抽屉原理

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 con
icts, 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 satis ed 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 rst 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 5
1 2 3 7 5
3 6
7 11 2 5 13 17
0 0
Sample Output
3 5
2 3 4

 

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
const int maxn=100000+5;
typedef long long LL;
LL a[maxn];
LL mod[maxn] ;

int main()
{
    int m,n;
    while(~scanf("%d%d",&m,&n))
    {
        if(m==0 && n==0)
        break;
        memset(a,0,sizeof(a));
        int k;
        for(int i=1; i<=n; i++)
        //方法一
         // scanf("%d",&a[i]);//,mod[i]=-2;
        //memset(mod, -1, sizeof(mod));
        //mod[0]=-1;
        /*LL sum=0;
        for(int i=0; i<n; i++)
        {
            sum+=a[i];
            if(mod[sum%m]!=-2)
            {
                for(int j=mod[sum%m]+1; j<=i; j++)
                {
                    printf("%d",j+1);
                    if(i!=j)
                    printf(" ");
                }
                cout<<endl;
                break;
            }
            mod[sum%m]=i;
        }*/
        {
            scanf("%d",&k);
            a[i]=a[i-1]+k;
        }
        memset(mod, 0, sizeof(mod));

        for(int i=1 ; i<=n; i++)
        {
            if(a[i]%m == 0)
            {
                for(int j=1; j<=i; j++)
                {
                    if(j == 1)
                        printf("%d",j);
                    else
                        printf(" %d",j);
                }
                cout<<endl;
                break;
            }
            if(mod[a[i]%m] == 0)
                 mod[a[i]%m]=i;
            else
            {
                for(int j=mod[a[i]%m]+1; j<=i; j++)
                {
                    if(j == i)
                        cout<<j<<endl;
                    else
                        cout<<j<<" ";
                }
                break;
            }
        }
    }
    return 0;
}

 

第二讲课件代码模式：