POJ 3370 Halloween treats（抽屜原理），poj3370

POJ 3370 Halloween treats（抽屜原理），poj3370

Halloween treats

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 6631		Accepted: 2448		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 , ... , an (1 ≤ ai ≤ 100000 ), where ai 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 ai 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

Source

Ulm Local 2007

題意：給出c和n，接下來n個數，求任意的幾個數的和為c的倍數，輸出任意一組答案（注意是任意的）

抽屜原理： 放10個蘋果到九個抽屜，最少有一個抽屜有大於1的蘋果

這個題為什麼說是抽屜原理呢？  你計算前n個數（一共有n個和）的和mod  c ，因為n大於c，所以你猜測會有多少個餘數，

最多有 n個，即 0~n-1，而0是滿足條件的，換而言之，這n個餘數中要麼有0，要麼最少有兩個相同的餘數，

現在看兩個餘數相同的情況，例如   假設sum[1]%c==sum[n]%c  那麼a[2]+a[3]+..+a[n]就是 c  的倍數，就說這麼多了，

看代碼吧：

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>using namespace std;#define N 100005int a[N];int vis[N];int c,n;int main(){    int i;    while(~scanf("%d%d",&c,&n))    {        memset(vis,-1,sizeof(vis));        for(i=1;i<=n;i++)        {            scanf("%d",&a[i]);        }        int temp=0,j;        for(i=1;i<=n;i++)        {            temp+=a[i];            temp%=c;            if(temp==0)            {                for(j=1;j<=i;j++)                    if(j==1)                       printf("%d",j);                    else                       printf(" %d",j);               printf("\n");                break;            }            if(vis[temp]!=-1)            {                for(j=vis[temp]+1;j<=i;j++)                  if(i==j)                      printf("%d",j);                   else                      printf("%d ",j);                      printf("\n");              break;            }            vis[temp]=i;        }    }    return 0;}