poj2356 Find a multiple(鸽巢原理)

Find a multiple

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 6302 Accepted: 2753 Special Judge

Description

The input contains N natural (i.e. positive integer) numbers ( N <= 10000 ). Each of that numbers is not greater than 15000. This numbers are not necessarily different (so it may happen that two or more of them will be equal). Your task is to choose a few of given numbers ( 1 <= few <= N ) so that the sum of chosen numbers is multiple for N (i.e. N * k = (sum of chosen numbers) for some natural number k).

Input

The first line of the input contains the single number N. Each of next N lines contains one number from the given set.

Output

In case your program decides that the target set of numbers can not be found it should print to the output the single number 0. Otherwise it should print the number of the chosen numbers in the first line followed by the chosen numbers themselves (on a separate line each) in arbitrary order. 

If there are more than one set of numbers with required properties you should print to the output only one (preferably your favorite) of them.

Sample Input

512341

Sample Output

223

Source

Ural Collegiate Programming Contest 1999

/*听着很高大上的鸽巢原理，其实就是加起来的那几个数对x个数取余，因为那个余数是循环的，如果得到的那个已经得到过一次，再得到的时候显然中间的差值就是x的倍数。。如果出现输入的n个数出现0的时候就会有找不到的情况，否则n个数中一定会有某几个数之和是n的倍数Time:2014-12-23 21:30*/#include<cstdio>#include<cstring>#include<algorithm>using namespace std;const int MAX=100000+10;int main(){    int sum,n;    int a[MAX],mod[MAX];    while(scanf("%d",&n)!=EOF){        memset(mod,-1,sizeof(mod));            sum=0;        for(int i=0;i<n;i++){            scanf("%d",&a[i]);        }        for(int i=0;i<n;i++){            sum+=a[i];            if(sum%n==0){                    printf("%d\n",i+1);//个数                for(int j=0;j<=i;j++){                    printf("%d\n",a[j]);                }                break;            }            if(mod[sum%n]!=-1){                printf("%d\n",i-mod[sum%n]);                for(int j=mod[sum%n]+1;j<=i;j++){                    printf("%d\n",a[j]);                }                break;            }            mod[sum%n]=i;        }    }return 0;}