POJ-3370-Halloween treats

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Halloween treats

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 8760 Accepted: 3173 Special Judge

Description

Every year there is the same problem at Halloween: Each neighbour is only willing to give a certain total number of sweets on that day, no matter how many children call on him, so it may happen that a child will get nothing if it is too late. To avoid conflicts, the children have decided they will put all sweets together and then divide them evenly among themselves. From last year's experience of Halloween they know how many sweets they get from each neighbour. Since they care more about justice than about the number of sweets they get, they want to select a subset of the neighbours to visit, so that in sharing every child receives the same number of sweets. They will not be satisfied if they have any sweets left which cannot be divided.

Your job is to help the children and present a solution.

Input

The input contains several test cases.
The first line of each test case contains two integers c and n (1 ≤ c ≤ n ≤ 100000), the number of children and the number of neighbours, respectively. The next line contains n space separated integers a₁ , ... , a_n (1 ≤ a_i ≤ 100000 ), where a_i represents the number of sweets the children get if they visit neighbour i.

The last test case is followed by two zeros.

Output

For each test case output one line with the indices of the neighbours the children should select (here, index i corresponds to neighbour i who gives a total number of a_i sweets). If there is no solution where each child gets at least one sweet print "no sweets" instead. Note that if there are several solutions where each child gets at least one sweet, you may print any of them.

Sample Input

4 51 2 3 7 53 67 11 2 5 13 170 0

Sample Output

3 52 3 4

题解：n个数求余数，用0~n种可能，当余数为0时，满足条件；当余数不为零时，有抽屉原理得，必有至少2个相同的余数，若A和B余数相同，即A mod k =B mod k，

那么可以理解为A+n*k = B，假设sum[i] mod k ==sum[j] mod k,那么i+1~j的和为k的倍数

#include<stdio.h>#include<algorithm>using namespace std;int main(){int c,p,a[100050],sum[100050];while(~scanf("%d%d",&c,&p)&&c&&p){sum[0]=0;int flag=0,l,i,j,r;for(i=1;i<=p;i++){scanf("%d",&a[i]);sum[i]=(sum[i-1]+a[i])%c;if(sum[i]==0)//余数和为0，说明能整除{flag=i;}}if(flag){for(i=1;i<=flag;i++)//从i=1~flag刚好能整除printf("%d%c",i,i==flag?'\n':' ');continue;}for(i=1;i<p;i++)//找出余数相同的{for(j=i+1;j<=p;j++)if(sum[i]==sum[j]){l=i+1;//开始下标r=j;//结束下标flag=j-i;//区间长度，即有多少个数。j+1-(i+1)break;}if(flag)break;}if(flag){for(i=l;i<=r;i++)printf("%d%c",i,i==r?'\n':' ');}} return 0;}