POJ 3370-Halloween treats（鴿巢原理）

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 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.

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.

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.

4 5
1 2 3 7 5
3 6
7 11 2 5 13 17
0 0

3 5
2 3 4

Ulm Local 2007

C個小孩去N戶人家索要糖果，第i家給出的糖果數為Ai。

孩子們想要能夠平分糖果，所以他們選擇其中一部分人家索要糖果。

若能夠平分，輸出孩子們選擇的住戶編號；若不能，輸出"no sweets"。

鴿巢原理。

對於一個正整數序列A1~Am，至少存在整數k，l（l≤k<l≤m），使得Ak~Al之和是m的倍數。

我的思路寫的程式碼不加輸入外掛就會超時…限時2000ms，加了之後1902ms險過…Orz

還有阿，這是特判的題…樣例給你的只是一種正確情況~

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <vector>
#include <queue>
#include <map>
#include <algorithm>
using namespace std;
#define maxn 100010
int a[maxn];
struct Remnant
{
    int h,r;//記錄前h個數和模上c後的餘數r
} re[maxn];
bool cmp(const Remnant &x,const Remnant &y)//結構體排序
{
    if(x.r==y.r) return (x.h<y.h);
    return (x.r<y.r);
}
int getint()//輸入外掛
{
    char c=getchar();
    int t=0;
    while (c<'0' || c>'9')
    {
        c=getchar();
    }
    while (c>='0' && c<='9')
    {
        t=t*10+c-'0';
        c=getchar();
    }
    return t;
}


int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    int c,n;
    while(cin>>c>>n,c||n)
    {
        //memset(a,0,sizeof(a));
        // memset(re,0,sizeof(Remnant)*maxn);
        int flag=-1,h;
        long long sum=0;
        for(int i=0; i<n; ++i)
        {
            a[i]=getint();//加了外掛之後1902ms險過…
            sum+=a[i];//累加前i個數
            re[i].r=sum%c;//儲存餘數
            re[i].h=i+1;//儲存前h個數下標h
            if(flag==-1&&re[i].r==0) flag=i;//當前存在數的和是c的倍數，即第一次掃描時糖果可分，並記錄位置
        }
        if(flag==-1)//第一次掃描糖果不可分
        {
            sort (re,re+n,cmp);//所有和的餘數遞增排序
            for(int i=0; i<n-1; ++i)
                if(flag==-1&&re[i].r==re[i+1].r)//若兩個餘數相等，說明它們原來的和是c的倍數
                {
                    flag=re[i].h;//記錄開始位置
                    h=re[i+1].h;//記錄結束位置
                }
            if(flag==-1) puts("no sweets");//糖果不可分
            else//第二次掃描發現糖果可分
            {
                for(int i=flag+1; i<h; ++i)
                    cout<<i<<" ";
                cout<<h<<endl;
            }
        }
        else//第一次掃描時糖果可分
        {
            for(int i=0; i<flag; ++i)
                cout<<i+1<<" ";
            cout<<flag+1<<endl;
        }
    }
    return 0;
}

/**
4 5
1 2 3 7 5
3 6
7 11 2 5 13 17
0 0
**/