[Huawei Machine test real problem] [2014]63. Equation transformation

Topic

输入一个正整数X，在下面的等式左边的数字之间添加+号或者-号，使得等式成立。1 2 3 4 5 6 7 8 9 = X比如：12-34+5-67+89 = 51+23+4-5+6-7-8-9 = 5请编写程序，统计满足输入整数的所有整数个数。输入：       正整数，等式右边的数字输出：       使该等式成立的个数样例输入：5样例输出：21

Code

/* ---------------------------------------* Date: 2015-07-06* sjf0115* title: Equation Transformation * Source: Huawei Machine Test real problem------------------------- ----------------*/#include <iostream>#include <string>#include <vector>#include <stack>#include <algorithm>using namespace STD;//int converted to stringstringINT2STR (intNUM) {stringStrif(num = =0) {str =' 0 ';returnStr }//if    intTMP = num; while(num) {Str.insert (Str.begin (), num%Ten+' 0 '); Num/=Ten; }//while    returnSTR;}/ * The number of equations that the current computed value result conforms to count equation result x number of neighbors merged result sequence 1 + 2 + 345 345 Is sequence * *   voidHelper vector<int>&num,intIndexintXintResultintSequenceint&count,stringOP) {if(Index = = num.size ()) {if(Result + sequence = = X | | result-sequence = = x) {++count;if(Result + sequence = = x) {OP + ="+"+INT2STR (sequence); }//if            Else{OP + ="-"+INT2STR (sequence); }//else            cout<<op<<endl; }//if        return; }//if    //Continuous numberHelper (num,index+1, X,result,sequence *Ten+ Num[index],count,op);//addition +    if(Op.size () >0){//With Num[index] as the starting point for sequenceHelper (num,index+1, X,result + sequence,num[index],count,op+"+"+INT2STR (sequence)); }//if    Else{//With Num[index] as the starting point for sequenceHelper (num,index+1, X,result + sequence,num[index],count,op+int2str (sequence)); }//else    if(Op.size () >0){//Subtraction-        //With Num[index] as the starting point for sequenceHelper (num,index+1, x,result-sequence,num[index],count,op+"-"+INT2STR (sequence)); }//if}intTransformationequation ( vector<int>Numintx) {intCount =0;stringOP =""; Helper (num,1X0, num[0],COUNT,OP);returnCount;}intMain () {intN,x;intNum//freopen ("c:\\users\\administrator\\desktop\\acm.in", "R", stdin);     while(Cin&GT;&GT;N&GT;&GT;X) { vector<int>Vec for(inti =0; i < n;++i) {Cin>>num;        Vec.push_back (num); }//for        cout<<transformationequation (vec,x) <<endl; }//while    return 0;}

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[Huawei Machine test real problem] [2014]63. Equation transformation