(Winter Camp) Mooo Moo (full backpack)

Mooo Moo

Time limit: 1 Sec memory limit: up to MB
Submissions: 5 Resolution: 4
Submitted State [Discussion Version]

Title DescriptionFarmer John had completely forgotten how many cows he owns!  He is too embarrassed to go to his fields to count the cows, since he doesn ' t want the cows to realize his mental lapse. Instead, he decides to count hisCows secretly by planting microphones in the fields in which his cows tend to gather, figuring that he can determine the Number of cows from the total volume of the mooing he hears.

FJ ' s n fields (1 <= n <=) is all arranged in a line along a long straight road. Each field might contain several types of cows; FJ owns cows that come from B different breeds (1 <= B <=), and a cow of breed I moos at a volume of V (i) (1 <  = V (i) <= 100). Moreover, there is a strong wind blowing down the road, which carries the sound of mooing in one direction from left to RI Ght:if the volume of mooing in some field was X, then under the next field this would contribute X-1 to the total mooing volum E (and X-2 in the field after, etc.). Otherwise stated, the mooing volume in a field was the sum of the contribution due to cows in that field, plus X-1, where X mooing volume in the preceding field.

Given the volume of mooing that FJ records in each field, please compute the minimum possible number of cows FJ might own.

The volume FJ records in any field are at most 100,000.

Input* Line 1:the integers N and B.
* Lines 2..1+b:line i+1 contains the integer V (i).
* Lines 2+b. 1+b+n:line 1+b+i contains the total volume of any mooing in field I.Output* Line 1:the minimum number of cows owned by FJ, or-1 if there are no configuration of cows consistent with the Inpu T.

Sample input

5 257017162019

Sample output

4

Tips

FJ owns 5, with mooing volumes 0,17,16,20,19. There is breeds of cows; The first Moos at a volume of 5, and the other at a volume of 7.There is 2 cows of breed #1 and 1 cow of breed #2 in Fiel D 2, and there is
Another cow of breed #1 in field 4.

"Analysis" by test instructions know, each area of the sound size in addition to itself only by the previous region, minus the impact of the previous area, only the sound of their own production, the minimum number of cattle, is a complete backpack.

#include <iostream>#include<cstring>#include<cstdio>#include<algorithm>#include<cmath>#include<string>#include<map>#include<stack>#include<queue>#include<vector>#defineINF 0x3f3f3f3f#defineMet (b) memset (a,b,sizeof a)#definePB Push_backtypedefLong Longll;using namespacestd;Const intN =1e3;Const intM =24005;intn,m,k,maxn=0, ans=0;intmx=-1;intv[n],a[n],b[n],dp[n*N];intMain () {scanf ("%d%d",&n,&k);  for(intI=1; i<=k;i++) scanf ("%d",&V[i]);  for(intI=1; i<=n;i++) scanf ("%d",&A[i]);  for(intI=1; i<=n;i++) {B[i]=a[i]-max (a[i-1]-1,0); MX=Max (mx,b[i]); }     for(intI=1; i<=mx;i++) dp[i]=inf;  for(intI=1; i<=k;i++){         for(intj=v[i];j<=mx;j++){            if(Dp[j-v[i]]!=inf) Dp[j]=min (dp[j],dp[j-v[i]]+1); }    }     for(intI=1; i<=n;i++){        if(dp[b[i]]==inf) {printf ("-1\n"); return 0; } ans+=Dp[b[i]]; } printf ("%d\n", ans); return 0;} 

(Winter Camp) Mooo Moo (full backpack)