epic 面试题+答案.pdf

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Write algorithm to return change in denominations of 10$, 5$, 1$, 25 cents, 10 cents double deno[]= {10, 5, 1, 0.25, 0.10, 0.05, 0.01}; //Scanner reader = new Scanner(System.in); float cost = 67.9500f; float cash = 100.00f; float change = cash-cost; System.out.println(change); for(int i=0;i=1) { } } #include #include #include #include using namespace std; /* Function declarations */ void giveTheChange(int dollar, int cents, vector &inDollar, vector &inCent); void process(int &sum,vector &denom,bool dollar); /* Function definitions */ void giveTheChange(int dollar, int cents, vector &inDollar, vector &inCent){ /* Find minmum number of dollars needed to give the change */ int sum=dollar; Business Unit/Tier 3 (Optional) | Practice Group/Tier 4 (Optional) Proprietary & Confidential (Optional) | Legal Disclaimer (Optional)

int centsLeft; int sumCent=cents; process(sum,inDollar,true); if(sum!=0){ centsLeft=sum*100; sumCent+=centsLeft; } process(sumCent,inCent,false); } void process(int &sum,vector &denom,bool dollar){ while(sum!=0 && denom.size()!=0) { if(denom.back()<= sum ) { sum-=denom.back(); if(dollar) cout<<"$"< inDollar(dollarArray,dollarArray+sizeof(dollarArray)/sizeof(int)); sort(inDollar.begin(), inDollar.end()); vector inCent(centArray, centArray+sizeof(centArray)/sizeof(int)); Business Unit/Tier 3 (Optional) | Practice Group/Tier 4 (Optional) Proprietary & Confidential (Optional) | Legal Disclaimer (Optional)

sort(inCent.begin(), inCent.end()); cout<<"Enter the amount that you wanna make change "<>dollar ; cout<<" "; cin >>cents; cout<>key; return 0; } __________________________________*********************************______________________ Find the no, of days from 1st Jan to the date entered // considering only 28 days in feb initial_date = 01; initial_month = 01; initial_year = 2011; // lets say target_date // target_month // target_year // days = 0; days += (target_year - initial_year )*365; day = {0,31,28,31,30,31,30,31,31,30,31,30,31}; while(target_month > initial_month) { days += day[initial_month]; initial_month++; } days += target_date -01; cout<public static int check(int n1,int n2,int n3) { count++; if(n1+n2==n3) { System.out.println(n1+"+"+n2+"="+n3); return 1; } else if(n2 + n3 == n1) { System.out.println(n2+"+"+n3+"="+n1); return 2; } else if(n1+n3==n2) { System.out.println(n1+"+"+n3+"="+n2); return 3; } else return 0; } public static void main(String args[]) { String f = "17512"; int n = f.length()/2; System.out.println(n); for(int i=0;i

_______________________________************************************______________________ Given a string...replace all occurances of a with the. #include #include using namespace std; int main() { string s = "I am a boy."; string temp; cout<

String[] ans = printAlphabet(410292); for (String str : ans) { } System.out.println(str); if (num >= 0 && num < 10){ static String[] printAlphabet(int num){ String[] retStr; if (num == 0 || num ==1){ } else { retStr = new String[]{""}; retStr = new String[keypad[num].length()]; for (int i = 0 ; i < keypad[num].length(); i++){ } retStr[i] = String.valueOf(keypad[num].charAt(i)); } return retStr; } String[] nxtStr = printAlphabet(num/10); int digit = num % 10; String[] curStr = null; if(digit == 0 || digit == 1){ } else { curStr = new String[]{""}; curStr = new String[keypad[digit].length()]; for (int i = 0; i < keypad[digit].length(); i++){ } curStr[i] = String.valueOf(keypad[digit].charAt(i)); } String[] result = new String[curStr.length * nxtStr.length]; int k=0; for (String cStr : curStr){ for (String nStr : nxtStr){ } result[k++] = nStr + cStr; } return result; } ---------------------------------------------------------------------*******************************************------------------------------------------ Business Unit/Tier 3 (Optional) | Practice Group/Tier 4 (Optional) Proprietary & Confidential (Optional) | Legal Disclaimer (Optional)

Print Fibonacci Series void main() { cin>>n; cout<>n; cout<

} int fibo(int n){ if(n==0) return 0; if(n==1) return 1; else return(fibo(n-1)+fibo(n-2)); } } ______________________________*************************__________________________________ _ Palindrome Date: A date is said to be a palindrome when it is expressed in MMDDYYYY format, it reads the same both ways. Given 2 years as input(ex: 2000, 2010), print out all the dates which are palindrome dates. public class PalindromeDate { public static void main(String[] args) { printDates(1000, 3010); } public static void printDates(int start, int end) { int[] daysInMonth = {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; for (int year = start; year <= end; year++) { String yyyy = Integer.toString(year); String mm = Integer.toString(year%10)+Integer.toString((year/10)%10); String dd = Integer.toString((year/100)%10)+Integer.toString((year/1000)%10); int m = Integer.parseInt(mm); int d = Integer.parseInt(dd); if (m > 0 && m < 13) { if (d > 0 && d <= ((m-1 != 2)? daysInMonth[m-1]: (year%4 != 0)?28:29)) { System.out.println(mm+dd+yyyy); } } } } } ________________________________*********************_________________________________ Business Unit/Tier 3 (Optional) | Practice Group/Tier 4 (Optional) Proprietary & Confidential (Optional) | Legal Disclaimer (Optional)

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