zoukankan      html  css  js  c++  java
  • poj 3370 Halloween treats(鸽巢原理)

    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 nspace 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

    Source

     
    和上一题 poj2356 差不多
     1 #pragma comment(linker, "/STACK:1024000000,1024000000")
     2 #include<iostream>
     3 #include<cstdio>
     4 #include<cstring>
     5 #include<cmath>
     6 #include<math.h>
     7 #include<algorithm>
     8 #include<queue>
     9 #include<set>
    10 #include<bitset>
    11 #include<map>
    12 #include<vector>
    13 #include<stdlib.h>
    14 using namespace std;
    15 #define max(a,b) (a) > (b) ? (a) : (b)  
    16 #define min(a,b) (a) < (b) ? (a) : (b)
    17 #define ll long long
    18 #define eps 1e-10
    19 #define MOD 1000000007
    20 #define N 100006
    21 #define inf 1e12
    22 ll n,m;
    23 ll sum[N];
    24 ll vis[N];
    25 ll a[N];
    26 ll tmp[N];
    27 int main()
    28 {
    29     while(scanf("%I64d%I64d",&n,&m)==2){
    30         if(n==0 && m==0){
    31             break;
    32         }
    33         memset(sum,0,sizeof(sum));
    34         for(ll i=1;i<=m;i++){
    35             //ll x;
    36             scanf("%I64d",&a[i]);
    37             sum[i]=sum[i-1]+a[i];
    38         }
    39         
    40         
    41         memset(vis,0,sizeof(vis));
    42         memset(tmp,0,sizeof(tmp));
    43         for(ll i=1;i<=m;i++){
    44             ll x=sum[i]%n;
    45             if(vis[x]){
    46                 ll y=tmp[x];
    47                 //printf("%d
    ",i-y);
    48                 for(ll j=y+1;j<i;j++){
    49                     printf("%I64d ",j);
    50                 }
    51                 printf("%d
    ",i);
    52                 break;
    53                 
    54             }
    55             if(x==0){
    56                 //printf("%d
    ",i);
    57                 for(ll j=1;j<i;j++){
    58                     printf("%I64d ",j);
    59                 }
    60                 printf("%d
    ",i);
    61                 break;
    62             }
    63             vis[x]=1;
    64             tmp[x]=i;
    65         }
    66         
    67     }
    68     return 0;
    69 }
    View Code
  • 相关阅读:
    ubuntu下在apache部署python站点
    MySQL设置从库只读模式
    mysql数据库,创建只读用户
    BUG处理流程说明
    bug的处理流程
    nginx 502 Bad Gateway 错误解决办法
    Linux下批量替换文件内容方法
    centos 安装pecl
    php学习资源
    Docker容器进入的4种方式(转)
  • 原文地址:https://www.cnblogs.com/UniqueColor/p/4811507.html
Copyright © 2011-2022 走看看