POJ 2356 Find a multiple
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Find a multiple
Time Limit: 1000ms
Memory Limit: 65536KB
This problem will be judged on
PKU. Original ID: 2356
64-bit integer IO format: %lld Java class name: Main
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
解题:抽屉原理
1 #include <iostream>
2 #include <cstdio>
3 #include <cstring>
4 using namespace std;
5 const int maxn = 10010;
6 int a[maxn],pos[maxn],n;
7 int main() {
8 while(~scanf("%d",&n)) {
9 memset(pos,-1,sizeof pos);
10 bool flag = true;
11 int sum = pos[0] = 0;
12 for(int i = 1; i <= n; ++i) {
13 scanf("%d",a + i);
14 sum = (sum + a[i])%n;
15 if(pos[sum] > -1 && flag) {
16 printf("%d\n", i - pos[sum]);
17 for(int j = pos[sum] + 1; j <= i; ++j)
18 printf("%d\n",a[j]);
19 flag = false;
20 }
21 pos[sum] = i;
22 }
23 }
24 return 0;
25 }
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