2016-2017 ACM-ICPC, NEERC, Southern Subregional Contest A. Toda 2
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A group of n friends enjoys playing popular video game Toda 2. There is a rating system describing skill level of each player, initially the rating of the i-th friend is ri.
The friends decided to take part in the championship as a team. But they should have equal ratings to be allowed to compose a single team consisting of all n friends. So the friends are faced with the problem: how to make all their ratings equal.
One way to change ratings is to willingly lose in some matches. Friends can form a party consisting of two to five (but not more than n) friends and play a match in the game. When the party loses, the rating of each of its members decreases by 1. A rating can't become negative, sori = 0 doesn't change after losing.
The friends can take part in multiple matches, each time making a party from any subset of friends (but remember about constraints on party size: from 2 to 5 members).
The friends want to make their ratings equal but as high as possible.
Help the friends develop a strategy of losing the matches so that all their ratings become equal and the resulting rating is maximum possible.
The first line contains a single integer n (2 ≤ n ≤ 100) — the number of friends.
The second line contains n non-negative integers r1, r2, ..., rn (0 ≤ ri ≤ 100), where ri is the initial rating of the i-th friend.
In the first line, print a single integer R — the final rating of each of the friends.
In the second line, print integer t — the number of matches the friends have to play. Each of the following t lines should contain n characters '0' or '1', where the j-th character of the i-th line is equal to:
- '0', if friend j should not play in match i,
- '1', if friend j should play in match i.
Each line should contain between two and five characters '1', inclusive.
The value t should not exceed 104, it is guaranteed that such solution exists.
Remember that you shouldn't minimize the value t, but you should maximize R. If there are multiple solutions, print any of them.
54 5 1 7 4
180101000011010101001000011110000001111000
21 2
021111
31 1 1
10
题目意思:有n个人,他们的分数分别是下面的数字,现在我们要把他们的分数降低为一样的,步骤是,选取2-5个人一起打比赛,被选出来打比赛的人都会掉一分,如果本身的分数已经为0则不会再降低。要求,最后大家的分数尽可能高。
思路:确实是一道贪心的题目,其实只需要两两配对最终肯定可以把全部人配成一样的分数,但是就保证不了最终的分数尽可能高了,比如1 4 4 4。所以我们需要把三个一起掉分的情况也要考虑进来。这道题其实分析起来,可以逆向来想的,比如1 4 4 4,我们假定最低分就是1,那么一共是要减去9分(三个4-1),这对2来说是做不到取整的,必须要用3,那么三次都把三个人放进来打比赛就变成1 1 1 1了。而且,除了1之外的任何一个正整数都是可以用2 3 凑出来的,如果这个数字是奇数,那么它减去3后剩下的就是偶数,后面就可以两两配对用常规贪心选两个最大就好,不用管了。
最后定型的方法为:判断是否全部相等,若不相等则找出最大值,找出最大值后再找出有多少个也是最大值的数字,记为cnt。如果cnt为1则再选取出第二大的数,两个都操作一次,如果cnt为奇数且不为1,那么就先取三个操作一次,剩下的也是最大值的数字则两两操作。如果cnt为偶数,那么就循环一下,两两操作。然后回到判定是否全部相等,循环下去。
代码如下:
#include<cstdio>#include<algorithm>using namespace std;int s[10005][105] = {0};int a[105];int n, i, j, cnt, ppp;bool cmp(int a, int b){return a > b;}void dfs(){int Max = 0, i, cnt = 0, k = 0, pos;int tmp[105];for(i = 0; i < n; i++)Max = max(a[i], Max);for(i = 0; i < n; i++)if(a[i] == Max){cnt++;tmp[k++] = i;}if(cnt == n)return;if(cnt == 1){Max = 0;for(i = 0; i < n; i++)if(i != tmp[0] && a[i] > Max){Max = a[i];pos = i;}s[ppp][tmp[0]] = s[ppp][pos] = 1;ppp++;if(a[tmp[0]])a[tmp[0]]--;if(a[pos])a[pos]--;}else if(cnt & 1){s[ppp][tmp[0]] = s[ppp][tmp[1]] = s[ppp][tmp[2]] = 1;ppp++;for(i = 3; i < cnt; i += 2, ppp++){s[ppp][tmp[i]] = s[ppp][tmp[i + 1]] = 1;}for(i = 0; i < cnt; i++)if(a[tmp[i]])a[tmp[i]]--;}else{for(i = 0; i < cnt; i += 2, ppp++){s[ppp][tmp[i]] = s[ppp][tmp[i + 1]] = 1;}for(i = 0; i < cnt; i++)if(a[tmp[i]])a[tmp[i]]--;}dfs();}int main(){scanf("%d", &n);for(i = 0; i < n; i++)scanf("%d", &a[i]);ppp = 0;dfs(); printf("%d\n", a[0]);printf("%d\n", ppp);for(i = 0; i < ppp; i++){for(j = 0; j < n; j++)printf("%d", s[i][j]);printf("\n");}return 0;}
这道题也不需要用sort了,因为快排n*logn还不如线性遍历一次快呢。
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