Find a multiple
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 4988 Accepted: 2159 Special Judge
Description
The input contains N natural (i.e. positive integer) numbers ( N <= 10000 ). Each of that numbers is not greater than 15000. This numbers are not necessarily different (so it may happen that two or more of them will be equal). Your task is to choose a few of given numbers ( 1 <= few <= N ) so that the sum of chosen numbers is multiple for N (i.e. N * k = (sum of chosen numbers) for some natural number k).
Input
The first line of the input contains the single number N. Each of next N lines contains one number from the given set.
Output
In case your program decides that the target set of numbers can not be found it should print to the output the single number 0. Otherwise it should print the number of the chosen numbers in the first line followed by the chosen numbers themselves (on a separate line each) in arbitrary order.
If there are more than one set of numbers with required properties you should print to the output only one (preferably your favorite) of them.
Sample Input
5
1
2
3
4
1
Sample Output
2
2
3
Source
题目大意： 给出n个数，选出连续的若干m个数，使得和为n的倍数。输出m，以及任意顺序的m个数。
#include <stdio.h>
#define MAX 10100
int s1[MAX];
int s2[MAX];
int main()
{
int n;
while(scanf("%d",&n)!=EOF)
{
int i,sum,begin,end;
sum=begin=end=0;
memset(s2,0,sizeof(s2));
for(i=1;i<=n;i++)
{
scanf("%d",&s1[i]);
}
for(i=1;i<=n;i++)
{
sum=(s1[i]+sum)%n;
if(sum==0) {begin=1;end=i;break;}
else if(!s2[sum])
{s2[sum]=i;}
else
{
begin=s2[sum]+1;
end=i;
break;
}
}
//printf("%d
",begin);
//printf("%d
",end);
//if(i>=n)
//printf("0
");
//else
//{
printf("%d
",end+1-begin);
for(i=begin;i<=end;i++)
printf("%d
",s1[i]);
//}
}
return 0;
}
//参考代码如下所示：
/*
#include<stdio.h>
#define MAX 10100
int a[MAX];
int locker[MAX];
int main()
{
int n,i;
while(scanf("%d",&n)!=EOF)
{
int sum=0,begin=0,end=0,num;
for(i=1; i<=n; i++)
scanf("%d",&a[i]);
memset(locker,0,sizeof(locker));
for(i=1; i<=n; i++)
{
sum=(a[i]+sum)%n;
if(sum==0)
{
begin=1;
end=i; break;
}
else if(!locker[sum]) //抽屉为空
{
locker[sum]=i; //抽屉保存的是该元素在数组中的地址
}
else //抽屉不为空
{
begin=locker[sum]+1;
end=i; break;
}
}
num=end+1-begin;
printf("%d
",num);
for(i=begin; i<=end; i++)
printf("%d
",a[i]);
}
return 0;
}
*/
组合数学之抽屉原理