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

2012-03-15 09:15 495 查看
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;
}
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