POJ 2356 Find a multiple / 3370 Halloween treats 鸽巢原理

题意：给定n个正整数，从中任意的选出一些数，使他们的和能够被n整除。

题解：《组合数学》P17

#include<cstdio>
#include<cstring>
using namespace std;

const int MAXN = 20000;
int sum[MAXN], num[MAXN], flag[MAXN];

void print ( int s, int t )
{
    printf("%d\n", t - s + 1);
    for ( int i = s; i <= t; i++ )
        printf("%d\n",num[i]);
}

int main()
{
    int n, i;
    scanf("%d",&n);
    for ( sum[0] = 0, i = 1; i <= n; i++ )
    {
        scanf("%d",&num[i]);
        sum[i] = sum[i-1] + num[i];
    }
    memset(flag,0,sizeof(flag));
    for ( i = 1; i <= n; i++ )
    {
        int t = sum[i] % n;
        if ( t == 0 ) { print ( 1, i ); break; }
        if ( flag[t] ) { print ( flag[t] + 1, i ); break; }
        flag[t] = i;
    }
    return 0;
}

下面是3370

#include<cstdio>
#include<cstring>
using namespace std;

#define lint __int64
const int MAXN = 110000;
lint sum[MAXN];
int num[MAXN], flag[MAXN];

void print ( int s, int t )
{
    for ( int i = s; i < t; i++ )
        printf("%d ",i);
    printf("%d\n",t);
}

int main()
{
    int c, n, i;
    while ( scanf("%d%d",&c,&n) )
    {
        if ( !c && !n ) break;
        for ( sum[0] = 0, i = 1; i <= n; i++ )
        {
            scanf("%d",&num[i]);
            sum[i] = sum[i-1] + num[i];
        }
        int start, end, t, f = 0;
        memset(flag,0,sizeof(flag));
        for ( i = 1; i <= n && !f; i++ )
        {
            t = sum[i] % c;
            if ( t == 0 && sum[i] >= c )
            {
                start = 1, end = i; f = 1; break;
            }
            if ( flag[t] && sum[i] - sum[flag[t]] >= c )
            {
                start = flag[t] + 1, end = i, f = 1; break;
            }
            flag[t] = i;
        }
        if ( f ) print (start, end);
        else printf("no sweets\n");
    }
    return 0;
}