Danganronpa

Danganronpa

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1017 Accepted Submission(s): 687

Problem Description
Chisa Yukizome works as a teacher in the school. She prepares many gifts, which consist of n kinds with a[i] quantities of each kind, for her students and wants to hold a class meeting. Because of the busy work, she gives her gifts to the monitor, Chiaki Nanami. Due to the strange design of the school, the students’ desks are in a row. Chiaki Nanami wants to arrange gifts like this:

1. Each table will be prepared for a mysterious gift and an ordinary gift.

2. In order to reflect the Chisa Yukizome’s generosity, the kinds of the ordinary gift on the adjacent table must be different.

3. There are no limits for the mysterious gift.

4. The gift must be placed continuously.

She wants to know how many students can get gifts in accordance with her idea at most (Suppose the number of students are infinite). As the most important people of her, you are easy to solve it, aren’t you?

Input
The first line of input contains an integer T(T≤10) indicating the number of test cases.

Each case contains one integer n. The next line contains n (1≤n≤10) numbers: a1,a2,…,an, (1≤ai≤100000).

Output
For each test case, output one line containing “Case #x: y” (without quotes) , where x is the test case number (starting from 1) and y is the answer of Chiaki Nanami’s question.

Sample Input
1
2
3 2

Sample Output
Case #1: 2

题意：有n种礼物，每种礼物的个数为a[i],每个桌子代表一个人，每个桌子上放一个神秘礼物放一个普通礼物，相邻的桌子普通礼物种类不同，问最多有多少人拿到礼物？
思路：
每种礼物让他们自己又当神秘礼物又当普通礼物放一个桌子上面
（1）最大的大于剩下的总和，此时肯定是（sum-max）*2+1，不难想出
（2）最大的小于剩下的总和，这是不足的需要后面的来补，答案肯定是sum/2，

贪心+优先对列也可以！

#include <iostream>#include <cstring>#include <cstdio>using namespace std;int main(){    int t, n, a, b, c;    cin>>t;    for(int j=1;j<=t;j++)    {        cin>>n;        b = 0, c = 0;        for(int i=0;i<n;i++)        {            cin>>a;            b = max(a, b);//找出最大值            c += a;//求和        }       //找出两者之间较小的        cout<<"Case #"<<j<<": "<<min(c/2, (c-b)*2+1)<<endl;    }    return 0;}