poj 3370 Halloween treats 解题报告 <鸽巢原理>
链接：http://poj.org/problem?id=3370
Description
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.
Input
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 , ... , 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
本题是一道组合数学题， 利用鸽巢原理就能解决；
首先简单说一下鸽巢原理，就是有n+1个球，放在n个盒子中，那么至少有一个盒子有两个或更多的球。
题意为从 N 个数中选 K 个使他们的和恰好能整除 M；
1 #include <stdio.h>
2 int a[100000], m[100000];
3 int main( )
4 {
5 int N, M;
6 while( scanf( "%d%d", &N, &M ) , N+M )
7 {
8 int sum=0, b=0, e=0;
9 for( int i=0; i<M; ++ i )
10 {
11 scanf( "%d", &a[i] );
12 m[i]=-2;
13 }
14 m[0]=-1;
15 for( int i=0; i<M; ++ i )
16 {
17 sum+=a[i];
18 sum%=N;
19 if( m[sum]!= -2 )
20 {
21 b=m[sum]+1,
22 e=i;
23 break;
24 }
25 m[sum]=i;
26 }
27 for( int i=b; i<=e; ++ i )
28 {
29 printf( i != e?"%d ":"%d\n" , i+1);
30 }
31 }
32 return 0;
33 }
poj 2356 Find a multiple 也是一样的
这是我最初写的代码
1 #include <stdio.h>
2 #include <stdlib.h>
3 #include <string.h>
4 int a[10005], b[10005], c[10005];
5 int main( )
6 {
7 int N;
8
9 while( scanf( "%d", &N ) != EOF )
10 {
11 memset( b, 0, sizeof( b ) );
12 memset( c, 0, sizeof( c ) );
13 for( int i=0; i<N; ++ i )
14
15 {
16 scanf( "%d",&a[i] );
17 b[i] = b[i-1]+a[i];
18 b[i] %= N;
19 c[i]=b[i];
20 //printf( "%d %d\n", b[i] , c[i]);
21
22 }
23 int f=0;
24 for( int i=0; i<N && !f; ++ i )
25 {
26 if( b[i]==0 )
27 {
28 if( a[i]%N )
29 {
30 printf( "%d\n", i+1 );
31 for( int k=0; k<=i; ++ k )
32 {
33 printf( "%d\n", a[k] );
34 }
35 break;
36 }
37 puts( "1" );
38 printf( "%d\n", a[i] );
39 break;
40 }
41 for( int j=i+1; j<N; ++ j )
42 {
43 if( b[i]==c[j] )
44 {
45 f=1;
46 printf( "%d\n",j-i );
47 for( int k=i+1; k <= j; ++ k )
48 {
49 printf( "%d\n", a[k] );
50 }
51 break;
52 }
53 }
54 }
55 }
56 }
修改版
1 #include <stdio.h>
2 int a[120000], m[120000], b[120000];
3 int main( )
4 {
5 int N;
6 //while( scanf( "%d",&N ) != EOF )
7 scanf( "%d",&N );
8 {
9 for( int i=0; i<N; ++ i )
10 scanf( "%d", &a[i] ), m[i]=-2, b[i]=a[i];
11 m[0]=-1;
12 int sum=0;
13 for( int i=0; i<N; ++ i )
14 {
15 sum+=a[i];
16 sum%=N;
17
18 if( m[sum] != -2 )
19 {
20 printf( "%d\n", i-m[sum] );
21 for( int j=m[sum]+1; j <= i; ++ j )
22 {
23 printf("%d\n",b[j] );
24 }
25 break;
26 }
27 m[sum]=i;
28 }
29 }
30 return 0;
31 }