整数对

 

整数对

考虑一个数A(首位不为0)，把它去掉一个数字以后得到另外一个数B。给出A与B的和N（1<=N<=10^9）编程计算出所有可能的数A，例如A＋B＝34，则可能是31或者27。

 

#include <iostream>#include <string>#include <vector>#include <iterator>using namespace std;int cnt = 0;int GenAfterVec(vector<int> SouVec, int del, const vector<int>& BefDelVec){if (BefDelVec.size() > 0 && BefDelVec[0] >= 5){int idx = del;while (idx > -1){if (--SouVec[idx] < 0) SouVec[idx] = 9;else break;--idx;}}int delcnt = 0;for (vector<int>::iterator it = SouVec.begin(); it < SouVec.end(); ){if (*it == 0){SouVec.erase(it);++delcnt;}else break;}for (int fir = 0; fir < 10; ++fir){vector<int> TmpSouVec = SouVec;vector<int> TgtVec;TgtVec.push_back(fir);for(int i = del - delcnt; i > -1 ; --i){int sum = TmpSouVec[i];int cur = TgtVec.front();if (i == 0 && cur == sum && cur != 0){cout<<"[ "<<cnt++<<" ] value, delete the "<<TgtVec.size() - 1<<" value: ";copy(TgtVec.begin(),TgtVec.end(),ostream_iterator<int>(cout, ""));copy(BefDelVec.begin(),BefDelVec.end(),ostream_iterator<int>(cout, ""));cout<<endl<<endl;}int pre = 0;if (cur <= sum)pre = sum - cur;else{pre = 10 + sum - cur;int idx = i - 1;while (idx > -1){if (--TmpSouVec[idx] < 0) TmpSouVec[idx] = 9;else break;--idx;}}TgtVec.insert(TgtVec.begin(), pre);}}return 0;}int GenBefVec(const vector<int>& SouVec, int inc, int del, int lay, vector<int>& BefDelVec){if (lay == del){GenAfterVec(SouVec, del, BefDelVec);return 0;}for (int t = 0; t < 2; ++t){int temp = SouVec[lay] - inc;if (temp % 2) continue;BefDelVec.insert(BefDelVec.begin(), (temp + t * 10) / 2);inc = t;GenBefVec(SouVec, inc, del, lay - 1, BefDelVec);BefDelVec.erase(BefDelVec.begin()); }}int main(){string StrNum;cout<<"Input a Number: ";cin >>StrNum;vector<int> SouVec;for (int i = 0; i < StrNum.length(); ++i){int temp = StrNum[i] - '0';if (temp < 0 || temp > 10){cout<<"Input error, Not a number!"<<endl;return 1;}SouVec.push_back(temp);}for(int del = SouVec.size() - 1; del > -1 ; --del){vector<int> BefDelVec;GenBefVec(SouVec, 0, del, SouVec.size() - 1, BefDelVec);}return 0;}