HDU Danganronpa
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Danganronpa
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 0 Accepted Submission(s): 0
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?
Each case contains one integer
123 2
Case #1: 2
#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
#include<math.h>
using namespace std;
const int N = 100;
int a[N];
int cmp(int x,int y)
{
return x>y;
}
int main()
{
int t, ncase=1;
scanf("%d", &t);
while(t--)
{
int n;
scanf("%d", &n);
memset(a,0,sizeof(a));
int sum=0;
for(int i=0;i<n;i++)
{
scanf("%d", &a[i]);
sum+=a[i];
}
sum/=2;
sort(a,a+n,cmp);
int ans=0, flag=a[0];
for(int i=1;i<n;i++)
{
if(a[i]>flag)
{
ans+=(2*flag);
flag=a[i]-flag;
}
else if(a[i]<flag)
{
ans+=(2*a[i]);
flag-=a[i];
}
else
{
sum+=(2*flag);
flag=a[i+1];
i++;
}
if(ans>=sum)
{
ans=sum;
break;
}
}
if(ans<sum&&flag>=2)
{
ans++;
}
printf("Case #%d: %d\n",ncase++,ans);
}
return 0;
}
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