C
来源:互联网 发布:ubuntu锐捷认证失败 编辑:程序博客网 时间:2024/06/18 10:29
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
和B—抽屉 是一道题。
#include<iostream>#include<cstdio>#include<cmath>#include<cstring>#include<algorithm>using namespace std;struct Node{ int num; int id;}p[1000001];int b[1000001];int main(){ int c,n; while(scanf("%d%d",&c,&n)!=EOF&&(c!=0||n!=0)) { long long sum=0; memset(b,-1,sizeof(b));//将b数组重置为-1; for(int i=0;i<n;i++) { scanf("%d",&p[i].num); p[i].id=i+1; } for(int i=0;i<n;i++) { sum+=p[i].num; if(sum%c==0) { for(int j=0;j<=i-1;j++) printf("%d ",p[j].id); printf("%d\n",p[i].id); break; } if(b[sum%c]!=-1) { for(int j=b[sum%c]+1;j<=i-1;j++) { printf("%d ",p[j].id); } printf("%d\n",p[i].id); break; } b[sum%c]=i; } } return 0;}
- c
- c
- c
- c
- C
- c
- c
- c
- C+
- c
- C
- c
- c
- c
- C
- C
- c
- C
- 正则表达式
- Java中List Set Map 是否有序等总结
- Linux内核中的互斥操作(1)——信号量
- java基础二分法查找
- 常用快捷键以及其对应的英文名称
- C
- 深入理解Java中的i++、++i语句
- Spring Web Flow 远程代码执行漏洞分析
- 【高级线程管理】线程池
- 深度学习中BP(Backpropagation)算法的工作流程
- CoordinatorLayout的AppBarLayout和CollapsingToolbarLayout
- JDK的网络资源地址
- 30道经典面试题
- SQLSTATE[42S02]: Base table or view not found: 1146 Table 'blog.user' doesn't exist (SQL: select * f