2016-2017 ACM-ICPC, NEERC, Southern Subregional Contest A Toda2 贪心

/*题目描述：给出n个人的分数，每轮选出2~5个人，把他们的分数减1，现要求把所有人的分数全部变成t，要求t最大，输出每一轮选人的策略。思路：注意两个性质：1、每轮选2个或3个人，因为如果有一轮选了4个人，可以分成两轮每轮选两个人；如果有一轮选了5个人，可以分成两轮，一轮选2个人，一轮选3个人2、如果最终所有人的分数是t（即保证t是可以实现的分数的前提下），所有人初始时的分数总和是sum，那么sum-n*t如果是偶数的话，一定可以通过每一轮选两个人来实现；如果是奇数的话，一定可以通过有一轮选3人，其余每轮选两人来实现。
*/
#pragma warning(disable:4786)
#pragma comment(linker, "/STACK:102400000,102400000")
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<stack>
#include<queue>
#include<map>
#include<set>
#include<vector>
#include<cmath>
#include<string>
#include<sstream>
#define LL long long
#define FOR(i,f_start,f_end) for(int i=f_start;i<=f_end;++i)
#define mem(a,x) memset(a,x,sizeof(a))
#define lson l,m,x<<1
#define rson m+1,r,x<<1|1
using namespace std;
const int INF = 0x3f3f3f3f;
const int mod = 1e9 + 7;
const double PI = acos(-1.0);
const double eps=1e-6;
const int maxn = 1e2 + 5 ;
int ans[maxn * maxn][maxn] , r[maxn] , n ;
struct node
{int val , num ;bool operator<(const node & rhs)const{return val < rhs.val ;}
};
int solve(int x)
{mem(ans ,0) ;priority_queue<node>Q;int sum = 0;for(int i = 1 ; i<= n ; i++){node st = {r[i] - x , i} ;if(st.val != 0){sum += st.val ;Q.push(st) ;}}int cnt = Q.size() , cur = 1 ;node y[4] ;if(sum & 1){if(cnt < 3)   return 0;for(int i = 1 ; i <= 3 ; i++){y[i] = Q.top();--y[i].val ;++ans[cur][y[i].num];Q.pop();}for(int i = 1 ; i<= 3 ; i++){if(y[i].val != 0){Q.push(y[i]);}}++cur;}while(!Q.empty()){for(int  i = 1 ; i<= 2 ; i++){y[i] = Q.top();Q.pop();++ans[cur][y[i].num];--y[i].val ;}for(int i = 1 ; i<= 2 ; i++){if(y[i].val != 0){Q.push(y[i] ) ;}}++cur;if(Q.size() == 1)return 0 ;}return (cur - 1 ) ;
}
int main()
{while(scanf("%d",&n)!=EOF){int maxv = -1 , minv = INF , tag = 0 ;for(int i = 1 ; i<= n ; i++){scanf("%d",&r[i]);if(i >1 && r[i] != r[i-1])      tag = 1;minv = min(minv , r[i]) ;maxv = max(maxv , r[i]) ;}if(!tag){printf("%d\n0\n",r[1]);continue ;}int flag = 0 , res , t ;for(int i = minv ; i > 0  ; --i){if(t = solve(i)){flag = 1 ;res = i ;break;}}if(!flag){printf("%d\n%d\n", 0 , n * maxv);for(int i = 1 ; i<= n ; i++){for(int j = 1 ; j<= maxv ; j++){for(int k = 1 ; k <= n ; k++){if(k == 1 || k == 2 || k == i)printf("1");elseprintf("0");}printf("\n") ;}}}else{printf("%d\n%d\n",res , t );for(int i = 1 ; i<= t ; i++){for(int j = 1 ; j<= n ; j++){printf("%d",ans[i][j]);}printf("\n") ;}}}return 0;
}

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