EASY_ZJU_PAT_ADVANCED LEVEL_1023 大数乘法
1023. Have Fun with Numbers (20)
时间限制
400 ms
内存限制
32000 kB
代码长度限制
16000 B
判题程序
Standard
作者
CHEN,Yue
Notice that the number 123456789 is a 9-digit number consisting exactly the numbers from 1 to 9,with no duplication. Double it we will obtain 246913578,which happens to be another 9-digit number consisting exactly the numbers from 1 to 9,only in a different permutation. Check to see the result if we double it again! Now you are suppose to check if there are more numbers with this property. That is,double a given number with k digits,you are to tell if the resulting number consists of only a permutation of the digits in the original number. Input Specification: Each input file contains one test case. Each case contains one positive integer with no more than 20 digits. Output Specification: For each test case,first print in a line "Yes" if doubling the input number gives a number that consists of only a permutation of the digits in the original number,or "No" if not. Then in the next line,print the doubled number. Sample Input:1234567899Sample Output: Yes 2469135798 /************************************************ @ AUTHOR: GAOMINQUAN @ DATA: 2014 - 2 - 24 @ MAIL: ENSOLEILLY@GMAI.COM @ HARD: EASY ** /************************************************/ #include<iostream> #include<vector> #include<string> #include<algorithm> using namespace std; vector<int> reverse_vector(vector<int> perVec){ int length = perVec.size() - 1; for(int i = 0; i<=length/2; i++){ ?//NOTICE IT'S LIKE THE PRIMER CACULATE,< AND <= swap(perVec[i],perVec[length-i]); } return perVec; } vector<int> change_to_bits(string inputNum){ vector<int> bits; for(int stringI = 0; stringI < inputNum.size(); stringI++){ int tempBits = inputNum[stringI] - '0'; bits.push_back(tempBits); } return bits; } vector<int> mutiply(vector<int> numbers){ bool addBit = false; vector<int> doubleNum; for(int i = numbers.size() - 1; i>=0; i--){ int tempDouble = numbers[i] * 2 + addBit; if(tempDouble>9){ tempDouble -= 10; addBit = true; }else{ addBit = false; } doubleNum.push_back(tempDouble); } if(addBit){ doubleNum.push_back(addBit); ?//NOTICE HERE !!!!! } return doubleNum; } bool checkSame(vector<int> v1,vector<int> v2){ sort(v1.begin(),v1.end()); sort(v2.begin(),v2.end()); bool same = (v1.size() == v2.size()); if(v1.size() == v2.size()){ for(int i = 0; i<v1.size(); i++){ if(v1[i] != v2[i]){ same = false; break; } } } return same; } int main(){ string input = "123456789"; cin>>input; string outputs[] = {"No","Yes"}; vector<int> bits = change_to_bits(input); //vector<int> doubleBits = mutiply(bits); vector<int> doubleBits = reverse_vector(mutiply(bits)); cout<<outputs[checkSame(bits,doubleBits)]<<endl; for(int i = 0; i<doubleBits.size(); i++){ cout<<doubleBits[i]; }cout<<endl; return 0; } (编辑:李大同) 【声明】本站内容均来自网络,其相关言论仅代表作者个人观点,不代表本站立场。若无意侵犯到您的权利,请及时与联系站长删除相关内容! |