FZU 1001 Duplicate Pair

Time Limit: 1000 mSec Memory Limit : 65536 KB
Problem Description
An array of length n, with address from 1 to n inclusive, contains entries from the set {1,2,...,n-1} and there's exactly two elements with the same value. Your task is to find out the value.



Input
Input contains several cases.
Each case includes a number n (1<n<=10^6), which is followed by n integers.
The input is ended up with the end of file.


Output
Your must output the value for each case, one per line.

Sample Input
2
1 1
4
1 2 3 2
Sample Output
1
2
Source

IBM Challenge 2004.1

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思路一：由于1<n<=10^6，所以可以创建一个数组a[1000005]来存储所有输入的数据，然后用快速排序sort，再遍历一遍，找到相邻的2个数相同的即可。

#include<iostream>#include<algorithm>using namespace std;int n,a[1000005];int main(){    while(scanf("%d",&n)!=EOF)    {        for(int i=0;i<n;i++)                   scanf("%d",&a[i]);        sort(a,a+n);        for(int i=0;i<n-1;i++)        {            if(a[i]==a[i+1])            {                printf("%d\n",a[i]);                break;            }        }    }    return 0;}

思路二：由于输入的数据只能是1,2,3....(n-1) , t 。共n个数， 其中 1  <= t  <= n-1。所以只要s = (1+2+…+n)   -  (1+2+…（n-1）+ t   )  =  n - t ，然后再  n -  s =  n - （n -  t ）=  t

#include <stdio.h>int main(){    int n;    while (scanf("%d", &n) != EOF) {        int i,val, s = 0;        for( i = 1; i <= n; i++) {            scanf("%d", &val);            s -= val;            s += i;        }        printf("%d\n", n - s);    }    return 0;
}