C
来源:互联网 发布:网络蜘蛛磁力链 编辑:程序博客网 时间:2024/06/05 02:21
点击打开链接
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 , ... , 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.
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.
4 51 2 3 7 53 67 11 2 5 13 170 0
3 52 3 4
题解与B-抽屉类似,参考(B-抽屉)
#include<cstdio>#include<cstring>#define LL long longLL a[100005],sum[100005],yu[100005];int main(){LL c,n,i,j;while(~scanf("%lld %lld",&c,&n)){if(c==0&&n==0) break;memset(yu,0,sizeof(yu));sum[0]=0;for(i=1;i<=n;i++){scanf("%lld",&a[i]);sum[i]=sum[i-1]+a[i];}for(i=1;i<=n;i++){if(sum[i]%c==0){for(j=1;j<=i;j++) printf("%lld%c",j,j==i?'\n':' ');break;}else if(yu[sum[i]%c]!=0){for(j=yu[sum[i]%c]+1;j<=i;j++) printf("%lld%c",j,j==i?'\n':' ');break;}else if(yu[sum[i]%c]==0) yu[sum[i]%c]=i;}}return 0;}
- c
- c
- c
- c
- C
- c
- c
- c
- C+
- c
- C
- c
- c
- c
- C
- C
- c
- C
- SpringMVC札集(08)——文件上传
- Unity快捷读取XML、JSON文件
- HDU
- JAVA设计模式之单利模式
- 没有被遣返的人生是不完整的
- C
- Dynamic log fileNames with log4net
- JavaDay02变量常量
- 论文写作词汇积累-不断更新中
- 求最大公约数和最小公倍数
- 基于Kerberos的NIFI单节点安全登陆配置
- scala中给集合创建懒加载view视图
- python写算法题:leetcode: 35. Search Insert Position
- JQuery对象和DOM对象的区别