hdu 5744 Keep On Movin （字符串）

最新推荐文章于 2024-08-13 10:26:25 发布

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本文探讨了如何利用多种字符构建多个回文串，并确保最短回文串长度最大化的算法实现。通过统计奇数数量字符类型来确定回文串数量，进而平衡分配字符以达到最优解。

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Keep On Movin
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 536 Accepted Submission(s): 389

Problem Description
Professor Zhang has kinds of characters and the quantity of the i-th character is ai. Professor Zhang wants to use all the characters build several palindromic strings. He also wants to maximize the length of the shortest palindromic string.

For example, there are 4 kinds of characters denoted as ‘a’, ‘b’, ‘c’, ‘d’ and the quantity of each character is {2,3,2,2} . Professor Zhang can build {“acdbbbdca”}, {“abbba”, “cddc”}, {“aca”, “bbb”, “dcd”}, or {“acdbdca”, “bb”}. The first is the optimal solution where the length of the shortest palindromic string is 9.

Note that a string is called palindromic if it can be read the same way in either direction.

Input
There are multiple test cases. The first line of input contains an integer T, indicating the number of test cases. For each test case:

The first line contains an integer n (1≤n≤105) – the number of kinds of characters. The second line contains n integers a1,a2,…,an (0≤ai≤104).

Output
For each test case, output an integer denoting the answer.

Sample Input

4
4
1 1 2 4
3
2 2 2
5
1 1 1 1 1
5
1 1 2 2 3

Sample Output

3
6
1
3

题意：给你n种字符，然后告诉你每种字符的数量，让你将他们组成几个回文串，求组成的回文串中最小长度的最优解。

思路：我们通过统计奇数个数的字符的种类数，求得需要分成的回文串的个数，然后将剩下的字符平分到回文串中，当分配的是奇数的时候，数量减一，最后，加上自己这个字符输出就行了。

ac代码：

#include<stdio.h>
#include<math.h>
#include<string.h>
#include<stack>
#include<set>
#include<map>
#include<queue>
#include<vector>
#include<iostream>
#include<algorithm>
#define MAXN 1010000
#define LL long long
#define ll __int64
#define INF 0x7fffffff
#define mem(x) memset(x,0,sizeof(x))
#define PI acos(-1)
#define eps 1e-8
using namespace std;
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll lcm(ll a,ll b){return a/gcd(a,b)*b;}
ll powmod(ll a,ll b,ll MOD){ll ans=1;while(b){if(b%2)ans=ans*a%MOD;a=a*a%MOD;b/=2;}return ans;}
double dpow(double a,ll b){double ans=1.0;while(b){if(b%2)ans=ans*a;a=a*a;b/=2;}return ans;}
//head
int a[MAXN];
int main()
{
    int t;scanf("%d",&t);
    while(t--)
    {
        int n;scanf("%d",&n);
        int cnt=0;int sum=0;
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
            if(a[i]%2) cnt++;
            sum+=a[i];
        }
        if(cnt<=1)
            printf("%d\n",sum);
        else
        {
            int now=sum-cnt;
            int ans=now/cnt;
            if(ans%2) ans--;
            if(ans==0)
                printf("1\n");
            else
                printf("%d\n",ans+1);
        }
    }
    return 0;
}