the method should return 0.If n=2: T(0) through T(10) do not contain the value 2.  If x=11, however:S(11)=1*1+1*1=2, so T(11) contains 2, and the method should return 11.If n=10: T(0) through T(6) do not contain 10.  If x=7:S(7)=49.S(49)=97.S(97)=130.S(130)=10.S(10)=1.and it starts to repeat... so T(7) is {49,97,130,10,1}, which contains 10, and the method should return 7.n=1 -> x=1 n=19 -> x=133n=85 -> x=5n=112 -> x=2666

Definition

 Class:SquareDigitsMethod:smallestResultParameters:intReturns:intMethod signature:int smallestResult(int param0)(be sure your method is public)  

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我的解法：

#include<iostream>class SquareDigits{public:  SquareDigits(){};  ~SquareDigits(){};  int smallestResult(int param0){    int i;     for( i = 0; ;i++){       int iNumber = setList(i);       if( isIn(param0,list,iNumber)){         break;       }          }     return i;  };private:  int bits[4];  int list[200];  int setList(int target){    int _a = s(target);    int i = 0;    list[i] = _a;    while(!isIn(_a = s(_a),list,++i)){      list[i] = _a;    }    return i;  };    int split(int target){    bits[0] = target / 1000;    target = target % 1000;    bits[1] = target / 100;    target = target % 100;    bits[2] =target / 10;    target = target % 10;    bits[3] = target;    return 0;  };    int s(int target){    split(target);    return bits[0] * bits[0] + bits[1] * bits[1] + bits[2] * bits[2] + bits[3] * bits[3];  };    bool isIn(int target,int * start,int length){    bool in = false;    int i;    for(i = 0; i < length;i++){      if(target == *(start + i)){       in = true;        break;        }    }    if(i == length )       in = false;    return in;  }; };