2016长城信息杯中国大学生程序设计竞赛中南邀请赛 xtu 1246 Heartstone

http://www.dengwenhuo.cn/?id=201

Bobo is playing Heartstone. There aren

minions
in the battlefield. Thei

-th
minion hash

i



hit
points (HP).

Bobo uses two kinds of magic. The one is Arcane Shot and the other is Frostbolt. Arcane Shot can deal twopoints damage to a minion (that is to decrease the minion's HP by two), while Frostbolt can deal three points damage.
A minion is killed when its HP is less or equal to zero.

Letf(a)

be
the minimum number of Frostbolt(s) required to kill all minions, if no more thana

Arcane
Shot(s) are used. Bobo would like to find outf(0)+f(1)+⋯+f(m)

modulo(10

9

+7)

.

Input

The input contains at most30

sets.
For each set:

The first line contains2

integersn,m

(1≤n≤10

5

,0≤m≤10

5



).

The second line containsn

integersh

1

,h

2

,…,h

n



(1≤h

i

≤10

4



).

Output

For each set, an integer denotesf(0)+f(1)+⋯+f(m)

modulo(10

9

+7)

.

Sample Input

3 2
1 2 3
3 2
2 2 2

Sample Output

6
6

参考了一下别人  自己写了一个 有挫的 没有写完 测试了几波输出都一样

用个优先队列维护~~~时间充足

#include<stdio.h>
#include<string.h>
#include<queue>
#include<algorithm>
#include<vector>
using namespace std;
#define ll long long
#define mod 1000000007

struct p
{
int x;
bool friend operator <(p a,p b)
{
return a.x%3<b.x%3 || a.x%3==b.x%3&&a.x<b.x;
}
};

int main()
{
int n,m;
ll ans,sum;
p t;
while(~scanf("%d %d",&n,&m))
{
priority_queue<p>q;
sum=0;
for(int i=0; i<n; i++)
{
scanf("%d",&t.x);
q.push(t);
sum+=t.x/3+(t.x%3==0?0:1);
}
ans=sum;printf("%lld\n",sum);
while(m--)
{
if(!q.empty())
{
t=q.top();
q.pop();
sum-=t.x/3+(t.x%3==0?0:1);
t.x-=2;
if(t.x>0)
{
sum+=t.x/3+(t.x%3==0?0:1);
q.push(t);
}
ans=(ans+sum)%mod;
}
}
printf("%lld\n",ans);
}
return 0;
}