UVA - 11100 The Trip, 2007

1.题面

http://acm.hust.edu.cn/vjudge/problem/viewProblem.action?id=21322

2.题意

给你n个数字,

1.把他们分成尽可能少的严格递增序列

2.在保证序列数尽可能少的情况下,希望能使元素最多的序列中的元素数量尽可能少

3.思路

排序后得到序列a

统计处每个元素的数量s[i],显而易见max(s[i])是第一个输出

设 k = max(i)

从0~k中的a[i]以k为间隔输出元素知道序列末尾

显而易见输出有k行

又因为k = max(i),所以序列必然是递增的

4.代码

/*****************************************************************    > File Name: Cpp_Acm.cpp    > Author: Uncle_Sugar    > Mail: [e-mail]    > Created Time: 2016年05月11日 星期三 21时36分19秒*****************************************************************/# include <cstdio># include <cstring># include <cctype># include <cmath># include <cstdlib># include <climits># include <iostream># include <iomanip># include <set># include <map># include <vector># include <stack># include <queue># include <algorithm>using namespace std;const int debug = 1;const int size  = 100000 + 10; const int INF = INT_MAX>>1;typedef long long ll;int arr[size];int main(){std::ios::sync_with_stdio(false);cin.tie(0);int i,j;int n,ncase=0;while (cin>>n&&n){if (ncase++)cout << '\n';for (i=0;i<n;i++){cin >> arr[i];}sort(arr,arr+n);int ans = 0,cnt = 1;for (i=1;i<n;i++){if (arr[i]==arr[i-1])cnt++;else cnt = 1;ans = max(ans,cnt);}cout << ans << endl;for (i=0;i<ans;i++){for (j=i;j<n;j+=ans){cout << arr[j] << (j+ans>=n?'\n':' ');}}}return 0;}