| Cent Savings |
| Time limit: 5000ms, special time limit:12500ms,Memory Limit:65536KB |
| Total Submit users: Accepted users:36 |
| problem 13345: No Special Judgement |
| Problem description |
To host a regional contest like NWERC a lot of preparation is necessary:organizing rooms and computers, making a good pro Blem set, inviting contestants, designing T-shirts, book-ing 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 Q Ueue in front of me is served. While waiting, I realize it supermarket recently started to round the total price of a purchase to the nearest mult Iple of cents (with 5 cents being rounded upwards). For example, 94 cents is rounded to + cents, while the while is rounded to 100.
It's possible to divide my purchase into groups and to pay for the parts separately. I managed to find D dividers to divide my purchase on up to D + 1 groups. I wonder where to place the dividers to minimize and the total cost of my purchase. As I am running out of time, I don't want to rearrange items on the belt.
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| Input |
The input consists of:
? One line with 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 to 1≤i≤n), the prices of the items in cents. The prices is given in the same order as the items appear on the belt.
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| Output |
Output the minimum amount of money needed to buy all the items with the using up to D dividers.
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| Sample Input |
5 113 21 55 60 425 21 1 1 1 1 |
| Sample Output |
1900
#include <stdio.h> #include <string.h>const int N = 2005;int Main () {int dp[n][25],n,d,sum[n]; while (scanf ("%d%d", &n,&d) >0) {for (Int. i=0; i<=n; i++) for (int j=1; j<=d; j + +) dp[i][j]=1<<29; for (int j=0; j<=d; j + +) dp[0][j]=0; sum[0]=0; for (int i=1; i<=n; i++) {scanf ("%d", &sum[i]); SUM[I]+=SUM[I-1]; } for (int i=1; i<=n; i++) if (sum[i]%10>4) dp[i][0]=sum[i]-sum[i]%10+10; else dp[i][0]=sum[i]-sum[i]%10; for (int i=1, i<=n; i++) for (int j=1; j<=d; j + +) for (int. ti=i-1; ti>=0; ti--) {in T Ad=sum[i]-sum[ti]; if (ad%10>4) {ad=ad-ad%10+10; } else ad=ad-ad%10; if (Dp[i][j]>dp[ti][j-1]+ad) Dp[i][j]=dp[ti][j-1]+ad; } printf ("%d\n", Dp[n][d]); } return0;}
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Hnu Cent Savings (DP)
