Hdoj-2095-find your Present (2) "Bit XOR"

Find your present (2)Time limit:2000/1000 MS (java/others) Memory limit:32768/32768 K (java/others)
Total submission (s): 18431 Accepted Submission (s): 7101


Problem DescriptionIn The New Year party, everybody'll get a "special present". Now it's your turn to get your special present, a lot of presents now putting on the desk, and only one of the them would be yo Urs. Each present have a card number on it, and your present ' s card number would be the one that different from all the others, a ndYou can assume this only one number appear odd times.For example, there is 5 present, and their card numbers are 1, 2, 3, 2, 1.so your present would be is the one with the card n Umber of 3, because 3 is the number, which different from all the others.
Inputthe input file would consist of several cases.
Each case would be presented by an integer n (1<=n<1000000, and n are odd) at first. Following that, n positive integers would be given in a line, all integers would smaller than 2^31. These numbers indicate the card numbers of the PRESENTS.N = 0 ends the input.
Outputfor each case, output an integer with a line, which is the card number of your present.
Sample Input

51 1 3 2 231 2 10

Sample Output

HintHintuse scanf to avoid time Limit exceeded

Author8600
Sourcehdu 2007-spring Programming Contest-warm up (1)
Recommend8600 | We have carefully selected several similar problems for you:20941597 1593 1595 1599
originally thought this method must be timed out, did not expect to be able to, visible: bit operation efficiency is still very high

	Exe.time	Exe.memory	Code Len.
	889MS	5324K	506 B

#include <stdio.h> #include <algorithm>using namespace  std;int a[1000010];int main () {    int n;    while (scanf ("%d", &n), n) {        int i,k;        for (i=0;i<n;++i) {            scanf ("%d", &a[i]);        }        Sort (a,a+n);        k=0;        int ok=0;        for (i=1;i<n;++i) {            if (A[i]^a[k]) {                //printf ("%d,%d\n", A[i],a[k]);                printf ("i-k=%d\n", i-k);                if ((i-k) &1) {                    ok=1;                    printf ("%d\n", A[k]);                    break;                }                else{                    k=i        ;        }}} if (!ok) printf ("%d\n", a[n-1]);            }    return 0;}

originally did not think can be directly with the difference or processing, because the number of other cards are even, any two of the same number is different or after 0, and different or satisfy the Exchange law, the order of operations independent, the result is the same

Exe.time	Exe.memory	Code Len.
748MS	1404K	187B

#include <stdio.h>int main () {    int n;    while (scanf ("%d", &n), n) {        int a,i,res=0;        for (i=0;i<n;++i) {            scanf ("%d", &a);            res=res^a;        }        printf ("%d\n", res);    }    return 0;}

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Hdoj-2095-find your Present (2) "Bit XOR"