Problem B : Ball Game
Description
There are 4 types of balls (A, B, C and D) and a huge box. The balls of the same type have the same weight.
Lily likes to play a game. She puts theses balls into the box one by one. Suppose she has already put some balls into the box and their total weight is w. When she continues putting a new ball into the box, she will receive a bonus score, denoted as c. The value of c equals to the product of the weight of the new ball and the last digit of w. Note that when the box is empty, the value of w is 0.
Now give you the number of each type of balls and their weights, can you tell Lily the maximum score she can get in total?
Input
There are several test cases. Each test case has one line containing 8 numbers.
Each of the first 4 numbers, A, B, C and D (0 ≤ A, B, C, D ≤ 30), indicates the number of the respective type of balls.
Each of the next 4 numbers, a, b, c and d (1 ≤ a, b, c, d ≤ 100,000), indicates the weight of the respective type of balls.
Output
For each test case, print one line with the maximum total score Lily can get.
Sample Input
1 1 1 1 2 3 4 5 3 6 8 8 25 66 18 12
Sample Output
71 5213
题意：很绕，慢慢读题吧。第一个样例 = 0 + 2*3 +5*4 +9*5 =71 ,即按顺序放。
解题思路：4维DP。跟那个从左上角走到走下角，求最大价值那题很像，这里变成4维了。递推公式是一样的。然后价值相等的时候要尽量使得最后一位最大。
#include<iostream>
#include<bitset>
#include<string.h>
#include<vector>
using namespace std;
typedef long long int ll;
const int MAXN=1200;
int dp[35][35][35][35];
int w[35][35][35][35];
int main(){
int A,B,C,D;
int a,b,c,d;
while(~scanf("%d%d%d%d",&A,&B,&C,&D)){
scanf("%d%d%d%d",&a,&b,&c,&d);
memset(dp,0,sizeof(dp));
memset(w,0,sizeof(w));
for(int i=0;i<=A;i++)
for(int j=0;j<=B;j++)
for(int k=0;k<=C;k++)
for(int l=0;l<=D;l++){
if(w[i][j][k][l] * a + dp[i][j][k][l] > dp[i+1][j][k][l]){
dp[i+1][j][k][l]=w[i][j][k][l] * a + dp[i][j][k][l];
w[i+1][j][k][l]=(w[i][j][k][l]+a)%10;
}
else{
if(w[i][j][k][l] * a + dp[i][j][k][l] == dp[i+1][j][k][l]){
w[i+1][j][k][l]=max((w[i][j][k][l]+a)%10,w[i+1][j][k][l]);
}
}
if(w[i][j][k][l] * b + dp[i][j][k][l] > dp[i][j+1][k][l]){
dp[i][j+1][k][l]=w[i][j][k][l] * b + dp[i][j][k][l];
w[i][j+1][k][l]=(w[i][j][k][l]+b)%10;
}
else{
if(w[i][j][k][l] * b + dp[i][j][k][l] == dp[i][j+1][k][l]){
w[i][j+1][k][l]=max((w[i][j][k][l]+b)%10,w[i][j+1][k][l]);
}
}
if(w[i][j][k][l] * c + dp[i][j][k][l] > dp[i][j][k+1][l]){
dp[i][j][k+1][l]=w[i][j][k][l] * c + dp[i][j][k][l];
w[i][j][k+1][l]=(w[i][j][k][l]+c)%10;
}
else{
if(w[i][j][k][l] * c + dp[i][j][k][l] == dp[i][j][k+1][l]){
w[i][j][k+1][l]=max((w[i][j][k][l]+c)%10,w[i][j][k+1][l]);
}
}
if(w[i][j][k][l] * d + dp[i][j][k][l] > dp[i][j][k][l+1]){
dp[i][j][k][l+1]=w[i][j][k][l] * d + dp[i][j][k][l];
w[i][j][k][l+1]=(w[i][j][k][l]+d)%10;
}
else{
if(w[i][j][k][l] * d + dp[i][j][k][l] == dp[i][j][k][l+1]){
w[i][j][k][l+1]=max((w[i][j][k][l]+d)%10,w[i][j][k][l+1]);
}
}
}
printf("%d\n",dp[A][B][C][D]);
}
return 0;
}