hdu-4811 ball&&zzulioj-2618 ACM-ICPC亚洲区域赛ZZULI站

题目：
Jenny likes balls. He has some balls and he wants to arrange them in a row on the table.
Each of those balls can be one of three possible colors: red, yellow, or blue. More precisely, Jenny has R red balls, Y yellow balls and B blue balls. He may put these balls in any order on the table, one after another. Each time Jenny places a new ball on the table, he may insert it somewhere in the middle (or at one end) of the already-placed row of balls.
Additionally, each time Jenny places a ball on the table, he scores some points (possibly zero). The number of points is calculated as follows:
1.For the first ball being placed on the table, he scores 0 point.
2.If he places the ball at one end of the row, the number of points he scores equals to the number of different colors of the already-placed balls (i.e. expect the current one) on the table.
3.If he places the ball between two balls, the number of points he scores equals to the number of different colors of the balls before the currently placed ball, plus the number of different colors of the balls after the current one.
What’s the maximal total number of points that Jenny can earn by placing the balls on the table?

最近给新生写zzulioj-2618: ACM-ICPC亚洲区域赛ZZULI站 的题解，顺便氵道博客。

TP：hdu-4811 zzulioj-2618

这题要求的是最大得分，考虑了考虑，如果先把两边的球放好，其他多余的球放中间可得分最多。具体做法就是检查每个球是否大于二，每种颜色取出最多两个球，模拟出这些球的得分，剩下的球数量乘以刚开始取出球的数目为剩下球的得分，最后得分相加即可。

hdu-4811 AC代码

#include
int main()
{
    long long a,b,c,d,e,f,z,h,i,x;
    while(~scanf("%lld%lld%lld",&a,&b,&c))
    {
        d=a>=2?2:a;
        e=b>=2?2:b;
        f=c>=2?2:c;
        z=d+e+f;
        for(i=0,x=0;i

zzulioj-2618 代码大同小异，就不贴出来了。