POJ 2356. Find a multiple 抽屉/鸽巢原理

Posted 2020-06-18 BlackStorm

Find a multiple Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 7192   Accepted: 3138   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 Ural Collegiate Programming Contest 1999

 

题目大意：

　　给定$N$个自然数，试找出其中的若干个数，它们的算术和是$N$的整数倍。

基本思路：

　　1、若存在$sum\\%N==0$，则$sum$一定是$N$的倍数（即$sum==kN,\\ k\\in [1, \\infty]$）。

　　2、若$sum[i]\\equiv sum[j]\\ mod\\ N$，则$(sum[j]-sum[i])\\%N==0$，即$sum[j]-sum[i]$是$N$的倍数。

　　3、（鸽巢原理）将$N$个物体放入$N-1$个盒子里，则一定至少有$1$个盒子放了$2$个以上的物体。这里我们将$N$个自然数分别记为$arr[i],\\ i\\in [1,\\ N]$，令$sum[i]=(\\sum\\limits^{N}_{i=1}arr[i])\\%N,\\ i\\in [1,\\ N]$，而$sum[i]$的取值范围为0 ~ N-1，因此：A.必然存在$i,\\ j\\in [1,\\ N]$使得sum[i]==sum[j]；B.可以存在$i\\in [1,\\ N]$有sum[i]==0。

　　4、建立一个标记数组$sgn[]$，标记$sum[]$的值，这样可以在读入预处理时找到相等的两个$sum$。同时根据第3点，我们得知答案必然存在，不会出现没有答案输出0的情况。

　　5、由于存在sum[i]==0的情况，所以要初始化sgn[0]=0。

代码：

 1 #include <stdio.h>
 2 #include <string.h>
 3 
 4 int N, arr[10010], sum[10010], sgn[10010];
 5 
 6 int main() {
 7     int l=0, r=-1;
 8     memset(sgn, 0xFF, sizeof(sgn)); sgn[0]=0;
 9     scanf("%d", &N);
10     for(int i=1; i<=N; i++) {
11         scanf("%d", arr+i);
12         sum[i]=(sum[i-1]+arr[i])%N;
13         if(!~sgn[sum[i]])
14             sgn[sum[i]]=i;
15         else {
16             l=sgn[sum[i]];
17             r=i;
18         }
19     }
20     printf("%d\\n", r-l);
21     for(int i=l+1; i<=r; i++)
22         printf("%d\\n", arr[i]);
23     return 0;
24 }

POJ 2356

 

另外，POJ3370也是与此题一样的做法，此题题解在这里。

 

——本文原创by BlackStorm，转载请注明出处。

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