uvalive6952 - Cent Sa dp

6952 Cent Savings

To host a regional contest like NWERC a lot of preparation

is necessary: organizing rooms and computers,

making a good problem set, inviting contestants, designing

T-shirts, booking hotel rooms and so on. I am

responsible for going shopping in the supermarket.

When I get to the cash register, I put all my n

items on the conveyor belt and wait until all the other

customers in the queue in front of me are served.

While waiting, I realize that this supermarket recently

started to round the total price of a purchase

to the nearest multiple of 10 cents (with 5 cents being

rounded upwards). For example, 94 cents are

rounded to 90 cents, while 95 are rounded to 100.

It is possible to divide my purchase into groups

and to pay for the parts separately. I managed to find d dividers to divide my purchase in up to d + 1

groups. I wonder where to place the dividers to minimize the total cost of my purchase. As I am

running out of time, I do not want to rearrange items on the belt.

Input

The input file contains several test cases, each of them as described below.

The input consists of:

• one line with two integers n (1 ≤ n ≤ 2000) and d (1 ≤ d ≤ 20), the number of items and the

number of available dividers;

• one line with n integers p1, …, pn (1 ≤ pi ≤ 10000 for 1 ≤ i ≤ n), the prices of the items in cents.

The prices are given in the same order as the items appear on the belt.

Output

For each test case, output the minimum amount of money needed to buy all the items, using up to d

dividers.

Sample Input

5 1

13 21 55 60 42

5 2

1 1 1 1 1

Sample Output

190

0**

题意：

用最多d个划分d+1个组使各组之和最小，和会四舍五入，如若和为15则付20,和为114则付110

思路：

dp[i][k+1],即在i点划分为第k+1组的最小和，就会发现dp[i][k+1]=dp[j][k]+(sum[j]-sum[i])/10*10或（sum[j]-sum[j]）/10*10+10,（判断条件就是如果(sum[j]-sum[i])%10<5就是前者，否则是后者）

j为前i-1位,dp[j][k]即j位划分为第k组的最小和

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
const int inf=99999999;
int p[2005];
int sum[2005];
int dp[2005][25];
int main()
{
int n,d;
while(scanf("%d%d",&n,&d)!=-1)
{
memset(sum,0,sizeof sum);
for(int i=1;i<=n;i++){
scanf("%d",&p[i]);
sum[i]=sum[i-1]+p[i];
}
memset(dp,0,sizeof dp);
for(int i=0;i<=n;i++)
for(int j=0;j<=d+1;j++)
dp[i][j]=inf;
for(int i=1;i<=n;i++){
int temp=sum[i]%10;
if(temp>=5)
dp[i][1]=sum[i]/10*10+10;
else
dp[i][1]=sum[i]/10*10;
}
for(int k=1;k<=d;k++)
for(int i=k;i<=n;i++){
for(int j=0;j<i;j++){
int res=(sum[i]-sum[j]);
if(res%10>=5) res=res/10*10+10;
else res=res/10*10;
//printf("res=%d\n",res);
dp[i][k+1]=min(dp[j][k]+res,dp[i][k+1]);
/*if(dp[i][k]>=inf)
puts("1-------");
else
printf("%d %d %d dp[j][k]=%d dp[i][k+1]=%d\n",k,i,j,dp[j][k],dp[i][k+1]);*/
}
}
int ans=dp
[d+1];
for(int i=1;i<=d+1;i++){
ans=min(dp
[i],ans);
}
printf("%d\n",ans);
}
}