J - Ball
Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u
Submit Status Practice HDU 4811
Appoint description:
System Crawler (2014-10-08)
Description
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?
Input
There are several test cases, please process till EOF.
Each test case contains only one line with 3 integers R, Y and B, separated by single spaces. All numbers in input are non-negative and won‘t exceed 10 9.
Output
For each test case, print the answer in one line.
Sample Input
2 2 2
3 3 3
4 4 4
Sample Output
15
33
51
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<cmath> 5 #include<algorithm> 6 #include<vector> 7 #define M(a,b) memset(a,b,sizeof(a)) 8 #include<map> 9 using namespace std; 10 11 long long R,Y,B; 12 13 int main() 14 { 15 while(scanf("%I64d%I64d%I64d",&R,&Y,&B)==3) 16 { 17 long long ans = 0; 18 if(R>=2&&Y>=2&&B>=2) //全大于1 19 ans = (R+Y+B-6)*6+15; 20 else 21 { 22 if(R==1&&Y>1&&B>1||Y==1&&R>1&&B>1||B==1&&Y>1&&R>1) //一个1其余大于1 23 { 24 ans = (R+Y+B-5)*5+10; 25 } 26 else if(R==1&&Y==1&&B>1||B==1&&Y==1&&R>1||R==1&&B==1&&Y>1) //两个1一个大于1 27 { 28 ans = (R+Y+B-4)*4+6; 29 } 30 else if(R==1&&Y==1&&B==1) //3个1 31 { 32 ans = 3; 33 } 34 else if(R==0&&Y>1&&B>1||B==0&&Y>1&&R>1||Y==0&&B>1&&R>1) //1个0 35 { 36 ans = (R+Y+B-4)*4+6; 37 } 38 else if(R==0&&Y==0&&B>1||B==0&&Y==0&&R>1||R==0&&B==0&&Y>1)//2个0 39 { 40 ans = (R+B+Y-2)*2+1; 41 } 42 else if(R==0&&Y==1&&B>1||R==1&&Y==0&&B>1||B==1&&Y==0&&R>1||B==0&&Y==1&&R>1||R==0&&B==1&&Y>1||B==0&&R==1&&Y>1) //1个0,1个1 43 { 44 ans = (R+B+Y-3)*3+3; 45 } 46 else if(R==0&&Y==1&&B==1||B==0&&Y==1&&R==1||Y==0&&B==1&&R==1)//1个0,2个1 47 { 48 ans = 1; 49 } 50 else ans = 0; //2个0,1个1,3个0 51 } 52 printf("%I64d\n",ans); 53 } 54 return 0; 55 }