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  • 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:

     

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  • 原文地址:https://www.cnblogs.com/chen9510/p/4702395.html
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