poj 2356 poj3370

Find a multiple

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 7705 Accepted: 3362 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

鸽巢原理入门题目：

题意：

求出n个数字里面，是否可以由几个数字组合成n的倍数，有就输出个数和这些数字分别是什么，没有就只输出0

题解：

由鸽巢原理可以知道，总会存在解，而且这个解我们可以看做是连续的数字段组合成的

sum【i】=sum【i-1】+a【i】

如果sum数组中有满足sum【i】%n==0的元素，那么就是答案

否则，sum数组中有n个元素，但是sum【i】%n最多得到n-1个不同的数字

            因此就一定存在模n得到相同结果的sum元素，这两个元素之间就是答案

            因为（sum【i】-sum【j】）%n==0；

#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;int n;int a[10005];int visit[10005];int main(){    //freopen("in.txt","r",stdin);    while(~scanf("%d",&n))    {        for(int i=1;i<=n;i++)            scanf("%d",&a[i]);        memset(visit,0,sizeof(visit));        int sum=0;        for(int i=1;i<=n;i++){            sum=(sum+a[i])%n;            if(sum==0){                printf("%d\n",i);                for(int j=1;j<=i;j++)                    printf("%d\n",a[j]);                break;            }            else if(visit[sum]){                printf("%d\n",i-visit[sum]);                for(int j=visit[sum]+1;j<=i;j++)                    printf("%d\n",a[j]);                break;            }            else                visit[sum]=i;        }    }    return 0;}

poj3370

题意：

给定m个数字，求其中某个子集是n的倍数

#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;int n,m;int a[100005];int visit[100005];int main(){    //freopen("in.txt","r",stdin);    while(scanf("%d%d",&n,&m)&&(n||m))    {        for(int i=1;i<=m;i++)            scanf("%d",&a[i]);        memset(visit,0,sizeof(visit));        int sum=0;        for(int i=1;i<=m;i++){            sum=(sum+a[i])%n;            if(sum==0){                for(int j=1;j<=i;j++)                    printf(i!=j?"%d ":"%d\n",j);                break;            }            else if(visit[sum]){                for(int j=visit[sum]+1;j<=i;j++)                    printf(i!=j?"%d ":"%d\n",j);                break;            }            else                visit[sum]=i;        }    }    return 0;}